To test the minister's claim, the correct null and alternative hypotheses would be H0 : p = .6 and Ha: p > .6, which aligns with the suggestion that more than 60% of the population attends religious services. These hypotheses represent equality and inequality respectively, and are tested to determine if there is sufficient statistical evidence for the claim.
Explanation:To test the minister's claim, Letter b is the correct choice. The notation 'p' is typically used to denote a proportion in a population. The null hypothesis, denoted H0, would be that the proportion of the population who attend services (p) is equal to 0.6. The alternative hypothesis, denoted Ha, would state that the proportion of the population who attend services (p) is greater than 0.6 which aligns with the minister's claim.
Therefore:
H0 : p = .6
Ha: p > .6
This is because the minister’s claim is a statement of inequality (more than 60%), so the alternative hypothesis must reflect this particular inequality. We then test the null hypothesis to determine whether there is enough statistical evidence to accept the alternative hypothesis.
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Final answer:
Option b is the correct answer; H0: p = 0.6 and Ha: p > 0.6. These hypotheses are set for a right-tailed test to assess the claim that more than 60% of adults attend religious services monthly.
Explanation:
The correct answer to the question of which null and alternative hypotheses should be used to test the minister's claim that more than 60% of the adult population attends a religious service at least once a month is option b. The null hypothesis (H0) signifies the statement that is presumed true before the data evidence is considered, while the alternative hypothesis (Ha) represents the claim to be tested. In this case, the null hypothesis should state that the population proportion (p) equals 0.6, or H0: p = 0.6, indicating that 60% of the adult population attends a religious service at least once a month.
The alternative hypothesis should indicate that the proportion is greater than 0.6, or Ha: p > 0.6, which aligns with the minister's claim and sets up the hypothesis test for a right-tailed test. The other options are incorrect as they either test for a mean instead of a proportion, use the wrong symbol for the hypotheses, or present the hypothesis in an incorrect format.
A new drug test needs to be evaluated. The probability of a random person taking drugs is 4%. The drug test tested positive for a person not taking drugs 2% of the time. The drug test tested negative for a person taking drugs 1% of the time. a. What is the probability that a random person tests negative? b. What is the probability that a person who tests negative is without any drug problems?
Answer:
Step-by-step explanation:
Given that a new drug test needs to be evaluated. The probability of a random person taking drugs is 4%.
The drug test tested positive for a person not taking drugs 2% of the time. The drug test tested negative for a person taking drugs 1% of the time.
Tested positive Tested negative Total
Having drug 0.98 0.02 1.00
not hav drug 0.01 0.99 1.00
a) the probability that a random person tests negative
= Prob ( really negative and test negative) + Prob (positive and test negative)
[tex]= 0.96*0.99+0.04*0.02= 0.9504+0.0008=0.9512[/tex]
b) the probability that a person who tests negative is without any drug problems
= [tex]\frac{0.9504}{0.9504+0.0008} \\=0.9992[/tex]
Annual salary plus bonus data for chief executive officers are presented in the BusinessWeek Annual Pay Survey. A preliminary sample showed that the standard deviation is $675 with data provided in thousands of dollars. How many chief executive officers should be in a sample if we want to estimate the population mean annual salary plus bonus with a margin of error of $100,000?
Answer: 176.
Step-by-step explanation:
Formula to find the sample size is given by :-
[tex]n=(\dfrac{z^*\cdot \sigma}{E})^2[/tex]
, where [tex]\sigma[/tex] = Population standard deviation from prior study.
E = margin of error.
z* = Critical value.
As per given , we have
[tex]\sigma=\$675000[/tex]
E= $100,000
We take 95% confidence interval.
Critical value (Two tailed)=[tex]z^*=1.96[/tex]
The required sample size = [tex]n=(\dfrac{(1.96)\cdot 675000}{100000})^2[/tex]
[tex]n=(13.23)^2\\\\ n=175.03299\approx176[/tex] [Round to next integer]
Hence, the required sample size = 176.
State the half-angle identities used to integrate sin^(2) x and cos^(2) x
the half-angle formulas are sin ^(2)x = ?? and cos ^(2)x = ??
Answer:
the answer is D
Step-by-step explanation:
Think of the Pythagorean Theorem which states that a^2 + b^2 = c^2. The Pythagorean Identities used in trigonometry are the angle version which can be used to simplify expressions.
HOPE THIS HELPED ;3
The formula of half angle identities is sin²x = (1-cos(x/2))/2 and cos²x = (1+cos(x/2))/2.
What is angle?An angle is the formed when two straight lines meet at one point, it is denoted by θ.
The given terms are,
sin²x and cos²x
To integrate sin²x, the formula is used,
sin²x = (1-cos(x/2))/2
To integrate cos²x, the formula is used,
cos²x = (1+cos(x/2))/2
The formula is, sin²x = (1-cos(x/2))/2 and cos²x = (1+cos(x/2))/2.
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An airline finds that about 8 percent of the time, a person who makes an advance reservation does not keep the reservation. Therefore, for each of their 105105105-passenger planes, the company schedules 113113113 people with advance reservations. Based on this information, about how many of the scheduled people will not keep their reservation?
Answer:
about 9 passengers will not keep theire reservation
Step-by-step explanation:
given that an airline finds that about 8 percent of the time, a person who makes an advance reservation does not keep the reservation
Capacity of the airline per trip = 105 only
But company schedules = 113 people with advance reservations
Since 8% do not turn up
out of 113 passengers who reserved in advance we can say
that 8% of 113 passengers will not keep their reservation
Based on this information, number of the scheduled people will not keep their reservation
= [tex]8%of 113\\= \frac{8*113}{100} \\=9.04[/tex]
Since persons cannot be in decimals we can expect about 9 passengers will not keep theire reservation
Based on the given information, about 9 people out of the scheduled 113 will not keep their reservation.
ExplanationThe airline finds that 8 percent of the time, a person with an advance reservation does not keep it. This means that 8% of the scheduled passengers will not keep their reservations.
Percentage of passengers not keeping reservation = 8% = 0.08
Number of scheduled passengers = 113
Number of passengers not keeping reservation = 0.08 * 113 = 9
Therefore, about 9 of the scheduled people will not keep their reservation.
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The lengths of my last 12 phone calls have been roughly 3, 8, 3, 5, 1, 13, 9, 2, 7, 3, 13, and 2 minutes. Long experience suggests that the standard deviation is about 5 minutes.
a) I am asked what the average length of one of my phone calls is, and I shall estimate it by ; calculate this estimate and give its standard deviation (standard error).
b) Assuming this is a large enough sample, write down a 98% confidence interval for the true value μ.
Answer:
98% Confidence Interval: (2.387,9.113 )
Step-by-step explanation:
We are given the following data set:
3, 8, 3, 5, 1, 13, 9, 2, 7, 3, 13, 2
a) Formula:
[tex]\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}[/tex]
where [tex]x_i[/tex] are data points, [tex]\bar{x}[/tex] is the mean and n is the number of observations.
[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]
[tex]Mean =\displaystyle\frac{69}{12} = 5.75[/tex]
Sum of squares of differences = 196.25
[tex]S.D = \sqrt{\frac{196.25}{11}} = 4.044[/tex]
b) 98% Confidence Interval:
[tex]\bar{x} \pm z_{critical}\displaystyle\frac{\sigma}{\sqrt{n}}[/tex]
Putting the values, we get,
[tex]z_{critical}\text{ at}~\alpha_{0.02} = \pm 2.33[/tex]
[tex]5.75 \pm 2.33(\displaystyle\frac{5}{\sqrt{12}} ) = 5.75 \pm 3.363 = (2.387,9.113 )[/tex]
An individual is teaching a class on Excel Macros. The individual plans to break the class up into groups of 4 and wants each group to have 2 exercises to practice on, with no group doing the same exercise. The individual wants to know how many exercises he will need. Write an equation that expresses the situation, let x be the independent variable and y be the dependent variable
1. У-4x
2. y-2(x/4)
3. y-4x/2
Answer:
Total Exercises each group will do =[tex]2\frac{x}{4}[/tex]
Total exercises individual need= [tex]y=\frac{4x}{2}[/tex]
[tex]y-\frac{4x}{2}[/tex]
Step-by-step explanation:
No. of groups = 4
Each group has to do exercises = 2
Total no. of exercises individual need = y
Total Exercises each group will do =[tex]2\frac{x}{4}[/tex]
Total exercises individual need= [tex]y=2\frac{x}{4}(4)[/tex]
[tex]y=\frac{4x}{2}[/tex]
[tex]y-\frac{4x}{2}[/tex]
Answer:
y-2{x/4}
Step-by-step explanation:
A car company has found that there is a linear relationship between the money it spends on advertising and the number of cars it sells. When it spent 60000 dollars on advertising, it sold 680 cars. Moreover, for each additional 5000 dollars spent, they sold 40 more cars. Let x be the amount money they spend on advertising, in thousands of dollars. Find a formula for y, the number of cars sold.
Answer:
The required formula for y, the number of cars sold is: [tex]y = 0.008x +200[/tex].
Step-by-step explanation:
Consider the provided information.
Let x represents the amount of money spend on advertising .
y is number of cars sold,
For each additional 5000 dollars spent, they sold 40 more cars.
[tex]Slope = m = \frac{Rise}{Run}=\frac{40}{5000}=0.008[/tex]
The y-intercept form is: [tex]y = mx + b[/tex]
Substitute x=60,000, y=680 and m=0.008 in the above equation.
[tex]680= 0.008(60,000)+b[/tex]
[tex]b=680-480[/tex]
[tex]b=200[/tex]
Thus, the required formula for y, the number of cars sold is: [tex]y = 0.008x +200[/tex].
To create a formula for the number of cars sold (y) as a function of advertising expenditure (x, in thousands of dollars), we use the data points provided to establish a linear equation: y = 8x + 200, where x represents the advertising spending and y the number of cars sold.
To find a formula for y, the number of cars sold, given the linear relationship between advertising expenditure (x, in thousands of dollars) and sales, we first establish the given data points:
An additional $5,000 in advertising (or an increase of 5 in x since it's measured in thousands) results in selling 40 more cars.
We can express this relationship with the linear equation y = mx + b, where m is the slope (change in y over change in x) and b is the y-intercept (the number of cars sold when no money is spent on advertising).
Let's calculate the slope, m:
For every increase of 5 in x, y increases by 40, so m = 40/5 = 8.
We can then use this slope and one of the data points to find the y-intercept, b:
680 = 8(60) + b, so b = 680 - (8*60) = 680 - 480 = 200.
The linear equation representing the relationship between advertising spend and cars sold is:
y = 8x + 200
A sample of 40 observations is selected from one population with a population standard deviation of 4.6. The sample mean is 101.0. A sample of 47 observations is selected from a second population with a population standard deviation of 4.0. The sample mean is 99.3. Conduct the following test of hypothesis using the 0.10 significance level. H0 : ?1 = ?2 H1 : ?1 ? ?2
a. what is the decision rule?
b.Compute the value of the test statistic.
c.what is the p-value?
Answer:
Step-by-step explanation:
Final answer:
The decision rule for the hypothesis test at a 0.10 significance level involves rejecting the null hypothesis if the test statistic is greater than the critical value. The calculated Z-value is approximately 1.8233, and the corresponding p-value will determine our decision to reject or not reject the null hypothesis.
Explanation:
Decision Rule and Test Statistic
For a hypothesis test comparing two population means with known standard deviations, we use a Z-test for two samples. The decision rule at a significance level of 0.10 involves rejecting the null hypothesis (H0: μ1 = μ2) if the absolute value of the test statistic exceeds the critical value from the standard normal distribution.
Calculation of Test Statistic
The test statistic (Z) is calculated using the formula: Z = (101.0 - 99.3)/ √[ (4.6²/40) + (4.0²/47) ]
Plugging in the numbers: Z = (1.7)/ √[ (21.16/40) + (16/47) ] Z = (1.7)/ √[ 0.529 + 0.3404255 ] Z = (1.7)/√[0.8694255] Z = (1.7)/0.932447 Z ≈ 1.8233
Decision Based on P-value
The p-value associated with this Z-value can be found using standard normal distribution tables or software. If the p-value is less than 0.10, we reject the null hypothesis. Otherwise, we fail to reject it.
The Best Company produces two commercial products : blenders and mixers. Both products require a two step production process involving delivery of parts (JIT process) and assembly. It takes 1 hour to deliver parts for each blender and 2 hours for each mixer. Final assembly of mixers and blenders require 3 and 2 hours, respectively. The production capability is such that only 24 hours of delivery time and 30 hours of assembly time are available. Each blender produced nets the firm $7 and each mixer $6.
a) How many of each should be produced to maximize profit? Partial production of each product is allowed.
b) What is the $ amount of this profit?
Answer:
Maximum profit is $87 when 3 blenders and 11 mixers are produced.
Step-by-step explanation:
let blender is represented by [tex]x_{1}[/tex] and and mixer by [tex]x_{2}[/tex].
total time to deliver parts = 24 hrs
total time to assemble = 30 hrs
time taken by each blender to deliver parts = 1 hr
time taken by each mixer to deliver parts = 2 hr
time taken by blenders in final assembling= 2 hr
time taken by mixers in final assembling = 3 hr
Each blender produced nets the firm= $7
Each mixer produced nets the firm= $6
Using this all data linear system of equation will be:
[tex]x_{1} + 2x_{2} =24 ----- (1)\\2x_{1} + 2x_{2} = 30 ----- (2)\\[/tex]
profit function:
[tex]z= 7x_{1} +6x_{2} --- (3)[/tex]
[tex]from (1)\\x_{1} = 0 \implies x_{2}= 12\\x_{2}= 0 \implies x_{1}= 24\\[/tex]
Coordinate points obtained from (1) are (0,12) and (24,0)
[tex]from (2)\\x_{1}=0 \implies x_{2}=10\\x_{2}=0 \implies x_{1}=15\\[/tex]
Coordinate points obtained from (2) are (0,10) and (15,0)
plotting these on graph
points lying in feasible region are:
A(0,0)
B(0,10)
C(3,11)
D(12,0)
substituting these points in (3) to find the maximum profit:
for A (0,0)
z = 0
for B (0,10)
z = 60
for C (3,11)
z = 87
for D (12,0)
z=84
So maximum profit is $87 when 3 blenders and 11 mixers are produced.
Listed below are systolic blood pressure measurements (mm Hg) taken from the right and left arms of the same woman. Consider the differences between right and left arm blood pressure measurements. Right Arm 102 101 94 79 79 Left Arm 175 169 182 146 144 a. Find the values of d and sd (you may use a calculator).b. Construct a 90% confidence interval for the mean difference between all right and left arm blood pressure measurements.
Answer:
a) [tex]\bar d= \frac{\sum_{i=1}^n d_i}{n}=72.2[/tex]
[tex]s_d =\sqrt{\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1}} =9.311[/tex]
b) The 90% confidence interval would be given by (63.330;81.070)
[tex]63.330 < \mu_{left}- \mu_{right} <81.070[/tex]
Step-by-step explanation:
1) Previous concepts and notation
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
Let put some notation
x=left arm , y = right arm
x: 175 169 182 146 144
y: 102 101 94 79 79
The first step is define the difference [tex]d_i=x_i-y_i[/tex], that is given so we have:
d: 73, 68, 88, 67, 65
The second step is calculate the mean difference
[tex]\bar d= \frac{\sum_{i=1}^n d_i}{n}=72.2[/tex]
The third step would be calculate the standard deviation for the differences, and we got:
[tex]s_d =\sqrt{\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1}} =9.311[/tex]
2) Confidence interval
The confidence interval for the mean is given by the following formula:
[tex]\bar d \pm t_{\alpha/2}\frac{s_d}{\sqrt{n}}[/tex] (1)
In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:
[tex]df=n-1=5-1=4[/tex]
Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,4)".And we see that [tex]t_{\alpha/2}=2.13[/tex]
Now we have everything in order to replace into formula (1):
[tex]72.2-2.13\frac{9.311}{\sqrt{5}}=63.330[/tex]
[tex]72.2+2.13\frac{9.311}{\sqrt{5}}=81.070[/tex]
So on this case the 90% confidence interval would be given by (63.330;81.070)
[tex]63.330 < \mu_{left}- \mu_{right} <81.070[/tex]
More people are using social media to network, rather than phone calls or e-mails (US News & World Report, October 20, 2010). From an employment perspective, jobseekers are no longer calling up friends for help with job placement, as they can now get help online. In a recent survey of 150 jobseekers, 67 said they used LinkedIn to search for jobs. A similar survey of 140 jobseekers, conducted three years ago, had found that 58 jobseekers had used LinkedIn for their job search. Use Table 1.
Let p1 represent the population proportion of recent jobseekers and p2 the population proportion of job seekers three years ago. Let recent survey and earlier survey represent population 1 and population 2, respectively.a. Set up the hypotheses to test whether there is sufficient evidence to suggest that more people are now using LinkedIn to search for jobs as compared to three years ago.
A. H0: p1 − p2 ≥ 0; HA: p1 − p2 < 0
B. H0: p1 − p2 ≤ 0; HA: p1 − p2 > 0
C. H0: p1 − p2 = 0; HA: p1 − p2 ≠ 0
b. Calculate the value of the test statistic. (Round intermediate calculations to 4 decimal places and final answer to 2 decimal places.)c. Calculate the critical value at the 5% level of significance. (Round your answer to 3 decimal places.)d. Interpret the results.
A. Do not reject H0; there is no increase in the proportion of people using LinkedIn
B. Do not reject H0; there is an increase in the proportion of people using LinkedIn
C. Reject H0; there is no increase in the proportion of people using LinkedIn
D. Reject H0; there is an increase in the proportion of people using LinkedIn
Answer:
B. H0: p1 − p2 ≤ 0; HA: p1 − p2 > 0
[tex]z=\frac{0.447-0.414}{\sqrt{0.431(1-0.431)(\frac{1}{150}+\frac{1}{140})}}=0.57[/tex]
[tex]z_{crit}=1.64[/tex]
A. Do not reject H0; there is no increase in the proportion of people using LinkedIn
Step-by-step explanation:
1) Data given and notation
[tex]X_{1}=67[/tex] represent the number of recent jobseekers
[tex]X_{2}=58[/tex] represent the number of job seekers three years ago.
[tex]n_{1}=150[/tex] sample of recent jobseekers selected
[tex]n_{2}=140[/tex] sample of job seekers three years ago selected
[tex]p_{1}=\frac{67}{150}=0.4468[/tex] represent the proportion of recent jobseekers
[tex]p_{2}=\frac{58}{140}=0.4143[/tex] represent the proportion of job seekers three years ago
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the value for the test (variable of interest)
[tex]\alpha=0.05[/tex] significance level given
2) Concepts and formulas to use
We need to conduct a hypothesis in order to check if "More people are using social media to network, rather than phone calls or e-mails", the system of hypothesis would be:
Null hypothesis:[tex]p_{1} - p_{2} \leq 0[/tex]
Alternative hypothesis:[tex]p_{1} - p_{2} > 0[/tex]
We need to apply a z test to compare proportions, and the statistic is given by:
[tex]z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}[/tex] (1)
Where [tex]\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{67+58}{150+140}=0.4310[/tex]
3) Calculate the statistic
Replacing in formula (1) the values obtained we got this:
[tex]z=\frac{0.4468-0.4143}{\sqrt{0.4310(1-0.4310)(\frac{1}{150}+\frac{1}{140})}}=0.5671[/tex]
In order to find the critical value since we have a right tailed test the we need to find a value on the z distribution that accumulates 0.05 of the area on the right tail, and this value is[tex]z_{crit}=1.64[/tex].
4) Statistical decision
Since is a right tailed test the p value would be:
[tex]p_v =P(Z>0.5671)= 0.285[/tex]
Comparing the p value with the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.
So the correct conclusion would be:
A. Do not reject H0; there is no increase in the proportion of people using LinkedIn
Find the volume of the solid generated by revolving the region bounded by the given lines and curves about the x-axis.
y = 2x, y = 2, x = 0
Answer:
V = 8π/3
Step-by-step explanation:
Haven't done calc in years so I might be wrong.
Graph out all the lines needed so you can have a better look at it.
I set y=2x = y=2 to find where the intersect each others so I can have my boundaries for integration.
You goal is to find the area so you can integrate around that area. We're revolving around the x-axis so the area will be a circle.
V = ∫A(x)dx = ∫(πr²)dr
Since we have two different radius, we subtract them from each others.
∫(πr₂² - πr₁²)dr
∫(π(2)² - π(2x)²)dr
∫(4π - 4πx²)dr
4π∫(1 - x²)dr
integrate from 0 to 1 since that's where our boundary is.
V = 4π∫(1 - x²)dr = 8π/3
To find the volume of the solid generated by revolving the region bounded by the given lines and curves about the x-axis, we can use the method of cylindrical shells and integrate the expression 2πx(2-2x) from 0 to 1.
Explanation:To find the volume of the solid generated by revolving the region bounded by the given lines and curves about the x-axis, we can use the method of cylindrical shells. The region is bounded by the line y = 2x, the horizontal line y = 2, and the vertical line x = 0. First, we need to determine the limits of integration by finding the points where the curves intersect. Setting y = 2x and y = 2 equal to each other, we find x = 1. Next, we integrate the expression 2πx(2-2x) with respect to x from 0 to 1. Simplifying and evaluating the integral gives us the volume of the solid generated as 2π/3 cubic units.
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An urn contains 4 blue balls and 6 orange balls. In how many ways can we select 2 blue balls and 5 orange balls from the urn? a) 24 b) 38 c) 36 d) 732 e) 8637 f) None of the above.
Answer:
b) 36
Step-by-step explanation:
We can use combinations to solve this problem.
The binomial coefficient [tex]\binom{n}{k}=\frac{n!}{k!(n-k)!}[/tex] counts the number of ways of choose k elements from a set of n elements.
The product rule from combinatorics says that if there are N ways of doing something and M ways of doing another thing, the number of ways of doing both things is equal to NM.
First, we choose the blue balls. The urn contains 4 blue balls and we select 2 so there are [tex]N=\binom{4}{2}=6[/tex] ways of doing this. Similarly, we choose the 5 orange balls from the set of 6 in the urn, which can be done in [tex]M=\binom{6}{5}=6[/tex] ways. By the product rule, there are MN=6(6)=36 ways of selecting all the balls.
The number of ways 2 blue balls and 5 orange balls can be selected from the urn is 36 number of ways: Option C is correct
Combination has to do with selection.
If r number is selected from n number, this is expressed using the formula:
[tex]nC_r=\frac{n!}{(n-r)!r!}\\[/tex]
If 2 blue balls are selected from 4 blue balls, this is expressed as:
[tex]4C_2=\frac{4!}{(4-2)!2!}\\4C_2=\frac{4\times 3 \times 2!!}{2!2!}\\4C_2=\frac{12}{2} = 6 ways[/tex]
Similarly, if 5 orange balls are selected from 5 orange balls, this is expressed as:
[tex]6C_5=\frac{6!}{(6-5)!5!}\\6C_5=\frac{6\times 5!}{1!5!}\\6C_5=\frac{6}{1} = 6 ways[/tex]
The number of ways 2 blue balls and 5 orange balls can be selected from the urn is 36 number of ways
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Errors in measuring the time of arrival of a wave front from an acoustic source sometimes have an approximate beta distribution. Suppose that these errors, measured in microseconds, have a approximately a beta distribution with α = 1 and β = 2
What is the probability that the measurement error in a randomly selected instance us less than 0.6 µs?
Give the mean and standard deviation of the measurement errors.
Answer:
a) P=0.84
b) Mean=0.33
Standard deviation=0.356
Step-by-step explanation:
The probabilty that the measurement error in a randomly selected instance us less than 0.6 µs is P=0.84.
The mean of a Beta(α = 1, β = 2) is
[tex]\mu=\frac{\alpha}{\alpha+\beta}=\frac{1}{1+2}=0.33[/tex]
The standard deviation of a Beta(α = 1, β = 2) is
[tex]\sigma=\sqrt{\frac{\alpha\beta}{(\alpha+\beta)^2*(\alpha+\beta+1)}}\\\\\\\sigma= \sqrt{\frac{1*2}{(1+2)^2*(1+2+1)}}=\sqrt{\frac{2}{(2)^2*(4)}}=\sqrt{\frac{2}{16} } =\sqrt{0.125}= 0.356[/tex]
Given a Beta Distribution with α = 1 and β = 2, probability calculations typically rely on integration of the probability density function, often executed with statistical software. The mean and standard deviation can be mathematically derived with the given α and β, giving us mean= 0.33 and standard deviation= 0.236.
Explanation:The question falls under the domain of probability distribution, specifically the Beta Distribution. The Beta distribution is a family of continuous probability distributions defined on the interval [0, 1] parameterized by two positive shape parameters, denoted by α and β. For a Beta distribution, the probability density function is given by f(x; α, β). When α = 1 and β = 2, the beta distribution becomes a decreasing linear function.
To find the probability that the measurement error is less than 0.6, we need to integrate the probability density function from 0 to 0.6. However, it's important to note that directly performing these integrations and calculations is a task typically performed by statistical software.
Concerning mean and standard deviation of the Beta Distribution, the mean (µ) is given by α / (α + β), and the variance (σ²) is given by (αβ) / [(α+β)²*(α+β+1)]. Here, given α=1 and β=2, we find µ=1/(1+2) = 0.33 and σ² = (1*2)/(3²*4)=0.0556, and taking square root of variance we get standard deviation σ = 0.236.
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Daniele has 33 quarters and dimes in her piggy bank. The piggy bank contains a total of $4.95. Write and solve a system of equations to find the number of dimes x and the number of quarters y. Solve the linear system by substitution.
Answer:the number of dimes is 22
the number of quarters is 11
Step-by-step explanation:
A dime is worth 10 cents. Converting to dollars, it becomes 10/100 = $0.1
A quarter is worth 25 cents. Converting to dollars, it becomes 25/100 = $0.25
Let x represent the number of dimes.
Let y represent the number of quarters.
Daniele has 33 quarters and dimes in her piggy bank. This means that
x + y = 33
The piggy bank contains a total of $4.95. This means that
0.1x + 0.25y = 4.95 - - - - - - - - -1
Substituting x = 33 - y into equation 1, it becomes
0.1(33 - y) + 0.25y = 4.95
3.3 - 0.1y + 0.25y = 4.95
- 0.1y + 0.25y = 4.95 - 3.3
0.15y = 1.65
y = 1.65/0.15 = 11
Substituting y = 11 into x = 33 - y, it becomes
x= 33 - 11 = 22
Final answer:
To find the number of dimes and quarters in Daniele's piggy bank, a system of equations can be solved by substitution method.
Explanation:
System of Equations:
Let x be the number of dimes and y be the number of quarters.
We have the equations: 0.10x + 0.25y = 4.95 and x + y = 33.
Solving by substitution, we first solve x + y = 33 for x to get x = 33 - y.
Substitute x = 33 - y into 0.10x + 0.25y = 4.95 and solve to find the values of x and y.
The solution is x = 15, y = 18, so there are 15 dimes and 18 quarters.
Definitions and Data Due Sun 02/03/2019 11:59 pm A researcher is interested in attitudes towards releasing prisoners with Alzheimer's who have a life sentence. 85 randomly selected Americans were asked, "Prisoners with Alzheimer's who have a life sentence should be released: Strongly Agree, Agree, Disagree, Strongly Disagree". Match the vocabulary word with its corresponding example. B All Americans The proportion of all Americans who strongly agree that prisoners with life sentences who have Alzheimer's should be released The proportion of the 85 Americans surveyed who strongly agree that prisoners with life sentences who have Alzheimer's should be released СВ The answer-Strongly Agree' or 'Agree, or Disagree, or Strongly Disagree, DThe 85 Americans who participated in the survey D The list of answers that the 85 Americans gave a. Data b. Variable c. Parameter d. Statistic e Sample f. Population 0 0
Answer:
Step-by-step explanation:
Matching each vocabulary word with its corresponding example:
(1). All Americans = f. Population (a group of items, units or subjects under reference of study e.g. inhabitants of a region, numbers of cars in a city, workers in a factory and so on)
(2). The proportion of all Americans who strongly agree that prisoners with life sentences who have Alzheimer's should be released = c. Parameter ( the number that summarizes some characteristic of a population)
(3). The proportion of the 85 Americans surveyed who strongly agree that prisoners with life sentences who have Alzheimer's should be released = d. Statistic (the sample characteristic corresponding to a population parameter used when a sample is used to make statistical inference about a population)
(4). The answer-Strongly Agree' or 'Agree, or Disagree, or Strongly Disagree = b. Variable (any quality, characteristic, quantity or number that can be counted or measured)
(5). The 85 Americans who participated in the survey = e. Sample (a part or fraction of a population selected on some basis)
(6). The list of answers that the 85 Americans gave = a. Data (the pieces of information collected to be used for the analysis)
Final answer:
In the research question provided, the key terms like Population, Parameter, Statistic, Sample, Variable, and Data are matched with examples based on a Likert-Response Scale survey on the release of prisoners with Alzheimer's who have life sentences.
Explanation:
A researcher inquiring about the public's view on releasing prisoners with Alzheimer's who have a life sentence used a Likert-Response Scale to understand the level of agreement with a statement. This method has several components we can define using the given options:
Population - (B) All Americans
Parameter - (C) The proportion of all Americans who strongly agree that prisoners with life sentences who have Alzheimer's should be released
Statistic - (D) The proportion of the 85 Americans surveyed who strongly agree that prisoners with life sentences who have Alzheimer's should be released
Variable - (CV) The answer 'Strongly Agree' or 'Agree, or Disagree, or Strongly Disagree'
Sample - (E) The 85 Americans who participated in the survey
Data - (F) The list of answers that the 85 Americans gave
In this study, the variable constitutes the different levels of agreement, while the data consists of the actual responses collected from the survey participants. The sample involves the subset of the population, which is the 85 Americans who were surveyed, and the statistic is the result derived from this sample. The population, on the other hand, includes all Americans, and the parameter is the value that would be obtained if all Americans could be surveyed.
A contractor claims that their soundproofing will remove 83% of the sound intensity inside the room. If 83% of the sound intensity inside the room is removed, the new sound level will sound what % less loud to people in the room? Round your answer to the nearest 1%.
Answer:
17%
Step-by-step explanation:
If 83% is removed than
total is always 100%
that's why
so the new sound level will sound
100%-83%=17% remains.
hence 17 % is correct answer .
You are driving on an asphalt road that had a 40 mi/h speed limit. A bicyclist veered into your lane so you slam on your brakes. Your tires left three skid marks of 69 ft, 70 ft, and 74 ft. The road had a drag factor of 0.95. Your brakes were operating at 98% efficiency. The police gave you a ticket for speeding. You insist that you were driving under the speed limit. Who is correct?
Answer:
Policeman is correct.
Step-by-step explanation:
Consider the provided information.
The tires left three skid marks of 69 ft, 70 ft, and 74 ft.
The road had a drag factor of 0.95. Your brakes were operating at 98% efficiency.
Thus, drag factor (f) = 0.95 and brakes efficiency (n) = 0.98
Compute the skid distance by finding the mean of length of 3 skids.
[tex]D=\frac{69+70+74}{3}\\D=\frac{213}{3}\\D=71[/tex]
Therefore, the skid distance is 71 ft.
Now find the speed of car using skid speed formula:
[tex]S=\sqrt{30Dfn}[/tex]
Where, S is the speed of car, D is the skid distance, f is the drag factor and n is the beak efficiency.
Substitute the respective values in above formula.
[tex]S=\sqrt{30\times 71\times0.95 \times0.98}[/tex]
[tex]S=\sqrt{1983.03}[/tex]
[tex]S\approx 44.5312[/tex]
Your speed was 44.5312 mi/h which is greater than 40 mi/hr. So policeman is correct.
A sign on the pumps at a gas station encourages customers to have their oil checked, and claims that one out of 8 cars needs to have oil added. If this is true, what is the probability of each of the following:
a) Exactly three out of the next eight cars needs oil. Probability =
b) At least three out of the next eight cars needs oil. Probability =
c) At most 6 out of the next 8 cars needs oil. Probability =
Answer:
0.0561,0.0673,0.999997
Step-by-step explanation:
Given that a sign on the pumps at a gas station encourages customers to have their oil checked, and claims that one out of 8 cars needs to have oil added.
X- no of cars that need oil in a sample of 8 cars is binomial with p = 1/8
since each car is independent of the other and there are only two outcomes.
a) Exactly three out of the next eight cars needs oil.
Probability = [tex]P(X=3) =0.0561[/tex]
b) At least three out of the next eight cars needs oil.
Probability = [tex]P(X\geq 3)\\= 0.0673[/tex]
c) At most 6 out of the next 8 cars needs oil.
Probability =[tex]P(X\leq 6)\\=0.999997[/tex]
Many couples believe that it is getting too expensive to host an "average" wedding in the United States. According to the website www.costofwedding, the average cost of a wedding in the U.S. in 2009 was $24,066. Recently, in a random sample of 40 weddings in the U.S. it was found that the average cost of a wedding was $23,224, with a standard deviation of $2,903. On the basis of this, a 95% confidence interval for the mean cost of weddings in the U.S. is $22,296 to $24,152.
Answer:
Step-by-step explanation:
Given that many couples believe that it is getting too expensive to host an "average" wedding in the United States.
Population mean =24066
Sample mean = 23224
Sample size = 40
Sample std dev = 2903
Since sample std dev is known, we use t critical value.
df =39
Sample mean follows a normal distribution with mean = 23224, and std dev = [tex]\frac{s}{\sqrt{n} } \\=\frac{2903}{\sqrt{40} } \\=459.005[/tex]
t critical value = 2.023
Margin of error = 2.023*459.005
Confidence interval[tex](22295.43, 24152.57)[/tex]
Final answer:
The question involves calculating a 95% confidence interval for the average cost of weddings in the U.S., resulting in a range of $22,296 to $24,152, based on recent sample data.
Explanation:
The question pertains to the formation of a confidence interval for the average cost of weddings in the U.S. based on a sample. The provided data asserts that the average cost of a wedding in 2009 was $24,066, while a more recent sample reveals an average of $23,224 with a standard deviation of $2,903. Calculating a 95% confidence interval results in a range of $22,296 to $24,152. This implies that we can be 95% confident that the true average cost of weddings in the whole population falls within this interval.
The mean hourly wage for employees in goods-producing industries is currently $24.57 (Bureau of Labor Statistics website, April, 12, 2012). Suppose we take a sample of employees from the manufacturing industry to see if the mean hourly wage differs from the reported mean of $24.57 for the goods-producing industries.
To determine if the mean hourly wage of employees in the manufacturing industry differs from the reported mean for the goods-producing industries, a hypothesis test can be conducted. Steps include setting up null and alternative hypotheses, selecting a significance level, collecting a sample, calculating a test statistic, determining critical values or p-values, and making a conclusion based on the results.
Explanation:To determine if the mean hourly wage of employees in the manufacturing industry differs from the reported mean of $24.57 for the goods-producing industries, we can conduct a hypothesis test.
Step 1: Set up the null and alternative hypotheses. The null hypothesis (H0) states that the mean hourly wage of employees in the manufacturing industry is equal to $24.57, while the alternative hypothesis (Ha) states that it is not equal to $24.57.Step 2: Select a significance level. This determines the threshold for rejecting the null hypothesis. Let's say we choose a significance level of 0.05 (5%).Step 3: Collect a sample of employees from the manufacturing industry and calculate their mean hourly wage.Step 4: Calculate the test statistic. In this case, we can use a t-test since we have sample data and want to compare it to a known population mean.Step 5: Determine the critical value(s) or p-value associated with the test statistic. If the test statistic falls in the rejection region (beyond the critical value(s)) or if the p-value is less than the significance level, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.Step 6: Make a conclusion based on the results. If we reject the null hypothesis, it suggests that the mean hourly wage of employees in the manufacturing industry differs from $24.57. If we fail to reject the null hypothesis, it suggests that there is not enough evidence to conclude a difference.Learn more about Hypothesis test here:https://brainly.com/question/34171008
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When a production process is operating correctly, thenumber of units produced per hour has a normal distri- bution with a mean of 92.0 and a standard deviation of3.6. A random sample of 4 different hours was taken.a. Find the mean of the sampling distribution of thesample means. b. Find the variance of the sampling distribution ofthe sample mean.c. Find the standard error of the sampling distribu-tion of the sample mean.d. What is the probability that the sample mean ex-ceeds 93.0 units?
Answer:
a) The mean of the sampling distribution of the sample means is 92.
b) The variance of the sampling distribution of the sample mean is 3.24.
c) The standard error of the sampling distribution of the sample mean is 1.8.
d) 28.77% probability that the sample mean ex-ceeds 93.0 units.
Step-by-step explanation:
The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\frac{\sigma}{\sqrt{n}}[/tex]
Normal probability distribution
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
[tex]\mu = 92, \sigma = 3.6[/tex].
a. Find the mean of the sampling distribution of thesample means.
The mean is the same as the mean of the population. So the mean of the sampling distribution of the sample means is 92.
b. Find the variance of the sampling distribution ofthe sample mean.
The standard deviation is [tex]s = \frac{\sigma}{\sqrt{n}} = \frac{3.6}{\sqrt{4}} = 1.8[/tex]
The variance is [tex]s^{2} = 1.8^{2} = 3.24[/tex]
c. Find the standard error of the sampling distribution of the sample mean.
This is the same as the standard deviation of the sample. So the standard error of the sampling distribution of the sample mean is 1.8.
d. What is the probability that the sample mean ex-ceeds 93.0 units?
This is 1 subtracted by the pvalue of Z when X = 93. So
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{93 - 92}{1.8}[/tex]
[tex]Z = 0.56[/tex]
[tex]Z = 0.56[/tex] has a pvalue of 0.7123.
So there is a 1-0.7123 = 0.2877 = 28.77% probability that the sample mean ex-ceeds 93.0 units.
The mean of the sampling distribution of the sample means is 92.0. The variance is 0.81. The standard error is 0.9.
Explanation:a. The mean of the sampling distribution of the sample means can be calculated by finding the mean of the original population, which is 92.0.
b. The variance of the sampling distribution of the sample mean can be calculated using the formula (standard deviation/original population size)^2. In this case, it is (3.6/4)^2 = 0.81.
c. The standard error of the sampling distribution of the sample mean can be calculated by taking the square root of the variance. So, in this case, it is √0.81 = 0.9.
d. To find the probability that the sample mean exceeds 93.0 units, you can calculate the z-score using the formula (sample mean - population mean)/standard error. Then, you can find the probability of the z-score using a standard normal distribution table or calculator.
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According to the Bureau of Labor Statistics it takes an average of 16 weeks for young workers to find a new job. Assume that the probability distribution is normal and that the standard deviation is two weeks. What is the probability that 20 young workers average less than 15 weeks to find a job? A. 0.0127 B. 0.0225 C. 0.0450 D. 0.0375
Answer:
Option A) 0.0127
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 16 weeks
Standard Deviation, σ = 2 weeks
Sample size = 20
We are given that the distribution of time taken to find a job is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
P(20 young workers average less than 15 weeks)
P(x < 15)
[tex]P( x < 15) = P( z < \displaystyle\frac{15-16}{\frac{2}{\sqrt{20}}}) = P(z < -2.236)[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(x < 15) =0.0127= 1.27\%[/tex]
Thus, 0.0127 is the probability that 20 young workers average less than 15 weeks to find a job.
The probability of a group of 20 workers finding jobs in less than 15 weeks, given the specific mean and standard deviation, was calculated using z-scores and the normal distribution. The result was approximately 0.0571, but this was an interpretation that wasn't presented as an answer choice.
Explanation:In this question, we're asked to calculate the probability that a group of 20 young workers on average find a job in less than 15 weeks, given that the mean is 16 weeks and standard deviation is 2 weeks. This is a problem of statistics, specifically involving the
normal distribution
and
z-scores
. The z-score is calculated using the formula Z = (X - μ) / (σ/√n), where X is the random variable (15 weeks in this case), μ is the population mean (16 weeks), σ is the standard deviation (2 weeks) and n is the sample size (20 workers). This gives us Z = (15 - 16) / (2/√20), which is approximately -1.58. Looking up this z-score in a standard normal distribution table gives us a probability of 0.0571, which isn't an option provided in the question. However, we may have made an error in calculation or interpretation. It's important to thoroughly understand and practice these statistical concepts.
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The Bernsteins, Hendersons, and Smiths each have five children. If the 15 children of these three families camp out in five different tents, where each tent holds three children, and the 15 children are randomly assigned to the five tents, what is the probability that every family has at least two of its children in the same tent?
Answer:
Probability=1-[tex]\frac{(5!)^{3} }{15!}[/tex]
Step-by-step explanation:
We have to solve this question with the help of complementary method. Observe the statement "Probability that every family has at least two of its children in the same tent". The complementary of this statement will be every family does not have atleast two of its children in the same tent ie. Each child of a family is not in the same tent as one from the same family.
P(A)=1-[tex]P(A)^{c}[/tex]
Therefore, we can get P(A) if we just take out the value of [tex]P(A)^{c}[/tex]
Probability=[tex]\frac{TotalNo.OfFavourableOutcomes}{TotalNo.ofOutcomes}[/tex]
Imagine that we have 15 locations to fill and 15 people for it, so the total no. of cases= 15!
Bernstein's children can be arranged in 5 different tents in 5! ways.
Similarly Henderson's and Smith's children can be arranged in 5 different tents in 5! ways only.
Therefore, [tex]P(A)^{c}[/tex]= [tex]\frac{(5!)^{3} }{15!}[/tex]
P(A)=1- [tex]\frac{(5!)^{3} }{15!}[/tex]
A man has S400,000 invested in three rental properties. One property earns 7.5% per year on the investment, the second earns 8%, and the third earns 9%. The total annual earnings from the three properties is S33,700, and the amount invested at 9% equals the sum of the first two investments. Let x equal the investment at 7.5%, y equal the investment at 8%, and z represent the investment at 9%. a. Write an equation that represents the sum of the three investments. b. Write an equation that states that the sum of the returns from all three investments is $33,700. c. Write an equation that states that the amount invested at 9% equals the sum of the other two investments. d. Solve the system of equations to find how much is invested in each property.
Answer: the amount of money invested in the first property is $140000
the amount of money invested in the second is property is $62353
the amount of money invested in the third is property is $20235
Step-by-step explanation:
Let x represent the amount of money invested in the first property.
Let y represent the amount of money invested in the second property.
Let z represent the amount of money invested in the third property.
The man has S400,000 invested in three rental properties. This means that
x + y + z = 400000 - - - - - - - - -1
The first property earns 7.5% per year on the investment, the second earns 8%, and the third earns 9%. The total annual earnings from the three properties is S33,700. This means that
7.5/100 × x + 8/100×y + 9/100×z = 33700
0.075x + 0.08y + 0.09z = 33700 - - - - - 2
The amount invested at 9% equals the sum of the first two investments. This means that
z = x + y - - - - - - - - - - -3
Substituting equation 3 into equation 1 and equation 2, it becomes
x + y + x + y = 400000
2x + 2y = 400000 - - - - - - - - 4
0.075x + 0.08y + 0.09(x + y) = 33700
0.075x + 0.08y + 0.09x + 0.09y = 33700
0.165x + 0.17y = 33700 - - - - - - - - 5
Multiplying equation 4 by 0.165 and equation by 2, it becomes
0.33x + 0.33y = 66000
0.33x + 0.34y = 67400
Subtracting
- 0.01x= - 1400
x = $140000
Substituting x = 140000 into equation 5, it becomes
0. 165(140000) + 0.17y = 33700
23100 + 0.17y = 33700
0.17y = 33700 - 23100 = 10600
y = 10600/0.17 = $62353
z = x + y = 140000 + 62353 = $202353
As the value of the multiple coefficient of determination increases,
a. the goodness of fit for the estimated multiple regression equation increases.
b. the value of the adjusted multiple coefficient of determination decreases.
c. the value of the regression equation's constant b0 decreases.
d. the value of the correlation coefficient decreases.
Answer:
a.the goodness of fit for the estimated multiple regression equation increases.
Step-by-step explanation:
As the value of the multiple coefficient of determination increases,
a. the goodness of fit for the estimated multiple regression equation increases.
As we know that the coefficient of determination measures the variability of response variable with the help of regressor. As we know that if the value of the coefficient of determination increases strength of fit also increases.
Hard times In June 2010, a random poll of 800 working men found that 9% had taken on a second job to help pay the bills. (www.careerbuilder) a) Estimate the true percentage of men that are taking on second jobs by constructing a 95% confidence interval. b) A pundit on a TV news show claimed that only 6% of work-ing men had a second job. Use your confidence interval to test whether his claim is plausible given the poll data.
Answer:
a) The 95% confidence interval would be given (0.0702;0.1098).
We can conclude that the true population proportion at 95% of confidence is between (0.0702;0.1098)
b) Since the confidence interval NOT contains the value 0.06 we have anough evidence to reject the claim at 5% of significance.
Step-by-step explanation:
Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
[tex]p[/tex] represent the real population proportion of interest
[tex]\hat p =0.09[/tex] represent the estimated proportion for the sample
n=800 is the sample size required
[tex]z[/tex] represent the critical value for the margin of error
Confidence =0.95 or 95%
Part a
The population proportion have the following distribution
[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]
The confidence interval would be given by this formula
[tex]\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
For the 95% confidence interval the value of [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2=0.025[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.
[tex]z_{\alpha/2}=1.96[/tex]
The margin of error is given by :
[tex]Me=z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
[tex]Me=1.96 \sqrt{\frac{0.09(1-0.09)}{800}}=0.0198[/tex]
And replacing into the confidence interval formula we got:
[tex]0.09 - 1.96 \sqrt{\frac{0.09(1-0.09)}{800}}=0.0702[/tex]
[tex]0.09 + 1.96 \sqrt{\frac{0.108(1-0.09)}{800}}=0.1098[/tex]
And the 95% confidence interval would be given (0.0702;0.1098).
We can conclude that the true population proportion at 95% of confidence is between (0.0702;0.1098)
Part b
Since the confidence interval NOT contains the value 0.06 we have anough evidence to reject the claim at 5% of significance.
The 95% confidence interval for the true percentage of men taking on second jobs is approximately 7.02% to 10.98%. The pundit's claim of 6% is not plausible.
To construct a 95% confidence interval for the true percentage of men taking on second jobs, we'll use the sample proportion (\( \hat{p} \)) and the margin of error formula. Then, we'll use this interval to test the pundit's claim.
Given:
- Sample size ( n ) = 800
- Sample proportion [tex](\( \hat{p} \))[/tex] = 9% = 0.09
a) Constructing a 95% confidence interval:
The margin of error ( E ) for a 95% confidence interval can be calculated using the formula:
[tex]\[ E = Z \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \][/tex]
Where:
- Z is the Z-score corresponding to the desired confidence level (95%)
- [tex]\( \hat{p} \)[/tex] is the sample proportion
- n is the sample size
We can find the Z-score corresponding to a 95% confidence level, which is approximately 1.96.
Now, let's calculate the margin of error and construct the confidence interval.
b) Testing the pundit's claim:
We'll compare the pundit's claim (6%) with the confidence interval constructed in part (a). If the pundit's claim falls within the confidence interval, it is plausible. Otherwise, it is not plausible.
Let's proceed with the calculations.
a) Constructing a 95% confidence interval:
First, let's calculate the margin of error ( E ):
[tex]\[ E = 1.96 \times \sqrt{\frac{0.09 \times (1 - 0.09)}{800}} \][/tex]
[tex]\[ E \approx 1.96 \times \sqrt{\frac{0.09 \times 0.91}{800}} \][/tex]
[tex]\[ E \approx 1.96 \times \sqrt{\frac{0.0819}{800}} \][/tex]
[tex]\[ E \approx 1.96 \times \sqrt{0.000102375} \][/tex]
[tex]\[ E \approx 1.96 \times 0.010118 \][/tex]
[tex]\[ E \approx 0.0198 \][/tex]
Now, let's construct the confidence interval:
Lower bound: [tex]\( \hat{p} - E = 0.09 - 0.0198 = 0.0702 \)[/tex]
Upper bound: [tex]\( \hat{p} + E = 0.09 + 0.0198 = 0.1098 \)[/tex]
So, the 95% confidence interval for the true percentage of men taking on second jobs is approximately [tex]\( (7.02\%, 10.98\%) \).[/tex]
b) Testing the pundit's claim:
The pundit's claim is 6%, which falls below the lower bound of the confidence interval (7.02%). Therefore, it is not plausible given the poll data.
A farmer wishes to test the effects of a new fertilizer on her tomato yield. She has four equal-sized plots of land-- one with sandy soil, one with rocky soil, one with clay-rich soil, and one with average soil. She divides each of the four plots into three equal-sized portions and randomly labels them A, B, and C. The four A portions of land are treated with her old fertilizer. The four B portions are treated with the new fertilizer, and the four C's are treated with no fertilizer. At harvest time, the tomato yield is recorded for each section of land. What type of experimental design is this? completely randomized design double-blind design matched-pairs design randomized block design
Answer:
Consider the following explanation
Step-by-step explanation:
Completely Randomized Design :-
A completely randomized design is probably the simplest experimental design, in terms of data analysis and convenience. With this design, subjects are randomly assigned to treatments.
This completely randomized design relies on randomization to control for the effects of extraneous variables. The experimenter assumes that, on average, extraneous factors will affect treatment conditions equally; so any significant differences between conditions can fairly be attributed to the independent variable.
Double Blind Design :-
In an experiment, if subjects in the control group know that they are receiving a placebo, the placebo effect will be reduced or eliminated; and the placebo will not serve its intended control purpose.
Blinding is the practice of not telling subjects whether they are receiving a placebo. In this way, subjects in the control and treatment groups experience the placebo effect equally. Often, knowledge of which groups receive placebos is also kept from analysts who evaluate the experiment. This practice is called double blinding. It prevents the analysts from "spilling the beans" to subjects through subtle cues; and it assures that their evaluation is not tainted by awareness of actual treatment conditions.
Matched Pairs Design :-
A matched pairs design is a special case of a randomized block design. It can be used when the experiment has only two treatment conditions; and subjects can be grouped into pairs, based on some blocking variable. Then, within each pair, subjects are randomly assigned to different treatments.
Randomized Block Design :-
With a randomized block design, the experimenter divides subjects into subgroups called blocks, such that the variability within blocks is less than the variability between blocks. Then, subjects within each block are randomly assigned to treatment conditions. Compared to a completely randomized design, this design reduces variability within treatment conditions and potential confounding, producing a better estimate of treatment effects.
The experimental design that is given in the problem, is an example of a Randomized Block Design. Here, the different type of soils can be considered different blocks. In these blocks, the variability among within the blocks is minimum and between the blocks, it is maximum. Also, the treatments ( fertilizers ) are assigned randomly to the plts in each block (different soil type ).
ureg placed 32 chairs in the auditorium. There are 8 chairs in each row. Which
equation could be used to represent this situation
A. 32 x 8
00
--
B. 32 + 8
00
8 = 32
D.
00
x 8 = 32
The equation to represent this situation is 8x = 32
Solution:
Given that ureg placed 32 chairs in the auditorium
There are 8 chairs in each row
To find: equation used to represent this situation
From given information,
1 row = 8 chairs
So let us find an expression to determine the number of rows to place 32 chairs
Let "x" be the number of rows required to place 32 chairs
Since 1 row contains 8 chairs, expression to determine the number of rows to place 32 chairs is given as:
[tex]8 \times \text{ number of row } = 32[/tex]
8x = 32
Thus the equation to represent this situation is 8x = 32
The net weights (in grams) of a sample of bottles filled by a machine manufactured by Edne, and the net weights of a sample filled by a similar machine manufactured by Orno, Inc., are; Edne 8 7 6 9 7 5 Orno 10 7 11 9 12 14 9 8 Testing the claim at the 0.05 level that the mean weight of the bottles filled by the Orno machine is greater than the mean weight of the bottles filled by the Edne machine, what is the critical value for this test?
Answer:
[tex]t=\frac{(10 -7)-(0)}{\sqrt{\frac{2.268^2}{8}+\frac{1.414^2}{6}}}=3.036[/tex]
[tex]p_v =P(t_{12}>3.036) =0.0051[/tex]
So with the p value obtained and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (Orno) is significantly higher than the mean for the group 2 (Edne).
Step-by-step explanation:
The system of hypothesis on this case are:
Null hypothesis: [tex]\mu_1 \leq \mu_2[/tex]
Alternative hypothesis: [tex]\mu_1 > \mu_2[/tex]
Or equivalently:
Null hypothesis: [tex]\mu_1 - \mu_2 \leq 0[/tex]
Alternative hypothesis: [tex]\mu_1 -\mu_2 > 0[/tex]
Our notation on this case :
[tex]n_1 =8[/tex] represent the sample size for group 1 (Orno)
[tex]n_2 =6[/tex] represent the sample size for group 2 (Edne)
We can calculate the sampel means and deviations with the following formulas:
[tex]\bar X =\frac{\sum_{i=1}^n X_i}{n}[/tex]
[tex]s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}[/tex]
[tex]\bar X_1 =10[/tex] represent the sample mean for the group 1
[tex]\bar X_2 =7[/tex] represent the sample mean for the group 2
[tex]s_1=2.268[/tex] represent the sample standard deviation for group 1
[tex]s_2=1.414[/tex] represent the sample standard deviation for group 2
If we see the alternative hypothesis we see that we are conducting a right tailed test.
On this case since the significance assumed is 0.05 and we are conducting a bilateral test we have one critica value, and we need on the right tail of the distribution [tex]\alpha/2 = 0.05[/tex] of the area.
The distribution on this case since we don't know the population deviation for both samples is the t distribution with [tex]df=8+6 -2=12[/tex] degrees of freedom.
We can use the following excel code in order to find the critical value:
"=T.INV(1-0.05,12)"
And the rejection zone is: (1.78,infinity)
The statistic is given by this formula:
[tex]t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{\sqrt{\frac{s^2_1}{n_1}+\frac{S^2_2}{n_2}}}[/tex]
And now we can calculate the statistic:
[tex]t=\frac{(10 -7)-(0)}{\sqrt{\frac{2.268^2}{8}+\frac{1.414^2}{6}}}=3.036[/tex]
The degrees of freedom are given by:
[tex]df=8+6-2=12[/tex]
And now we can calculate the p value using the altenative hypothesis:
[tex]p_v =P(t_{12}>3.036) =0.0051[/tex]
So with the p value obtained and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (Orno) is significantly higher than the mean for the group 2 (Edne).