Answer: The correct answer is C.
Step-by-step explanation: The y-intercept is the point on the graph at which the line crosses over the y-axis. In this case, your y-intercept is 1, because the line crosses over the y-axis at 1. The slope formula is rise/run, or y2 -y1/x2-x1. However since you were not supplies with any coordinates, you'll have to go with rise (up/down) over run (left/right). The slope of the line falls one cube down ward, making the slope negative 1. The slope "runs" to the right 4 cubes, making the slope -1 (falls 1 cube)/4 (moves to the left/right 4 cubes).
What is the equation of a line that passes through the point (8, 1) and is perpendicular to the line whose equation is y=−23x+5 ?
Answer:
Step-by-step explanation:
The equation of a straight line can be represented in the slope-intercept form, y = mx + c
Where c = intercept
m = slope
The equation of the given line is
y = - 23x+5
Comparing it with the slope intercept equation, slope, m = -23
If a line is perpendicular to another line, the slope of the line is the negative reciprocal of the given line. This means that the slope of the line passing through the point (8, 1) is 1/23
We would determine the intercept, c by substituting m = 1/23, x = 8 and y = 1 into y = mx + c. It becomes
1 = 1/23 × 8 + c
1 = 8/23 + c
c = 1 - 8/23 = 15/23
The equation becomes
y = x/23 + 15/23
Using the logistic model f(x)=1501+9e−2x, evaluate the function at f(4). Round your answer to the nearest tenth.
Answer:
[tex]f(4) = 1501 + 9e^{-2*4} = 1501.00[/tex]
Step-by-step explanation:
f(4) is the value of f when x = 4.
We have that
[tex]f(x) = 1501 + 9e^{-2x}[/tex]
So
[tex]f(4) = 1501 + 9e^{-2*4} = 1501.00[/tex]
Answer:1501
Step-by-step explanation:
The given logistic model is expressed as
f(x)=1501+9e−2x
To evaluate the function at f(4), we would substitute x = 4 into the given logistic model. It becomes
f(4)=1501+9e−2 × 4
f(4)=1501+9e−8
Input 1501 plus 9 plus shift Ln plus -8 in a calculator. It becomes
1501 + 0.00302 = 1501.00302
Approximating to the nearest tenth, it becomes I501
The radioactive isotope carbon-14 is present in small quantities in all life forms, and it is constantly replenished until the organism dies, after which it decays to stable carbon-12 at a rate proportional to the amount of carbon-14 present, with a half-life of 5595 years. Suppose C(t) is the amount of carbon-14 present at time t.(a) Find the value of the constant k in the differential equation.(b) In 1988 three teams of scientists found that the Shroud of Turin, which was reputed to be the burial cloth of Jesus, contained about 91 percent of the amount of carbon-14 contained in freshly made cloth of the same material. How old is the Shroud of Turin, according to these data?
Answer:
a) [tex]k = 0.000124[/tex]
b) According to these data, the Shroud of Turin has around 760 years.
Step-by-step explanation:
The amount of carbon-14 is modeled by the following equation:
[tex]C(t) = C_{0}e^{-kt}[/tex]
In which [tex]C_{0}[/tex] is the initial amount and k is the rate of decrease.
(a) Find the value of the constant k in the differential equation.
Half-life of 5595 years.
So [tex]C(5595) = 0.5C_{0}[/tex]
[tex]C(t) = C_{0}e^{-kt}[/tex]
[tex]0.5C_{0} = C_{0}e^{-5595k}[/tex]
[tex]e^{-5595k} = 0.5[/tex]
Applying ln to both sides
[tex]-5595k = -0.69[/tex]
[tex]k = 0.000124[/tex]
b) In 1988 three teams of scientists found that the Shroud of Turin, which was reputed to be the burial cloth of Jesus, contained about 91 percent of the amount of carbon-14 contained in freshly made cloth of the same material. How old is the Shroud of Turin, according to these data?
This is t when [tex]C(t) = 0.91C_{0}[/tex]
[tex]C(t) = C_{0}e^{-kt}[/tex]
[tex]0.91C_{0} = C_{0}e^{-0.000124t}[/tex]
[tex]e^{-0.000124t} = 0.91[/tex]
Applying ln to both sides
[tex]-0.000124t = -0.094[/tex]
[tex]t = 760.57[/tex]
According to these data, the Shroud of Turin has around 760 years.
Use the theoretical method to determine the probability of the following event. A randomly selected person has a birthday in November
Answer:
There is an 8.22% probability that a randomly selected person has a birthday in November.
Step-by-step explanation:
The theoretical method to find the probability is the division of the number of desired outcomes by the number of total outcomes.
A randomly selected person has a birthday in November
There are 365 days in a year, so the number of total outcomes is 365.
There are 30 days in november, so the number of desired outcomes is 30.
So the probability is
[tex]P = \frac{30}{365} = 0.0822[/tex]
There is an 8.22% probability that a randomly selected person has a birthday in November.
Answer:
The probability of a randomly selected person has a birthday in November is [tex]8.2\%[/tex]
Explanation
A randomly selected person has a birthday in November
The probability of the event is
[tex]$P(B)=\frac{\text { No }of \text { days in november }}{No\ of \text { days in year }}$[/tex]
[tex]$P(B)=\frac{30}{365}$[/tex]
[tex]P(B)=8.2\%[/tex]
Therefore, the probability of a randomly selected person has a birthday in November is [tex]8.2\%[/tex]
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https://brainly.com/question/13683667https://brainly.com/question/4538149Assume X and Y are independent random variables with the following distributions:
Col1 X -1 10 1 2
Col2 P(X) 0.3 0.1 0.5 0.1
Col1 Y 2 3 5
Col2 P(Y) 0.6 0.3 0.1 18.
1. Find the mean, variance, and standard deviation of X.
2. Find the mean, variance, and standard deviation of Y.
3. Let W = 3 + 2 X. Find the mean, variance, and standard deviation of W.
Answer:
Step-by-step explanation:
Given that X and Y are independent random variables with the following distributions:
x -1 10 1 2 Total
p 0.3 0.1 0.5 0.1 1
xp -0.3 1 0.5 0.2 1.4
x^2p 0.3 10 0.5 0.4 11.2
Mean of X = 1.4
Var(x) = 11.2-1.4^2 = 9.24
y 2 3 5
p 0.6 0.3 0.1 1
yp 1.2 0.9 0.5 0 2.6
y^2p 2.4 2.7 2.5 0 7.6
Mean of Y = 2.6
Var(Y) = 11.2-1.4^2 = 0.84
3) W=3+2x
Mean of w =3+2*Mean of x = 7.2
Var (w) = 0+2^2 Var(x)= 36.96
True or False? Tell whether the pair of ratios form a proportion. 49/21 and 28/12 Please explain why you chose what you chose. Need answered asap:-)
Answer:
True, they form a proportion because their cross-multiplication products are equal.
Step-by-step explanation:
[tex]\frac{49}{21} =\frac{28}{12}[/tex] True or False?
If they are a proportion, the products you get from cross-multiplying are the same.
Multiply the left numerator by right denominator:
49 X 12 = 588
Multiply the left denominator by right numerator:
21 X 28 = 588
588 = 588
Therefore it is true, 49/21 and 28/12 form a proportion.
Assume that the weights of Chinook Salmon in the Columbia River are normally distributed. You randomly catch and weigh 40 such salmon. The mean weight from your sample is 23.6 pounds with a standard deviation of 3.5 pounds. Test the claim that the mean weight of Columbia River salmon is greater than 23 pounds. Test this claim at the 0.10 significance level.
(a) What type of test is this?i) This is a right-tailed test.ii) This is a two-tailed test. iii) This is a left-tailed test.(b) What is the test statistic? Round your answer to 2 decimal places.
Answer:
a) i) This is a right-tailed test.
b)
[tex]\text{Test statistic} = 1.08[/tex]
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 23 pounds
Sample mean, [tex]\bar{x}[/tex] = 23.6 pounds
Sample size, n = 40
Alpha, α = 0.10
Sample standard deviation, s = 3.5 pounds
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 23\text{ pounds}\\H_A: \mu > 23\text{ pounds}[/tex]
This is a one tailed(right).
Formula:
[tex]\text{Test statistic} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]\text{Test statistic} = \displaystyle\frac{23.6 - 23}{\frac{3.5}{\sqrt{40}} } = 1.08[/tex]
One study reports that 34% of newly hired MBAs are confronted with unethical business practices during their first year of employment. One business school dean wondered if her MBA graduates had similar experiences. She surveyed recent graduates from her school's MBA program to find that 28% of the 116 graduates from the previous year claim to have encountered unethical business practices in the workplace. Can she conclude that her graduates' experiences are different?
Answer:
[tex]z=\frac{0.28 -0.34}{\sqrt{\frac{0.34(1-0.34)}{116}}}=-1.364[/tex]
[tex]p_v =2*P(Z<-1.364)=0.173[/tex]
The p value obtained was a very high value and using the significance level assumed [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIl to reject the null hypothesis, and we can said that at 5% of significance the proportion of graduates from the previous year claim to have encountered unethical business practices in the workplace is not significant different from 0.34.
Step-by-step explanation:
1) Data given and notation
n=116 represent the random sample taken
X represent the number graduates from the previous year claim to have encountered unethical business practices in the workplace
[tex]\hat p=0.28[/tex] estimated proportion of graduates from the previous year claim to have encountered unethical business practices in the workplace
[tex]p_o=0.34[/tex] is the value that we want to test
[tex]\alpha=0.05[/tex] represent the significance level
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value (variable of interest)
2) Concepts and formulas to use
We need to conduct a hypothesis in order to test the claim that the proportion is 0.34 or no.:
Null hypothesis:[tex]p=0.34[/tex]
Alternative hypothesis:[tex]p \neq 0.34[/tex]
When we conduct a proportion test we need to use the z statistic, and the is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].
3) Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
[tex]z=\frac{0.28 -0.34}{\sqrt{\frac{0.34(1-0.34)}{116}}}=-1.364[/tex]
4) Statistical decision
It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.
The next step would be calculate the p value for this test.
Since is a bilateral test the p value would be:
[tex]p_v =2*P(Z<-1.364)=0.173[/tex]
The p value obtained was a very high value and using the significance level assumed [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIl to reject the null hypothesis, and we can said that at 5% of significance the proportion of graduates from the previous year claim to have encountered unethical business practices in the workplace is not significant different from 0.34.
Robin can clean 72 rooms in 666 days.
How many rooms can Robin clean in 999 days?
Will give brainiest to correct answer!!!!!!!!!!!!!!
Answer:
108 rooms
Step-by-step explanation:
Robin can clean 72 rooms in 666 days
means in 666 days rooms that can be clean is 72 rooms
in 1 days rooms that can be clean is 72/666 rooms
in 999 days rooms can be clean is 999\times 72\div 666
after solving we get 108 answer
the standard deviation of a set of 5 different integers each of which is between 0 and 10
Answer:
Mean = 3.8, std dev = 3.06
Step-by-step explanation:
First let us select any 5 integers without repitition from 0 to 10
Let this be 0,2,3,5,9
To calculate mean and standard deviation
x (x-3.8)^2
0 14.44
2 3.24
3 0.64
5 1.44
9 27.04
total 19 46.8
Mean 3.8 9.36 Variance
Variance 9.36
Std dev 3.059411708
Please note that this mean and std deviation would vary according to our selection of 5 integers.
The standard deviation of the set {2, 4, 6, 8, 10} is approximately 2.83.
To calculate the standard deviation of a set of 5 different integers, each of which is between 0 and 10, you'll first need to find the mean (average) of the set and then calculate the squared differences from the mean. Here are the steps:
Find the mean (average) of the set:
Add up all the integers and divide by the number of integers (in this case, 5).
Calculate the squared differences from the mean:
For each integer in the set, subtract the mean and then square the result. Do this for all 5 integers.
Find the variance:
Add up all the squared differences from step 2 and divide by the number of integers (5 in this case).
Calculate the standard deviation:
Take the square root of the variance to find the standard deviation.
Let's go through an example:
Suppose you have the following set of 5 different integers between 0 and 10: {2, 4, 6, 8, 10}
Find the mean:
(2 + 4 + 6 + 8 + 10) / 5 = 30 / 5 = 6
Calculate the squared differences from the mean:
(2 - 6)^2 = 16
(4 - 6)^2 = 4
(6 - 6)^2 = 0
(8 - 6)^2 = 4
(10 - 6)^2 = 16
Find the variance:
(16 + 4 + 0 + 4 + 16) / 5 = 40 / 5 = 8
Calculate the standard deviation:
√8 ≈ 2.83 (rounded to two decimal places)
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Jhons garden has a big planter that mensures 18 2/3 in by 8 5/6 in. What Is the area of Jhons planter
Area of Jhons planter is 164.89 square inches
Solution:
Given that,
Jhons garden has a big planter that measures [tex]18\frac{2}{3}[/tex] in by [tex]8 \frac{5}{6}[/tex] inches
To find: area of Jhons planter
From given information,
[tex]\text{ length } = 18\frac{2}{3} = \frac{3 \times 18 + 2}{3} = \frac{56}{3} \text{ inches }[/tex]
[tex]\text{ width } = 8\frac{5}{6} = \frac{6 \times 8 + 5}{6} = \frac{53}{6} \text{ inches }[/tex]
The area of planter is given as:
[tex]area = length \times width[/tex]
[tex]area = \frac{56}{3} \times \frac{53}{6} = \frac{2968}{18} = 164.89[/tex]
Thus area of Jhons planter is 164.89 square inches
Considered safe for agricultural use. A well in Texas is used to water crops. This well is tested on a regular basis for arsenic. A random sample of 36 tests gave a sample mean of = 7.3 ppb arsenic, with s = 1.9 ppb. Does this information indicate that the mean level of arsenic in this well is less than 8 ppb? Use a 0.01 level of signifcance..
Answer:
This information indicates that the mean level of arsenic in this well is less than 8 ppbat 0.01 level of signifcance..
Step-by-step explanation:
Given that a well in Texas is used to water crops.
This well is tested on a regular basis for arsenic.
A random sample of 36 tests gave a sample mean of = 7.3 ppb arsenic, with s = 1.9 ppb
H0: Sample mean = 8
Ha: Sample mean <8
(left tailed test at 1% level)
Mean difference =-0.70
Std error of mean = [tex]\frac{1.9}{\sqrt{36} } \\=0.3167[/tex]
Test statistic t = -2.204
df = 35
p value = 0.0171
Since p >0.01 we accept H0
This information indicates that the mean level of arsenic in this well is less than 8 ppbat 0.01 level of signifcance..
Factor x3 + 2x2 + x completely. (x + 1)2 x(x2 + 1) x(x + 1)2
Answer:
[tex]x(x+1)^2[/tex]
Step-by-step explanation:
Given:
The expression to factor is given as:
[tex]x^3+2x^2+x[/tex]
In order to factor it, we write the factors of each of the terms of the given polynomial. So,
The factors of the three terms are:
[tex]x^3=x\times x\times x\\\\2x^2=2\times x\times x\\\\x=x[/tex]
Now, 'x' is a common factor for all the three terms. So, we factor it out. This gives,
[tex]x(\frac{x^3}{x}+2\frac{x^2}{x}+\frac{x}{x})\\\\x(x^2+2x+1)[/tex]
Now, we know a identity which is given as:
[tex](a+b)^2=a^2+2ab+b^2[/tex]
Here, [tex]x^2+2x+1[/tex] can be rewritten as [tex]x^2+2(1)(x)+1^2[/tex]
So, [tex]a=x\ and\ b=1[/tex]
Thus, [tex]x^2+2(1)(x)+1^2= (x+1)^2[/tex]
Therefore, the complete factorization of the given expression is:
[tex]x^3+2x^2+x=x(x+1)^2[/tex]
Answer:
x(x+1)^2
Step-by-step explanation:
Two hundred randomly selected people were asked to state their primary source for news about current events: (1) Television, (2) Radio, (3) Internet, or (4) Other. We wish to determine whether the percent distribution of responses is 10%, 30%, 50%, and 10% for news sources 1 through 4, respectively. What is the null hypothesis? a. H0: p1= .10, p2 = .30, p3 = .50, p4 = .10 b. H0: p1 = .25, p2 = .25, p3 = .25, p4 = .25 c. H0: p = .50 d. H0: the proportions are not all equal.
Answer:
H0 as the proportions are not as per the given estimation
Step-by-step explanation:
Given that two hundred randomly selected people were asked to state their primary source for news about current events: (1) Television, (2) Radio, (3) Internet, or (4) Other.
The hypothesised distribution is with 10%, 30%, 50%, and 10% for news sources 1 through 4, respectively
To check whether this is correct, we need not do test for each proportion
Instead we can do hypothesis testing for chi square goodness of fit with
H0 as the proportions are not as per the given estimation
It is believed that as many as 23% of adults over 50 never graduated from high school. We wish to see if this percentage is the same among the 25 to 30 age group. Question 1. How many of this younger age group must we survey in order to estimate the proportion of non-grads to within .10 with 90% confidence? Use the value of p from the over-50 age group. (Round up to the nearest integer.) n = Question 2. Suppose we still want 90% confidence but we want to cut the margin of error to .04. What is the necessary sample size? (Round up to the nearest integer.) n = Question 3. What sample size is needed to estimate the proportion of non-grads to within .04 with 95% confidence? (Round up to the nearest integer.) n =
Answer:
1) n=48
2) n=298
3) n=426
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[/tex] represent the estimated proportion for the sample
n is the sample size required (variable of interest)
[tex]z[/tex] represent the critical value for the margin of error
The population proportion have the following distribution
[tex]p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})[/tex]
Part 1
In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 90% of confidence, our significance level would be given by [tex]\alpha=1-0.90=0.10[/tex] and [tex]\alpha/2 =0.05[/tex]. And the critical value would be given by:
[tex]z_{\alpha/2}=-1.64, z_{1-\alpha/2}=1.64[/tex]
The margin of error for the proportion interval is given by this formula:
[tex] ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex] (a)
And on this case we have that [tex]ME =\pm 0.1[/tex] and we are interested in order to find the value of n, if we solve n from equation (a) we got:
[tex]n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}[/tex] (b)
We can assume that the estimated proportion is 0.23 for the 25 to 30 group. And replacing into equation (b) the values from part a we got:
[tex]n=\frac{0.23(1-0.23)}{(\frac{0.1}{1.64})^2}=47.63[/tex]
And rounded up we have that n=48
Part 2
The margin of error on this case changes to 0.04 so if we use the same formula but changing the value for ME we got:
[tex]n=\frac{0.23(1-0.23)}{(\frac{0.04}{1.64})^2}=297.7[/tex]
And rounded up we have that n=298
Part 3
In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2 =0.025[/tex]. And the critical value would be given by:
[tex]z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96[/tex]
The margin of error for the proportion interval is given by this formula:
[tex] ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex] (a)
And on this case we have that [tex]ME =\pm 0.04[/tex] and we are interested in order to find the value of n, if we solve n from equation (a) we got:
[tex]n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}[/tex] (b)
We can assume that the estimated proportion is 0.23 for the 25 to 30 group. And replacing into equation (b) the values from part a we got:
[tex]n=\frac{0.23(1-0.23)}{(\frac{0.04}{1.96})^2}=425.22[/tex]
And rounded up we have that n=426
The probability shows that the number of younger age group that must will be surveyed will be 48.
How to compute the probability?From the information given, the confidence level is 90%, margin or error is 0.1. The critical value from the z table is given as 1.645.
The number of samples will be:
= 0.23 × (1 - 0.23) × (1.645/0.1)²
= 0.23 × 0.77 × (1.645/0.1)²
= 48
The sample size when we are 90% confidence but we want to cut the margin of error to .04 will be computed thus:
= (1.645/.004)² × 0.23 × (1 - 0.23)
= (41.125)² × 0.1771
= 300
The sample size that is needed to estimate the proportion of non-grads to within .04 with 95% confidence will be:
= (1.96/0.04)² × 0.23 × (1 - 0.23)
= 48.99 × 0.1771
= 426
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For the line = 36.5 + .48x, how do you interpret the value .48?rnrm.gif A. It's the amount of change in y when x increases by one unit.rnrm.gif B. It's the value of y when x = 0.rnrm.gif C. It's the value of x when y = 0.rnrm.gif D. It's the mean of the distribution for the response variable (y) around the explanatory variable (x).rnrm.gif E. It's the amount of change in x when y increases by one unit.
Answer:
A. It's the amount of change in y when x increases by one unit
Step-by-step explanation:
For this case we have the following linear model adjusted:
[tex]\hat y = 36.5 +0.48 x[/tex]
Where y is the dependent variable, x the independent variable, 36.5 represent the intercept and 0.48 the slope.
We can analyze one by one the options to select the most appropiate.
A. It's the amount of change in y when x increases by one unit
True, for this case the slope is defined as:
[tex]m =0.48 \frac{\Delta y}{\Delta x}[/tex]
And is defined as the amount of change in y when x increase 1 unit.
B. It's the value of y when x = 0
False the value of y when x=0 is y= 36.5+0.48(0) = 36.5
C. It's the value of x when y = 0
False, when y=0 we have this:
[tex] 0=36.5 +0.48 x[/tex]
[tex]x= -\frac{36.5}{0.48}=-76.04[/tex]
D. It's the mean of the distribution for the response variable (y) around the explanatory variable (x)
False, the value 0.48 represent the slope obtained from an estimation of least squares and not represent the mean for the response variable y.
E. It's the amount of change in x when y increases by one unit.
False, is defined inverse as: the amount of change in y when x increase 1 unit.
The value .48 in the given linear equation represents the rate of change in y for each unit increase in x. It signifies the slope of the line, implying that the value of y increases by .48 for every unit increase in x.
Explanation:In the equation y = 36.5 + .48x, the value .48 represents the slope of the line in the context of linear equation. The slope indicates the rate of change in y with respect to x. In this case,
.48 means that for each unit increase in the value of x (independent variable), the value of y (dependent variable) will increase by .48 units.
Therefore, the correct answer is A: It's the amount of change in y when x increases by one unit. This is a fundamental component of understanding linear equations, slopes and how they represent the relationship between two variables.
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A physical therapist wanted to know whether the mean step pulse of men was less than the mean step pulse of women. She randomly selected 56 men and 80 women to participate in the study. Each subject was required to step up and down a 6-inch platform. The pulse of each subject was then recorded. The following results were obtained.Two sample T for Men vs WomenN Mean StDev SE MeanMen 56 112.5 11.1 1.5Women 80 118.7 14.2 1.695% CI for men – mu WomenT-Test mu Men = wu Women (vs<)T = 2.85 P = 0.0025 DF = 132State the null and alternative hypotheses. Which of the following is correct?a)A. H0: u1 = u2; Ha :u1>u2B. H0:u1 = u2;Ha::u1 not equal u2C. H0:u1 = u2;Ha:u1
Answer:
Null hypothesis:[tex]\mu_{m} \geq \mu_{w}[/tex]
Alternative hypothesis:[tex]\mu_{m} < \mu_{w}[/tex]
[tex]p_v =P(t_{134}<-2.85)=0.0025[/tex]
D. Reject H0, there is sufficient evidence to conclude that the mean step pulse of men was less than the mean step pulse of women.
So on this case the 95% confidence interval would be given by [tex]-10.502 \leq \mu_{m} -\mu_w \leq -1.898[/tex]
We are 95% confident that the mean difference is in the confidence interval.
Step-by-step explanation:
1) Data given and notation
[tex]\bar X_{m}=112.5[/tex] represent the mean for the sample of men
[tex]\bar X_{w}=118.7[/tex] represent the mean for the sample women
[tex]s_{m}=11.1[/tex] represent the sample standard deviation for the sample men
[tex]s_{w}=14.2[/tex] represent the sample standard deviation for the sample eomen
[tex]n_{m}=56[/tex] sample size for the group men
[tex]n_{w}=80[/tex] sample size for the group women
z would represent the statistic (variable of interest)
2) Concepts and formulas to use
We need to conduct a hypothesis in order to check if the means for the two groups are the same, the system of hypothesis would be:
Null hypothesis:[tex]\mu_{m} \geq \mu_{w}[/tex]
Alternative hypothesis:[tex]\mu_{m} < \mu_{w}[/tex]
We don't have the population standard deviation's, so for this case is better apply a t test to compare means, and the statistic is given by:
[tex]t=\frac{\bar X_{m}-\bar X_{w}}{\sqrt{\frac{s^2_{m}}{n_{m}}+\frac{s^2_{w}}{n_{w}}}}[/tex] (1)
t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.
3) Calculate the statistic
With the info given we can replace in formula (1) like this:
[tex]t=\frac{112.5-118.7}{\sqrt{\frac{11.1^2}{56}+\frac{14.2^2}{80}}}}=-2.85[/tex]
4) Statistical decision
The degrees of freedom are given by:
[tex]df=n_m +n_w -2= 56+80-2=134[/tex]
Since is a left tailed test the p value would be:
[tex]p_v =P(t_{134}<-2.85)=0.0025[/tex]
D. Reject H0, there is sufficient evidence to conclude that the mean step pulse of men was less than the mean step pulse of women.
5) Confidence interval
The confidence interval for the difference of means is given by the following formula:
[tex](\bar X_m -\bar X_w) \pm t_{\alpha/2}\sqrt{(\frac{s^2_m}{n_m}+\frac{s^2_w}{n_w})}[/tex] (1)
The point of estimate for [tex]\mu_1 -\mu_2[/tex] is just given by:
[tex]\bar X_m -\bar X_w =112.5-118.7=-6.2[/tex]
Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,134)".And we see that [tex]z_{\alpha/2}=1.98[/tex]
Now we have everything in order to replace into formula (1):
[tex]-6.2-1.98\sqrt{\frac{11.1^2}{56}+\frac{14.2^2}{80}}}=-10.502[/tex]
[tex]-6.2+1.98\sqrt{\frac{11.1^2}{56}+\frac{14.2^2}{80}}}=-1.898[/tex]
So on this case the 95% confidence interval would be given by [tex]-10.502 \leq \mu_{m} -\mu_w \leq -1.898[/tex]
We are 95% confident that the mean difference is in the confidence interval.
In the context of the study conducted by the physical therapist, the null hypothesis states that the mean step pulse of men equals the mean step pulse of women, and the alternative hypothesis states that the mean step pulse of men is less than that of women. With a t-test result that has a P-value less than 0.05, we reject the null hypothesis and accept the alternative hypothesis, indicating a significant difference between the mean step pulses of men and women.
Explanation:The physical therapist's study is essentially a hypothesis testing problem. The null hypothesis (σ_0) is that the mean step pulse of men is equal to the mean step pulse of women, and the alternative hypothesis (σ_a) is that the mean step pulse of men is less than the mean step pulse of women.
For this problem, we can write the hypotheses as follows:
σ_0: μ1 = μ2
σ_a: μ1 < μ2
The student's available choices seem to reflect this. Choosing between 'H0: μ1 = μ2; Ha :μ1>μ2' and 'H0:μ1 = μ2;Ha::μ1 not equal μ2'. The correct choice is not listed because the alternative hypothesis is incorrectly defined in both cases. The correct alternative hypothesis would be 'μ1 < μ2', pointing that the mean step pulse of men is less than that of women.
The t-test result given in the question 'T = 2.85 P = 0.0025 DF = 132' indicates that the mean step pulse of men is not equal to that of women, with the P-value being less than 0.05. Thus, with this information, you would typically reject the null hypothesis and accept the alternative hypothesis, which suggests a significant difference between the mean step pulses of men and women.
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A minister claims that more than 60% of the adult population attends a religious service at least once a month. The null and alternative hypotheses you'd use to test this claim would be:
a. H0 : μ = .6, Ha: μ > .6
b. H0 : p = .6, Ha: p > .6
c. H0 : p = .6, Ha: p ≠ .6
d. H0 : pˆ = .6, Ha: pˆ > .6
e. H0 : x = .6, Ha: x > .6
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.
Learn more about Null and Alternative Hypotheses here:https://brainly.com/question/33444525
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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.
You test calories for a food item. The brand name has a mean of 158.706 and a sample standard deviation = 25.236, when seventeen are tested. The generic item has a mean of 122.471 and a sample standard deviation = 25.183, when seventeen are tested. Which is a confdence interval of 95%?
Multiple Choice: (Show work)
A) 17.21 to 55.26
B) 18.12 to 54.35
C) 17.79 to 54.67
D) 18.622 to 53.848
Answer:
option C
Step-by-step explanation:
given,
[tex]\bar{x_1} = 158.706, \sigma_1 = 25.36 , n_1= 17[/tex]
[tex]\bar{x_1} = 122.471, \sigma_1 = 25.183 , n_2= 17[/tex]
α = 1 - 0.95 = 0.05
degree of freedom (df) = 17 -1 = 16
critical value[tex]= t_{\alpha/2},df = t_{0.025},16 =2.120[/tex] (from t-table)
margin of error = [tex]t_{\alpha/2}\sqrt{\dfrac{s_1^2}{n_1}+\dfrac{s_2^2}{n_2}}[/tex]
=2.120\times \sqrt{\dfrac{25.36^2}{17}+\dfrac{25.183^2}{17}}[/tex]
= 2.120 x 8.6467
= 18.33
Margin of error = 18.33
Point estimation of difference = [tex]\bar{x_1} - \bar{x_2}[/tex]
= 36.235
lower limit = 36.235 - 18.33 = 17.91
upper limit = 36.235 + 18.33 = 54.57
hence, the nearest option near to answer is option C
Consider the expression 7x2 - 4 - 9x. What is the coefficient of x?
A) -4
B) 7
C) -9
D) 9
Answer:
D) 9
Step-by-step explanation:
7 x 2 - 4 - 9x
14 - 4 - 9x
10 - 9x
You could easily see at first that the coefficient of x is 9 because the coefficient is the number that is before the variable, but you can also do the whole process.
Hope this was helpful :)
Skye’s gross annual income is $36,192. She is paid weekly and has 4% deducted from her paychecks for her 403(b). Her employer matches her deduction, up to 3%. How much is deposited into Skye’s 403(b) each payday?
Answer:
$48.72
Step-by-step explanation:
Step 1: identify the given parameters
Sky's gross annual income = $36,192
Step 2: calculate Sky's weekly salary
= 36,192/52
=$696/week
Step 3: calculate Sky's weekly 4% deduction
= $696 X 4%
=$696 X 0.04 = $27.84
Step 4: calculate Sky's weekly 3% deduction
= $696 X 3%
=$696 X 0.03 = $20.88
Step 5: calculate how much is deposited into Skye’s 403(b) each payday
= $27.84 + $20.88
=$48.72
Answer: 48.72
Step-by-step explanation:
The success rate of a freshman graduating at the same college they started at is 73%. For a group of 15 students, what is the probability that exactly 8 graduate? (round your answer to the nearest hundredth)
Probability of graduating is 73% = 0.73
Probability of not graduating = 1-73% = 0.27
Number of students (n) = 15
Number of graduates (x) = 8
Use the binomial probability formula:
P(x) = (n/x) *p^x * (1-p)^n-x
P(8) = 15/8 * 0.73^8 * 0.27^7
P(8) = 0.0543
Rounded to nearest hundredth = 0.05
Final answer:
To find the probability, use the binomial formula: [tex]\( P(X = 8) = \binom{15}{8} \times (0.73)^8 \times (0.27)^7 \).[/tex] Calculate to get approximately 0.195.
Explanation:
To calculate the probability of exactly 8 out of 15 freshmen graduating, we can use the binomial probability formula:
[tex]\[ P(X = k) = \binom{n}{k} \times p^k \times (1-p)^{n-k} \][/tex]
Where:
( n = 15 ) (total number of students)
( k = 8 ) (number of successes, i.e., number of students graduating)
( p = 0.73 ) (probability of success, i.e., the probability that a student graduates)
Using these values, we can plug them into the formula:
[tex]\[ P(X = 8) = \binom{15}{8} \times (0.73)^8 \times (1-0.73)^{15-8} \][/tex]
[tex]\[ P(X = 8) = \binom{15}{8} \times (0.73)^8 \times (0.27)^7 \][/tex]
Using a calculator or statistical software to compute the binomial coefficient, we find:
[tex]\[ \binom{15}{8} = 6435 \][/tex]
Plugging this into the formula:
[tex]\[ P(X = 8) = 6435 \times (0.73)^8 \times (0.27)^7 \][/tex]
[tex]\[ P(X = 8) \approx 0.195 \][/tex]
So, the probability that exactly 8 out of 15 freshmen graduate is approximately 0.195, rounded to the nearest hundredth.
There are 7 cars in an amusement park ride. There are
42 people in the cars. An equal number of people ride in
each car. How many people ride in one car?
Answer:
6 people ride in each car.
Step-by-step explanation:
42 people / 7 cars = 6 people/car
In this high school level mathematics question, we determine that 6 people ride in one car in the amusement park ride with 7 cars and 42 people. The calculation involves dividing the total number of people by the total number of cars.
Mathematics - In this question, we are determining how many people ride in one car in an amusement park ride where there are 7 cars and 42 people in total. Since the number of people is divided equally among the cars, we can calculate the number of people in one car by dividing the total number of people by the total number of cars.
Calculation:
Total People = 42
Total Cars = 7
Number of People in One Car = Total People / Total Cars = 42 / 7 = 6
Therefore, 6 people ride in one car in this amusement park ride.
Need Help ASAP!! 100 points andbranliest plz pleople plz help.
Answer:
what grade are you in so then I could get the answer
Answer:
[tex] |78| + |12| = |12| + |78| [/tex]
We are conducting a hypothesis test to determine if fewer than 80% of ST 311 Hybrid students complete all modules before class. Our hypotheses are H0 : p = 0.8 and HA : p < 0.8. We looked at 110 randomly selected students and found that 97 of these students had completed the modules before class. What is the appropriate conclusion for this test?
Answer:
[tex]z=\frac{0.881 -0.8}{\sqrt{\frac{0.8(1-0.8)}{110}}}=2.124[/tex]
Null hypothesis:[tex]p\geq 0.8[/tex]
Alternative hypothesis:[tex]p < 0.8[/tex]
Since is a left tailed test the p value would be:
[tex]p_v =P(Z<2.124)=0.983[/tex]
So the p value obtained was a very high value and using the significance level assumed [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.
Be Careful with the system of hypothesis!
If we conduct the test with the following hypothesis:
Null hypothesis:[tex]p\leq 0.8[/tex]
Alternative hypothesis:[tex]p > 0.8[/tex]
[tex]p_v =P(Z>2.124)=0.013[/tex]
So the p value obtained was a very low value and using the significance level assumed [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.
Step-by-step explanation:
1) Data given and notation
n=110 represent the random sample taken
X=97 represent the students who completed the modules before class
[tex]\hat p=\frac{97}{110}=0.882[/tex] estimated proportion of students who completed the modules before class
[tex]p_o=0.8[/tex] is the value that we want to test
[tex]\alpha[/tex] represent the significance level
z would represent the statistic (variable of interest)
[tex]p_v{/tex} represent the p value (variable of interest)
2) Concepts and formulas to use
We need to conduct a hypothesis in order to test the claim that the true proportion is less than 0.8 or 80%:
Null hypothesis:[tex]p\geq 0.8[/tex]
Alternative hypothesis:[tex]p < 0.8[/tex]
When we conduct a proportion test we need to use the z statistic, and the is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].
3) Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
[tex]z=\frac{0.881 -0.8}{\sqrt{\frac{0.8(1-0.8)}{110}}}=2.124[/tex]
4) Statistical decision
It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.
The significance level assumed [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.
Since is a left tailed test the p value would be:
[tex]p_v =P(Z<2.124)=0.983[/tex]
So the p value obtained was a very high value and using the significance level assumed [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.
Be Careful with the system of hypothesis!
If we conduct the test with the following hypothesis:
Null hypothesis:[tex]p\leq 0.8[/tex]
Alternative hypothesis:[tex]p > 0.8[/tex]
[tex]p_v =P(Z>2.124)=0.013[/tex]
So the p value obtained was a very low value and using the significance level assumed [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.
Based on the given data, we do not reject the null hypothesis because the p-value (0.1308) is greater than the alpha level (0.05). Therefore, we conclude that there is insufficient evidence to suggest that fewer than 80% of ST 311 Hybrid students complete all modules before class.
To determine whether fewer than 80% of ST 311 Hybrid students complete all modules before class, we conducted a hypothesis test with the following hypotheses:
Null Hypothesis (H0): p = 0.8Alternative Hypothesis (HA): p < 0.8We collected data from a sample of 110 students, where 97 had completed the modules before class. The test resulted in a p-value of 0.1308.
Alpha: 0.05Decision: Do not reject the null hypothesis.Reason for decision: The p-value is greater than 0.05.Conclusion: There is insufficient evidence to conclude that fewer than 80% of the students complete all modules before class.
The effects of a weight loss drug are standard normally distributed where negative data values represent weight loss. What is the probability a person loses 1.5 pounds or more? (round your answer to the nearest thousandth)My math:Weight loss of 1.5lbs, Z-score = -1.5Probability a person loses 1.5+ lbs = P(x > -1.5)1 – P(X > -1.5)1 - 0.0668 = 0.9332, or 0.932 (this answer was labeled as WRONG)Comment from online quiz: What is the z-score? How can you find the probability from the z-table?Please help clarify what I did wrong. Thanks! -Michelle
Answer:
[tex]P(X\leq -1.5) = P(X < -1.5)=P(Z<-1.5)=0.067[/tex]
Step-by-step explanation:
For this case we know that our random variable X="weight loss or gain" is distributed on this way:
[tex]X \sim N (\mu =0, \sigma=1)[/tex]
And we want the probability a person loses 1.5 pounds or more. If we interpret this an individual person losses 1.5 pounds or more if our random variable is equal or lower than 1.5. That means this:
[tex]P(X\leq -1.5) = P(X < -1.5)=P(Z<-1.5)[/tex]
And for this case we can use the normal standard distribution or excel with the following code:
"=NORM.DIST(-1.5,0,1,TRUE)"
And we got:
[tex]P(X\leq -1.5) = P(X < -1.5)=P(Z<-1.5)=0.067[/tex]
We need to remember that if the negative number decrease on the weight loss we are increasing the loss. For this reason we just need to find P(X<-1.5).
Use this information when answering this question:
n = 14
s = 20
H0: σ2 ≤ 500 and Ha: σ2 > 500
The null hypothesis:
a. should be revised
b. should not be rejected
c. should be rejected
d. None of these alternatives is correct.
Answer: c. should be rejected
Step-by-step explanation:
n = 14
s = 20
H0: σ2 ≤ 500 and Ha: σ2 > 500
Verify that P = Ce^t /1 + Ce^t is a one-parameter family of solutions to the differential equation dP dt = P(1 − P).
Answer:
See verification below
Step-by-step explanation:
We can differentiate P(t) respect to t with usual rules (quotient, exponential, and sum) and rearrange the result. First, note that
[tex]1-P=1-\frac{ce^t}{1+ce^t}=\frac{1+ce^t-ce^t}{1+ce^t}=\frac{1}{1+ce^t}[/tex]
Now, differentiate to obtain
[tex]\frac{dP}{dt}=(\frac{ce^t}{1+ce^t})'=\frac{(ce^t)'(1+ce^t)-(ce^t)(1+ce^t)'}{(1+ce^t)^2}[/tex]
[tex]=\frac{(ce^t)(1+ce^t)-(ce^t)(ce^t)}{(1+ce^t)^2}=\frac{ce^t+ce^{2t}-ce^{2t}}{(1+ce^t)^2}=\frac{ce^t}{(1+ce^t)^2}[/tex]
To obtain the required form, extract a factor in both the numerator and denominator:
[tex]\frac{dP}{dt}=\frac{ce^t}{1+ce^t}\frac{1}{1+ce^t}=P(1-P)[/tex]
The Sutton police department must write, on average, 6 tickets a day to keep department revenues at budgeted levels.
Suppose the number of tickets written per day follows a Poisson distribution with a mean of 6.5 tickets per day. Interpret the value of the mean.
-The mean has no interpretation.
-The expected number of tickets written would be 6.5 per day.
-Half of the days have less than 6.5 tickets written and half of the days have more than 6.5 tickets written.
-The number of tickets that is written most often is 6.5 tickets per day.
Answer:
The expected number of tickets written would be 6.5 per day.
Step-by-step explanation:
Given that the Sutton police department must write, on average, 6 tickets a day to keep department revenues at budgeted levels.
Suppose the number of tickets written per day follows a Poisson distribution with a mean of 6.5 tickets per day.
This means population parameter i.e. expected value of tickets a day is 6.5
Option I is wrong, because 6.5 is the parameter of Poisson distribution for number of tickets
Option III is wrong because mean is the overall average and hence need not be balanced by exactly half.
Option IV is wrong because mean is the expected value and so we cannot say more than 6.5 tickets
Only correct option is option 2.
The expected number of tickets written would be 6.5 per day.
A study was made comparing the cost of a one-bedroom apartment in philadelphia with the cost of similar apartments in Baltimore. A sample of 30 apartments in Philadelphia showed a sample mean of $950 with a standard devation of $50. A sample of 25 apartments in Baltimore showed a sample mean of $915 and a sample stand deviation of $45. Test to see if there is a significant difference in mean rental rate between the two cities. Use a 5% leve of significance.What is your conclusion?
Answer:
[tex]t=\frac{(950 -915)-(0)}{\sqrt{\frac{50^2}{30}}+\frac{45^2}{25}}=2.73[/tex]
[tex]p_v =2*P(t_{53}>2.73) =0.0086[/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 (Philadelphia) is significantly different than the mean for the group 2 (Baltimore).
Step-by-step explanation:
The system of hypothesis on this case are:
Null hypothesis: [tex]\mu_1 = \mu_2[/tex]
Alternative hypothesis: [tex]\mu_1 \neq \mu_2[/tex]
Or equivalently:
Null hypothesis: [tex]\mu_1 - \mu_2 = 0[/tex]
Alternative hypothesis: [tex]\mu_1 -\mu_2\neq 0[/tex]
Our notation on this case :
[tex]n_1 =30[/tex] represent the sample size for group 1 (Philadelphia)
[tex]n_2 =25[/tex] represent the sample size for group 2 (Baltimore)
[tex]\bar X_1 =950[/tex] represent the sample mean for the group 1
[tex]\bar X_2 =915[/tex] represent the sample mean for the group 2
[tex]s_1=50[/tex] represent the sample standard deviation for group 1
[tex]s_2=45[/tex] represent the sample standard deviation for group 2
If we see the alternative hypothesis we see that we are conducting a bilateral test or two tail.
Critical values
On this case since the significance level is 0.05 and we are conducting a bilateral test we have two critical values, and we need on each tail of the distribution [tex]\alpha/2 = 0.025[/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=n_1+n_2 -2= 30+25-2=53[/tex] degrees of freedom.
We can use the following excel codes in order to find the critical values:
"=T.INV(0.025,53)", "=T.INV(1-0.025,53)"
And we got: (-2.01, 2.01)
Calculate th statistic
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{(950 -915)-(0)}{\sqrt{\frac{50^2}{30}}+\frac{45^2}{25}}=2.73[/tex]
The degrees of freedom are given by:
[tex]df=30+25-2=53[/tex]
P value
And now we can calculate the p value using the altenative hypothesis:
[tex]p_v =2*P(t_{53}>2.73) =0.0086[/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 (Philadelphia) is significantly different than the mean for the group 2 (Baltimore).