Answer:
We conclude that the residents of Wilmington, Delaware, have higher income than the national average
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = $44,500
Sample mean, [tex]\bar{x}[/tex] = $52,500
Sample size, n = 16
Alpha, α = 0.05
Sample standard deviation, s = $9,500
a) First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 44,500\text{ dollars}\\H_A: \mu > 44,500\text{ dollars}[/tex]
We use one-tailed(right) t test to perform this hypothesis.
c) Formula:
[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]t_{stat} = \displaystyle\frac{52500 - 44500}{\frac{9500}{\sqrt{16}} } = 3.37[/tex]
b) Rejection rule
If the calculated t-statistic is greater than the t-critical value, we fail to accept the null hypothesis and reject it.
Now,
[tex]t_{critical} \text{ at 0.05 level of significance, 15 degree of freedom } = 1.753[/tex]
Since,
[tex]t_{stat} > t_{critical}[/tex]
We fail to accept the null hypothesis and reject it. We accept the alternate hypothesis. We conclude that the residents of Wilmington, Delaware, have higher income than the national average
A doctor wants to estimate the mean HDL cholesterol of all 20- to 29-year-old females. The number of subjects needed to estimate the mean HDL cholesterol within 3 points with 99% confidence is 160 subjects. Suppose the doctor decides that he could be content with only 90% confidence. Assuming s=14.7 based on earlier studies, how would this decrease in confidence affect the sample size required?
Answer:
[tex]n=(\frac{1.64(14.7)}{3})^2 =64.57 \approx 65[/tex]
So the answer for this case would be n=65 rounded up to the nearest integer
And the sample size would decrease by 160-65=95 subjects
Step-by-step explanation:
1) 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]\bar X[/tex] represent the sample mean for the sample
[tex]\mu[/tex] population mean (variable of interest)
s represent the sample standard deviation
n represent the sample size
ME represent the margin of error
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
2) Solution to the problem
Since the new 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: "=-NORM.INV(0.05,0,1)".And we see that [tex]z_{\alpha/2}=1.64[/tex]
Assuming that the deviation is known we can express the margin of error is given by this formula:
[tex] ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (a)
And on this case we have that ME =\pm 3 and we are interested in order to find the value of n, if we solve n from equation (a) we got:
[tex]n=(\frac{z_{\alpha/2} \sigma}{ME})^2[/tex] (b)
Replacing into formula (b) we got:
[tex]n=(\frac{1.64(14.7)}{3})^2 =64.57 \approx 65[/tex]
So the answer for this case would be n=65 rounded up to the nearest integer
And the sample size would decrease by 160-65=95 subjects
The probability a guest at a downtown hotel arrived via taxi is 0.42. The probability a guest arriving at a downtown hotel requested parking for a car is 0.29. Assume a guest arriving by taxi is mutually exclusive of a guest requesting parking for a car. What is the probability of a guest arriving by taxi and requesting parking for a car?
Answer:
0% probability of a guest arriving by taxi and requesting parking for a car.
Step-by-step explanation:
Two events are said to be mutually exclusive if one event precludes the other. That means that if A and B are mutually exclusive, the probability of A and B happening at the same time is 0%.
In this problem, we have that a guest arriving via taxi and requesting parking at the hotel are mutually exclusive.
So there is a 0% probability of a guest arriving by taxi and requesting parking for a car.
The vector u = 3224, 1429, 2275 gives the numbers of hamburgers, chicken sandwiches, and cheeseburgers, respectively, sold at a fast-food restaurant in one week. The vector v = 1.50, 2.50, 1.90 gives the prices (in dollars) per unit for the three food items. Find the dot product u · v. (Round your answer to two decimal places.) $
Answer:
u.v = $12,731.
Step-by-step explanation:
The dot product between two vectors, a and b, in which
a = (c,d,e)
b = (f,g,h)
Is given by the following formula
a.b = (c,d,e).(f,g,h) = cf + dg + eh.
In this problem, we have that:
u = (3224, 1429, 2275)
v = (1.50, 2.50, 1.90)
So
u.v = (3224, 1429, 2275).(1.50, 2.50, 1.90) = 3224*1.50 + 1429*2.50 + 2275*1.90 = $12,731.
In a recent public opinion poll of 750 adults, 40% said that, if they had to decide between watching a rerun of Gilligan's Island and watching the State of the Union address, they'd choose the rerun. The poll reported a margin of error of 5%. Which of the following best describes what is meant by a 5% margin of error?A. Of those polled, 5% were unclear as to which broadcast they would rather watch.B. The true population percentage is likely to be within 5% of the sample percentage.C. The 750 adults are not a random sample from the adult population.D. Of those polled, 40% will not watch the State of the Union address.E. Of those polled, 5% will probably change their minds and watch the State of the Union address.
Answer:
B. The true population percentage is likely to be within 5% of the sample percentage.
Step-by-step explanation:
These samples have a confidence level of x%, which leads us to a confidence interval with a margin of error is M%.
This means that we are x% sure that the true population percentage is within M% of the sample percentage.
In this problem, the true population percentage is likely to be within 5% of the sample percentage.
So the correct answer is:
B. The true population percentage is likely to be within 5% of the sample percentage.
In the slideshow Simplifying Radicals, under media in Chapter 9, one specific condition to satisfy is to use the quotient rule.
A. True
B. False
Answer:
True
Step-by-step explanation:
In Mathematics, A radical symbol is defined as √
It is an expression that is uses a root, such as the square root (√ ) or cube root ( ∛ )
To fully show a radical expression:
[tex]\sqrt{A}[/tex]
A is the radicand
√ is the radical symbol
Here, the degree is 2.
Quotient Rule for Radicals
For non- negative real numbers, x and y:
[tex]\frac{\sqrt[n]{x}}{\sqrt[n]{y} }[/tex] = [tex]\sqrt[n]{\frac{x}{y}}[/tex]
Example :
[tex]\sqrt{\frac{36}{9}}[/tex]= [tex]\frac{\sqrt{36}}{\sqrt{9} }[/tex] = [tex]\frac{6}{3}[/tex] = 2
So, in simplifying radicals, quotient rule is used as demonstrated above. So the condition is true
Can someone help me with this!
Answer:
Please read the answers below.
Step-by-step explanation:
Let's recall that in a square all its sides are equal length and all the four internal angles measure 90 °
5. If LN = 46, then we have:
OM = 46 (Same length than LN)
PN = LN/2 = 46/2 = 23
ON = √LN²/2 = √ 46²/2 = √ 2,116/2 = √ 1,058 = 32.53 (Rounding to two decimal places)
MN = ON = 32.53
6. m ∠EFG = 90°
m ∠GDH = ∠GDH/2 = 90/2 = 45°
m ∠FEG = ∠DEF/2 = 90/2 = 45°
m ∠DHG = 180 - (∠GDH + ∠DGH) = 180 - (45 + 45)= 180 - 90 = 90°
7. Solve for x
6x - 21 = ∠PQR/2
6x - 21 = 90/2
6x - 21 = 45
6x = 45 +21
6x = 66
x = 11
1. Alexander’s grandparents get him a puppy for his thirteenth birthday, and he names the puppy Boomer. When Boomer is 1 month old, he weighs 4 pounds. When he is 2 months old, he weighs 7 pounds. In this relationship, x represents Boomer’s age (in months), and y represent her weight (in pounds). 1a.) Graph the two points for this relationship and the line passing through them on the coordinate plane. (2 points). 1b). How do you know the graph shows a proportional or non-proportional relationship? (1pt)
Answer:
1a) Plot on the graph.
1b) Its proportional if y (weight) varies directly with variation in x(age) and fits the form y=k*x
Step-by-step explanation:
1a)
draw graph with x axis in increments of 1 month
draw y-axis in increments of 1 pound.
put 1 dot at (1, 4)
put 1 dot at (2,7)
connect the 2 dots, and continue the line to hit one of the axes.
It will not hit the origin.
1b) show that y=kx
4=k*1 , so k=4
7=k*2 , so k=3.5
Since we have 2 different values for k (4 & 3.5), it's not proportional.
Random samples of 50 women and 50 men are taken at Norwich University. They are asked their reaction to increased tuition fees. Of the women, 23 favored the increase. Of the men, 19 favor the increase. At a 10% significance level, does this indicate that a larger proportion of women favor the increase than men?
Answer:
[tex]z=\frac{0.46-0.38}{\sqrt{0.42(1-0.42)(\frac{1}{50}+\frac{1}{50})}}=0.8104[/tex]
[tex]p_v =P(Z>0.8104)=0.209[/tex]
If we compare the p value and using any significance level for example [tex]\alpha=0.1[/tex] always [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the the proportion of women favored the increase is not significantly higher than the proportion of men favored the increase at 10% of significance.
Step-by-step explanation:
1) Data given and notation
[tex]X_{1}=23[/tex] represent the number of women favored the increase
[tex]X_{2}=19[/tex] represent the number of men favored the increase
[tex]n_{1}=50[/tex] sample 1 selected
[tex]n_{2}=50[/tex] sample 2 selected
[tex]p_{1}=\frac{23}{50}=0.46[/tex] represent the proportion of women favored the increase
[tex]p_{2}=\frac{19}{50}=0.38[/tex] represent the proportion of men favored the increase
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the value for the test (variable of interest)
[tex]\alpha=0.1[/tex] represent the significance level
2) Concepts and formulas to use
We need to conduct a hypothesis in order to check if that a larger proportion of women favor the increase than men, the system of hypothesis would be:
Null hypothesis:[tex]p_{1} \leq p_{2}[/tex]
Alternative hypothesis:[tex]p_{1} > p_{2}[/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{23+19}{50+50}=0.42[/tex]
3) Calculate the statistic
Replacing in formula (1) the values obtained we got this:
[tex]z=\frac{0.46-0.38}{\sqrt{0.42(1-0.42)(\frac{1}{50}+\frac{1}{50})}}=0.8104[/tex]
4) Statistical decision
We can calculate the p value for this test.
Since is a one right side test the p value would be:
[tex]p_v =P(Z>0.8104)=0.209[/tex]
If we compare the p value and using any significance level for example [tex]\alpha=0.1[/tex] always [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the the proportion of women favored the increase is not significantly higher than the proportion of men favored the increase at 10% of significance.
The manager of a grocery store has taken a random sample of 100 customers. The average length of time it took the customers in the sample to check out was 3.1 minutes. The population standard deviation is known to be 0.5 minute. We want to test to determine whether or not the mean waiting time of all customers is significantly more than 3 minutes.
a. .0500 b. .0228 c. .0456 d. .0250
Answer:
The correct option is B) 0.0228
Step-by-step explanation:
Consider the provided information.
The manager of a grocery store has taken a random sample of 100 customers. The average length of time it took the customers in the sample to check out was 3.1 minutes. The population standard deviation is known to be 0.5 minute.
Thus, the value of n = 100, [tex]\bar x=3.1[/tex] and σ = 0.5
We want to test to determine whether or not the mean waiting time of all customers is significantly more than 3 minutes.
The null and alternative hypotheses are,
[tex]H_0:\mu\leq3[/tex] and [tex]H_a:\mu>3[/tex]
Compute the test statistic [tex]z=\frac{\bar x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]z=\frac{3.1-3}{\frac{0.5}{\sqrt{100}}}[/tex]
[tex]z=\frac{0.1}{0.05}[/tex]
[tex]z=2[/tex]
By using the table.
P value = P(Z>2) = 0.0228
Hence, the correct option is B) 0.0228
The number of years a Bulldog lives is a random variable with mean 9 and standard deviation 3 , while for Chihuahuas, the mean is 15 and the standard deviation is 4 . Approximate the probability the that in a kennel of 100 Bulldogs and 100 Chihuahuas, the average Chihuahua lives at least 7 years longer than the average Bulldog.
Answer:
[tex]P(\bar X_C -\bar X_B > 7)=P(Z> \frac{(15-9)-7}{0.5})=P(Z>-2) =1-P(Z<-2) = 1-0.02275= 0.97725[/tex]
Step-by-step explanation:
Previous concepts
The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".
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".
Solution to the problem
From the central limit theorem since the sample size for both cases are >30 we can assume that the average follows a normal distribution.
From the central limit theorem we know that the distribution for the sample mean [tex]\bar X[/tex] is given by:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
And let's put some notation
B= represent the bulldogs, C= Chihuahuas
[tex]\bar X_B \sim N(\mu_B =9, \sigma_{\bar x_B}=\frac{3}{\sqrt{100}}=0.3)[/tex]
[tex]\bar X_C \sim N(\mu_C =15, \sigma_{\bar x_C}=\frac{4}{\sqrt{100}}=0.4)[/tex]
And the distribution for the difference of averages would be given by:
[tex]\bar X_C -\bar X_B \sim N(\mu_D = 15-9=6, \sigma_D=\sqrt{\frac{3^2}{100}+\frac{4^2}{100}}=0.5)[/tex]
And for this case we want this probability:
[tex]P(\bar X_C -\bar X_B > 7)[/tex]
And for this can use the z score given by:
[tex] z=\frac{\bar X_D - \mu_D}{\sigma_D}[/tex]
And after replace we got:
[tex]P(Z> \frac{(15-9)-7}{0.5})=P(Z>-2)[/tex]
And we can use the complement rule and we got this:
[tex]P(Z>-2) =1-P(Z<-2) = 1-0.02275= 0.97725[/tex]
Final answer:
To approximate the probability that the average Chihuahua lives at least 7 years longer than the average Bulldog in a kennel of 100 Bulldogs and 100 Chihuahuas, calculate the difference in means and the standard deviation of the difference. Then, calculate the z-score to find the probability using a z-table or calculator. The approximate probability is less than 0.001.
Explanation:
To approximate the probability that the average Chihuahua lives at least 7 years longer than the average Bulldog in a kennel of 100 Bulldogs and 100 Chihuahuas, we need to compare the means of the two populations and determine the difference in years. The mean for Bulldogs is 9 years and the mean for Chihuahuas is 15 years, so the difference in means is 15 - 9 = 6 years.
Now, we need to find the standard deviation of the difference in means. Since we're dealing with independent samples, we can use the formula:
Standard deviation of the difference in means = sqrt((standard deviation of Bulldogs^2) / 100 + (standard deviation of Chihuahuas^2) / 100) = sqrt((3^2) / 100 + (4^2) / 100) = sqrt(0.09 + 0.16) = sqrt(0.25) = 0.5.
Next, we calculate the z-score using the formula:
z-score = (difference in means - 7) / standard deviation of the difference in means = (6 - 7) / 0.5 = -2 / 0.5 = -4.
Finally, we can find the probability using a z-table or calculator. The probability of the average Chihuahua living at least 7 years longer than the average Bulldog is extremely small, as the z-score of -4 corresponds to a very low probability value. Therefore, the approximate probability is less than 0.001.
A producer of fine chocolates believes that the sales of two varieties of truffles differ significantly during the holiday season. The first variety is milk chocolate while the second is milk chocolate filled with mint. It is reasonable to assume that truffle sales are normally distributed with unknown but equal population variances. Two independent samples of 18 observations each are collected for the holiday period. A sample mean of 12 million milk chocolate truffles sold with a sample standard deviation of 2.5 million. A sample mean of 13 million truffles filled with mint sold with a sample standard deviation of 2.3 million. Use milk chocolate as population 1 and mint chocolate as population 2. Assuming the population variances are equal, which of the following is the value of the appropriate test statistic?
Answer:
[tex]\S^2_p =\frac{(18-1)(2.5)^2 +(18 -1)(2.3)^2}{18 +18 -2}=5.77[/tex]
[tex]S_p=2.402[/tex]
[tex]t=\frac{(12 -13)-(0)}{2.402\sqrt{\frac{1}{18}+\frac{1}{18}}}=-1.249[/tex]
[tex]p_v =2*P(t_{34}<-1.249) =0.2201[/tex]
Step-by-step explanation:
When we have two independent samples from two normal distributions with equal variances we are assuming that
[tex]\sigma^2_1 =\sigma^2_2 =\sigma^2[/tex]
And the statistic is given by this formula:
[tex]t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}[/tex]
Where t follows a t distribution with [tex]n_1+n_2 -2[/tex] degrees of freedom and the pooled variance [tex]S^2_p[/tex] is given by this formula:
[tex]\S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}[/tex]
This last one is an unbiased estimator of the common variance [tex]\sigma^2[/tex]
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 =18[/tex] represent the sample size for group 1
[tex]n_2 =18[/tex] represent the sample size for group 2
[tex]\bar X_1 =12[/tex] represent the sample mean for the group 1
[tex]\bar X_2 =13[/tex] represent the sample mean for the group 2
[tex]s_1=2.5[/tex] represent the sample standard deviation for group 1
[tex]s_2=2.3[/tex] represent the sample standard deviation for group 2
First we can begin finding the pooled variance:
[tex]\S^2_p =\frac{(18-1)(2.5)^2 +(18 -1)(2.3)^2}{18 +18 -2}=5.77[/tex]
And the deviation would be just the square root of the variance:
[tex]S_p=2.402[/tex]
And now we can calculate the statistic:
[tex]t=\frac{(12 -13)-(0)}{2.402\sqrt{\frac{1}{18}+\frac{1}{18}}}=-1.249[/tex]
Now we can calculate the degrees of freedom given by:
[tex]df=18+18-2=34[/tex]
And now we can calculate the p value using the altenative hypothesis:
[tex]p_v =2*P(t_{34}<-1.249) =0.2201[/tex]
If we compare the p value obtained 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.
To find the test statistic for comparing the sales of two varieties of truffles, we use the formula for a two-sample t-test with assumed equal population variances. The test statistic is calculated using the sample means, sample standard deviations, and sample sizes for both types of truffles.
Explanation:The student is asking how to find the value of the test statistic when comparing the means of two independent samples with unknown but equal population variances. Using the provided sample means and standard deviations for the two varieties of milk chocolate truffles—one plain and one filled with mint—we can apply the formula for the test statistic in a two-sample t-test where population variances are assumed to be equal.
The formula is given by:
t = (X₁ - X₂) / S_p * sqrt(1/n₁ + 1/n₂)
where:
First, calculate the pooled standard deviation (S_p) using the formula:
S_p = sqrt(((n₁-1) * S₁² + (n₂-1) * S₂²) / (n₁ + n₂ - 2))
Then, plug the values into the above formula to find the t-statistic.
A production manager at a wall clock company wants to test their new wall clocks. The designer claims they have a mean life of 16 years with a standard deviation of 4 years. If the claim is true, in a sample of 42 wall clocks, what is the probability that the mean clock life would be greater than 15.1 years? Round your answer to four decimal places.
Answer:
92.79% probability that the mean clock life would be greater than 15.1 years.
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 = 16, \sigma = 4, n = 42, s = \frac{4}{\sqrt{42}} = 0.6172[/tex]
What is the probability that the mean clock life would be greater than 15.1 years?
This probability is 1 subtracted by the pvalue of Z when [tex]X = 15.1[/tex]. So:
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{15.1 - 16}{0.6172}[/tex]
[tex]Z = -1.46[/tex]
[tex]Z = -1.46[/tex] has a pvalue of 0.0721.
So there is a 1-0.0721 = 0.9279 = 92.79% probability that the mean clock life would be greater than 15.1 years.
Final answer:
Using the Central Limit Theorem and the given standard deviation and mean, the standard error was calculated, and then the z-score was determined. After finding the z-score, the area to the right of it on the standard normal distribution curve gives the probability that the mean clock life is greater than 15.1 years, which is 0.9274.
Explanation:
To find the probability that the mean clock life of a sample of 42 wall clocks is greater than 15.1 years, we can use the Central Limit Theorem which tells us that the sampling distribution of the sample mean will be normally distributed if the sample size is large enough. Since we are dealing with a sample of 42, which is generally considered sufficiently large, the sampling distribution of the sample mean is normally distributed. Given the population mean (μ) is 16 years, the standard deviation (σ) is 4 years, the sample size (n) is 42, and we want to find the probability of the sample mean (μ_x) > 15.1 years, we first need to find the standard error (SE) which is σ/sqrt(n). Then we can find the z-score which is (15.1 - μ) / SE. Lastly, we use the z-score to find the probability from z-tables or a calculator with normal distribution functions.
The standard error (SE) is:
SE = 4 / sqrt(42) ≈ 0.6172
The z-score (z) is:
z = (15.1 - 16) / 0.6172 ≈ -1.4573
To find the probability of the sample mean being greater than 15.1 years, we want the area to the right of the z-score on the standard normal distribution curve. Looking up the z-score of -1.4573 on the z-table or using a normal distribution calculator gives us the probability to the left of the z-score. We subtract this value from 1 to find the probability to the right:
P(X > 15.1) = 1 - P(Z < -1.4573) ≈ 1 - 0.0726 = 0.9274
Therefore, the probability that the mean clock life is greater than 15.1 years is 0.9274 when rounded to four decimal places.
A research article reports the results of a new drug test. The null hypothesis is that the drug has no effect. The alternative hypothesis is that the drug decreases vision loss in people with Macular Degeneration (i.e., the drug is effective). The article gives a p-value of 0.04 in the analysis section. Indicate which of the following interpretations of the p-value are correct. NOTE: There could be more than one answer. A. The probability of getting results at least as extreme as the ones in the study if the drug is actually not effective. B. The probability that the drug is effective. C. The probability that the drug is not effective.
Answer:
A.
Step-by-step explanation:
Hello!
First a little reminder:
The p-value is defined as the probability corresponding to the calculated statistic if possible under the null hypothesis (i.e. the probability of obtaining a value as extreme as the value of the statistic under the null hypothesis).
In the example, the results of a drug test were reported, being the null hypothesis "the drug has no effect" and the alternative hypothesis "the drug decreases vision loss in people with Macular Degeneration"
After conducting the test, the researchers obtained the p-value 0.004
Taking the previous definition of the p-value, the correct answer is A.
To tell whether B. and C. are correct or incorrect there should be specified what signification level was used in the analysis. Remember, the p-value is the probability of the statistic value under the null hypothesis and to use it to make a decision ver the null hypothesis you have to compare it with the signification level. If the p-value is greater than the level of significance, then you don't reject the null hypothesis (Then you can conclude that the drug has no effect) If the p-value is equal or less than the level of significance, then you reject the null hypothesis. (Then you can conclude that the drug decreases the vision loss)
Using a level of signification of 0.01 then the decision is to not reject the null hypothesis but with levels of 0.05 or 0.1 then the decision is to reject it. This is why it is important to know what level of significance was used in the test when interpreting the p-value.
I hope it helps!
n = 14 s = 20 H0: σ2 ≤ 500 Ha: σ2 ≥ 500 The test statistic for this problem equals _____.
a. 12.68
b. 13.33
c. 13.66
d. .63
The test statistic for a chi-square test with a sample size of 14, sample standard deviation of 20, and null hypothesis variance of 500 should be 10.40. However, since this number does not match any of the options provided, there may be an error in the question or the options.
Explanation:The subject in question involves performing a hypothesis test to determine whether the variance of a population is greater than or equal to a specified value. Given a sample size (n = 14), sample standard deviation (s = 20), and the null hypothesis being (σ² ≤ 500), we need to calculate the test statistic for a chi-square distribution with degrees of freedom df = n - 1, which in this case is df = 13. The test statistic is calculated using the formula:
chi-square test statistic = χ² = (n - 1)s² / σ²
Plugging in the values, we get:
chi-square test statistic = (14 - 1) * 20² / 500 = 13 * 400 / 500 = 5200 / 500 = 10.40
However, since this value does not match any of the answer options provided in the initial question, it appears that there may be an error in the question itself or in the provided options. It's important to carefully review the calculation and ensure the values and hypotheses given are as intended.
In a survey of 1000 women age 22 to 35 who work full time, 540 indicated that they would be willing to give up some personal time in order to make more money. The sample was selected in a way that was designed to produce a sample that was representative of women in the targeted age group. (a) Do the sample data provide convincing evidence that the majority of women age 22 to 35 who work full-time would be willing to give up some personal time for more money? Test the relevant hypotheses using α = 0.01. (Round your test statistic to two decimal places and your P-value to four decimal places.)
Answer:
[tex]z=\frac{0.540-0.5}{\sqrt{\frac{0.5(1-0.5)}{1000}}}=2.53[/tex]
[tex]p_v =P(Z>2.53)=0.0057[/tex]
So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.01[/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 1% of significance the true proportion is not significantly higher than 0.5.
Step-by-step explanation:
1) Data given and notation
n=1000 represent the random sample taken
X=540 represent the people indicated that they would be willing to give up some personal time in order to make more money
[tex]\hat p=\frac{540}{1000}=0.54[/tex] estimated proportion of people indicated that they would be willing to give up some personal time in order to make more money
[tex]p_o=0.5[/tex] is the value that we want to test
[tex]\alpha=0.01[/tex] represent the significance level
Confidence=99% or 0.99
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 majority of women age 22 to 35 who work full-time would be willing to give up some personal time for more money :
Null hypothesis:[tex]p\leq 0.5[/tex]
Alternative hypothesis:[tex]p > 0.5[/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].
Check for the assumptions that he sample must satisfy in order to apply the test
a)The random sample needs to be representative: On this case the problem no mention about it but we can assume it.
b) The sample needs to be large enough
[tex]np_o =100*0.5=500>10[/tex]
[tex]n(1-p_o)=1000*(1-0.5)=500>10[/tex]
3) Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
[tex]z=\frac{0.540-0.5}{\sqrt{\frac{0.5(1-0.5)}{1000}}}=2.53[/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 provided [tex]\alpha=0.01[/tex]. The next step would be calculate the p value for this test.
Since is a right tailed test the p value would be:
[tex]p_v =P(Z>2.53)=0.0057[/tex]
So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.01[/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 1% of significance the true proportion is not significantly higher than 0.5.
The sample data provides convincing evidence that the majority of women age 22 to 35 who work full-time would be willing to give up some personal time for more money.
Explanation:To test whether the majority of women age 22 to 35 who work full-time would be willing to give up some personal time for more money, we need to conduct a hypothesis test. The null hypothesis is that the proportion of women willing to give up personal time is 0.50, and the alternative hypothesis is that it is greater than 0.50. We can use a one-sample proportion test.
In this case, the sample proportion is 540/1000 = 0.54. The sample size is large enough for conducting a test as the expected count for both categories is greater than 10. The test statistic can be calculated as (sample proportion - hypothesized proportion) / sqrt((hypothesized proportion * (1 - hypothesized proportion)) / sample size). The Z-score can then be compared to the critical Z-value at a significance level of 0.01 to determine if we reject or fail to reject the null hypothesis. The p-value can also be calculated using the Z-score.
In this specific case, the test statistic is (0.54 - 0.50) / sqrt((0.50 * (1 - 0.50)) / 1000) = 7.21. The corresponding p-value is less than 0.0001, which is smaller than the significance level of 0.01. Therefore, we reject the null hypothesis and conclude that the majority of women age 22 to 35 who work full-time would be willing to give up some personal time for more money.
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Directions: Estimate the sum or difference using front end estimation.
1) 97 – 39
2) 812,344 – 187,675
3) 321 + 79
Directions: Estimate the product using front end estimation.
4) 315 x 821
5) 562 x 791
6) 82 x 156
7) 711 x 884
8) 126 x 952
9) 824 x 541
10) 4027 x 78
11) 796 x 123
Directions: Estimate the quotient using front end estimation.
12) 817 19
13) 3615 72
14) 232 64
15) 559 81
16) 2986 222
17) 10275 232
18) 7428 286
19) 7143 369
20) 628,597 1525
Answer:
Step-by-step explanation:
1) 1) 97 – 39. It becomes
100 - 40 = 60
2)2) 812,344 – 187,675. It becomes
800000 - 200000 = 600000
3)321 + 79. It becomes
300 + 80 = 410
4) 315 x 821. It becomes
300 + 800 = 1100
5) 562 x 791. It becomes
600× 800 = 480000
6) 82 x 156. It becomes
80×200 = 16000
7) 711 x 884. It becomes
700×900 = 630000
8) 126 x 952. It becomes
100×1000 = 100000
9) 824 x 541. It becomes
800× 500 = 400000
10) 4027 x 78. It becomes
4000×80 = 320000
11) 796 x 123. It becomes
800×100 = 80000
12) 817 19. It becomes
800/20 = 40
13) 3615 72. It becomes
4000/70 = 57.1 = 60
14) 232 64. It becomes
200/60 = 33.3 = 30
15) 559 81. It becomes
600/80 = 7.5 = 8
16) 2986 222. It becomes
3000/200 = 15
17) 10275 232. It becomes
10000/200 = 50
18) 7428 286. It becomes
7000/300 = 23.3 = 20
19) 7143 369. It becomes
7000/400 = 17.5 = 18
20) 628,597 1525, it becomes
600000 - 2000 = 300
Answer:
1) 100 - 40 = 60
2) 800000 - 200000 = 600000
3) 300 + 80 = 410
4) 3000 + 800 = 1100
5) 600× 800 = 480000
6) 80×200 = 16000
7) 700×900 = 630000
8) 100×1000 = 100000
9) 800× 500 = 400000
10) 4000×80 = 320000
11) 800×100 = 80000
12) 800/20 = 40
13) 4000/70 = 57.1 = 60
14) 200/60 = 33.3 = 30
15) 600/80 = 7.5 = 8
16) 3000/200 = 15
17) 10000/200 = 50
18) 7000/300 = 23.3 = 20
19) 7000/400 = 17.5 = 18
20) 600000 / 2000 = 300
Step-by-step explanation:
1) 1) 97 – 39. It becomes
100 - 40 = 60
2)2) 812,344 – 187,675. It becomes
800000 - 200000 = 600000
3)321 + 79. It becomes
300 + 80 = 410
4) 315 x 821. It becomes
300 + 800 = 1100
5) 562 x 791. It becomes
600× 800 = 480000
6) 82 x 156. It becomes
80×200 = 16000
7) 711 x 884. It becomes
700×900 = 630000
8) 126 x 952. It becomes
100×1000 = 100000
9) 824 x 541. It becomes
800× 500 = 400000
10) 4027 x 78. It becomes
4000×80 = 320000
11) 796 x 123. It becomes
800×100 = 80000
12) 817 19. It becomes
800/20 = 40
13) 3615 72. It becomes
4000/70 = 57.1 = 60
14) 232 64. It becomes
200/60 = 33.3 = 30
15) 559 81. It becomes
600/80 = 7.5 = 8
16) 2986 222. It becomes
3000/200 = 15
17) 10275 232. It becomes
10000/200 = 50
18) 7428 286. It becomes
7000/300 = 23.3 = 20
19) 7143 369. It becomes
7000/400 = 17.5 = 18
20) 628,597 1525, it becomes
600000 - 2000 = 300
You go to a local mechanic to get your tires changed. The tires cost $300. There is a 6% sales tax, but you get a 10% discount. There is also a $10 non-taxable disposal fee for your old tires, which the mechanic tells you is not subject to discount.a. Write a function, t(x) for the total purchase amount after taxes but before discounts and fees.b. Write a function, d(x) to account for the total after discounts on purchase amount x.c. Does it matter whether the mechanic adds the tax first or takes the discount first?
Answer:
a) [tex]T(x)=300x+18x=318x[/tex]
b) [tex]D(x)=318x-(0.1)(318)+10x\\=296.2x[/tex]
c) Yes.
Step-by-step explanation:
We have that:
[tex]Tires=$300\\Tax=0.006\\Discount=0.1\\[/tex]
And we have $10 free of taxes.
Making x= number of tires to buy, then we have that the total cost of tires is:
[tex]Total_{Tires}=300x[/tex]
So, what we pay for taxes is given by:
[tex]Taxes=(300x)(0.06)=18x[/tex]
a) Then, according to the above, we can write down the total cost before the discount as:
[tex]T(x)=300x+18x=318x[/tex]
b) And the total cost after discounts, is then given by:
[tex]D(x)=318x-(0.1)(318)+10x\\=296.2x[/tex]
c) If the discount is added first, then less tax will be paid because the amount on which it is paid is lower. If the discount is added later, then the taxes will have been taxed on a higher amount, so it does matter whether they are added first or later.
please help!!!! I don't have alot of time!! 100 points and will mark brainliest
At x = -2, there is a blue dot, which is a local minimum, because it is below the x axis.
The dot is located at y = -4
The answer is Yes there is a local minimum at x = -2
The local minimum is y = -4
Answer:
The answer is Yes there is a local minimum at x = -2
The local minimum is y = -4
Step-by-step explanation:
Found answer online so no work!!!
The reported average cost per workbook at a large college is $27.50. A professor claims that the actual average cost per workbook is higher than $27.50. A sample of 44 different workbooks has an average cost of $28.90. The population standard deviation is known to be $5.00. Can the null hypothesis be rejected at alpha= 0.05?
Answer:
We conclude that the actual average cost per workbook is higher than $27.50.
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = $27.50
Sample mean, [tex]\bar{x}[/tex] = $28.90
Sample size, n = 44
Alpha, α = 0.05
Population standard deviation, σ = $5.00
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 27.50\text{ dollars}\\H_A: \mu > 27.50\text{ dollars}[/tex]
We use one-tailed z test to perform this hypothesis.
Formula:
[tex]z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]z_{stat} = \displaystyle\frac{28.90 - 27.50}{\frac{5.00}{\sqrt{44}} } = 1.8573[/tex]
Now, [tex]z_{critical} \text{ at 0.05 level of significance } = 1.64[/tex]
Since,
[tex]z_{stat} > z_{critical}[/tex]
We reject the null hypothesis and accept the alternate hypothesis.
Thus, we conclude that the actual average cost per workbook is higher than $27.50.
In the Lotka-Volterra predator-prey model dx/dt=-ax+bxy,dy/dt=ey-cxy, where x(t) is the predator population and y(t) is the prey population, the coefficient c represents which of the following:A) the predator die-off rateB) the prey growth rateC) the increase in the predator population due to interactions with the preyD) the decrease in the prey population due to interactions with the predator
Answer:
D) the decrease in the prey population due to interactions with the predator
Step-by-step explanation:
We have that:
x is the predator population
y is the prey population.
The coefficient c appears in the equation of dy/dt. So this coefficient is related to the population of the prey. It appears with a minus sign, this means that the prey population decreases. It also multiplies x, so it means that it is related to the predator population.
The correct answer is:
D) the decrease in the prey population due to interactions with the predator
Suppose there are three balls in a box. On one of the balls is the number 1, on another is the number 2, and on the third is the number 3. You select two balls at random and without replacement from the box and note the two numbers observed. The sample space S consists of the three equally likely outcomes ((1,2), (1,3), (2,3)) (disregarding order). Let X be the sum of the two balls selected. (a) What is the distribution for X? (b) What is the probability that the sum is at least 4? (c) What is the mean of X?
The random variable X represents the sum of numbers on the balls chosen. The probability distribution for X is X=3 with 1/3 probability, X=4 with 1/3 probability and X=5 with 1/3 probability. The probability that the sum is at least 4 is 2/3. The expected value or mean of X is 4.
Explanation:In this problem, the random variable X represents the sum of the numbers on the two balls we pick from the box. The possible sum values, excluding the order in which we pick the balls, could be (1+2)=3, (2+3)=5, or (1+3)=4.
(a) Thus, the probability distribution for X would be:
X = 3 with probability 1/3 (occurs when we pick balls 1,2)
X = 4 with probability 1/3 (occurs when we pick balls 1,3)
X = 5 with probability 1/3 (occurs when we pick balls 2,3)
(b) A sum of at least 4 can occur either when X=4 or X=5. Because each of these two outcomes is equally likely with a probability of 1/3, we sum these probabilities to find that the probability that the sum is at least 4 is 2/3.
(c) The mean of X is the expected value, which is calculated by multiplying each outcome by its probability and summing these products. In this case, that would be (3*(1/3)) + (4*(1/3)) + (5*(1/3)) = 4
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Draw a structured flowchart or write pseudocode that describes the process of guessing a number between 1 and 100. After each guess, the player is told that the guess is too high or too low. The process continues until the player guesses the correct number. Pick a number and have a fellow student try to guess it by following your instructions
Pseudocode is below
Step-by-step explanation:
random_number = genRandomInt[1, 100]
get_input = input(“Select a number between 1 and 100: ")
while get_input<100
if get_input >random_number:
print(“the number you selected is high”)
else if get_input < random_number
print(“the number you selected is low”)
else:
print(“correct number!”)
end
The pseudocode provides a structured process for the task of guessing a number between 1 and 100. It directs the user through guessing, receiving feedback and adjustment for subsequent guesses, until correctly guessing the number.
Explanation:The process of guessing a number can be represented in a structured flowchart or pseudocode as follows:
Step 1: Start
Step 2: Define a random number between 1 and 100
Step 3: The player enters a guess
Step 4: Check if the guess is equal to, greater than or less than the random number. If equal, go to Step 6. If the guess is greater, then output 'Guess is too high.' If the guess is lower, output 'Guess is too low.'
Step 5: The player enters a new guess, then return to Step 4.
Step 6: Output 'Congratulations! Correct guess.'
Step 7: End
This pseudocode explains a structured process of guessing a number. The user is given feedback after each guess, allowing them to make an informed guess for the next round. This sequence continues until the player guesses the correct number, finally resulting in a congratulatory message and termination of the process.
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A political scientist believes that she can show that the average age of the voters in a presidential election is less than 40 years. To this end an exit poll of 100 voters is obtained. The data obtained in the poll has a sample average age of 39.2 years with a sample standard deviation of 8 years. We establish a 1% level of significance.
a. Referring to Political Scientist, give the alternative hypothesis.
b. Referring to Political Scientist, describe the rejection region. Be sure to label the endpoint(s).
c. Referring to Political Scientist, what is the conclusion?
Answer:
a) Null hypothesis:[tex]\mu \geq 40[/tex]
Alternative hypothesis:[tex]\mu < 40[/tex]
b) The rejection zone for this test would be:
[tex] (-\infty, -2.36)[/tex]
c) If we compare the p value and the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is significantly less than 40 years at 1% of signficance.
Step-by-step explanation:
1) Data given and notation
[tex]\bar X=39.2[/tex] represent the sample mean
[tex]s=8[/tex] represent the sample standard deviation
[tex]n=100[/tex] sample size
[tex]\mu_o =40[/tex] represent the value that we want to test
[tex]\alpha=0.01[/tex] represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
Part a
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the true mean is less than 40, the system of hypothesis would be:
Null hypothesis:[tex]\mu \geq 40[/tex]
Alternative hypothesis:[tex]\mu < 40[/tex]
If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]t=\frac{39.2-40}{\frac{8}{\sqrt{100}}}=-1[/tex]
Part b: Rejection zone
On this case we are conducting a left tailed test so we need to find first the degrees of freedom for the distribution given by:
[tex]df=n-1=100-1=99[/tex]
Now we need to find a critical value on th t distribution with 99 degrees of freedom that accumulates 0.01 of the area on the right tail and this value is given by [tex]t_{crit}=-2.36[/tex]
And we can find it using the following excel code: "=T.INV(0.01,99)"
So then the rejection zone for this test would be:
[tex] (-\infty, -2.36)[/tex]
Part c: P-value and conclusion
Since is a one left side test the p value would be:
[tex]p_v =P(t_{(99)}<-1)=0.160[/tex]
Conclusion
If we compare the p value and the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is significantly less than 40 years at 1% of signficance.
At Frucht Orchards, apple trees produce an average of pounds of fruit per tree with a standard deviation of pounds. Their mango trees produce an average of pounds of fruit per tree with a standard deviation of pounds. Frucht Orchards is trying two new fertilizers. An apple tree fed Amazing Apples Fertilizer yielded pounds of fruit, while a mango tree fed Amazing Mangos Fertilizer yielded pounds of fruit. Which new fertilizer appears to be more "amazing"?
Answer:
We can conclude that new fertilizer appears to be more effective on apple trees.
Step-by-step explanation:
Mean and standard deviation parameters are missing in the question, I will assume amounts as below:
At Frucht Orchards, apple trees produce an average of 840 pounds of fruit per tree with a standard deviation of 120 pounds. Their mango trees produce an average of 350 pounds of fruit per tree with a standard deviation of 190 pounds. Frucht Orchards is trying two new fertilizers. An apple tree fed Amazing Apples Fertilizer yielded 940 pounds of fruit, while a mango tree fed Amazing Mangos Fertilizer yielded 400 pounds of fruit. Which new fertilizer appears to be more “amazing”?
We need to calculate standardized values (z-scores) of fertilized apple and mango trees to decide which fertilizer appears to be more "amazing".
z score for an unique value can be calculated using the equation:
z=[tex]\frac{X-M}{s}[/tex] where
X is the weight of the specific fruit that the fertilized trees produce. M is the average weight of the specific fruits that the trees produce.s is the standard deviation of the weights of specific fruit (583)For fertilized apple
z=[tex]\frac{940-840}{120}[/tex]≈0.83
For fertilized mango
z=[tex]\frac{400-350}{190}[/tex]≈0.26
Since 0.83>0.26 we can conclude that new fertilizer appears to be more effective on apple trees.
Ann has 30 jobs that she must do in sequence, with the times required to do each of these jobs being independent random variables with mean 50 minutes and standard deviation 10 minutes. Bob has 30 jobs that he must do in sequence, with the times required to do each of these jobs being independent random variables with mean 52 minutes and standard deviation 15 minutes. Ann's and Bob's times are:
Answer:
[tex]P(T_A < T_B) = P(T_A -T_B<0)=P(Z<0.608) =0.728[/tex]
Step-by-step explanation:
Assuming this problem: "Ann has 30 jobs that she must do in sequence, with the times required to do each of these jobs being independent random variables with mean 50 minutes and standard deviation 10 minutes. Bob has 30 jobs that he must do in sequence, with the times required to do each of these jobs being independent random variables with mean 52 minutes and standard deviation 15 minutes. Ann's and Bob's times are independent. Find the approximate probability that Ann finishes her jobs before Bob finishes his jobs".
Previous concepts
The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".
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".
Solution to the problem
For this case we can create some notation.
Let A the values for Ann we know that n1 = 30 jobs solved in sequence and we can assume that the random variable [tex]X_i[/tex] the time in order to do the ith job for [tex]i=1,2,....,n_1[/tex]. We will have the following parameters for A.
[tex]\mu_A = 50, \sigma_A =10[/tex]
W can assume that B represent Bob we know that n2 = 30 jobs solved in sequence and we can assume that the random variable [tex[X_i[/tex] the time in order to do the ith job for [tex]i=1,2,....,n_2[/tex]. We will have the following parameters for A
[tex]\mu_B = 52, \sigma_B =15[/tex]
And we can find the distribution for the total, if we remember the definition of mean we have:
[tex]\bar X= \frac{\sum_{i=1}^n X_i}{n}[/tex]
And [tex]T =n \bar X[/tex]
And the [tex]E(T) = n \mu[/tex]
[tex]Var(T) = n^2 \frac{\sigma^2}{n}=n\sigma^2 [/tex]
So then we have:
[tex]E(T_A)=30*50 =1500 , Var(T_A) = 30*10^2 =3000[/tex]
[tex]E(T_B)=30*52 =1560 , Var(T_B) = 30 *15^2 =6750[/tex]
Since we want this probability "Find the approximate probability that Ann finishes her jobs before Bob finishes his jobs" we can express like this:
[tex]P(T_A < T_B) = P(T_A -T_B<0)[/tex]
Since we have independence (condition given by the problem) we can find the parameters for the random variable [tex]T_A -T_B [/tex]
[tex]E[T_A -T_B] = E(T_A) -E(T_B)=1500-1560=-60[/tex]
[tex]Var[T_A -T_B]= Var(T_A)+Var(T_B) =3000+6750=9750[/tex]
And now we can find the probability like this:
[tex]P(T_A < T_B) = P(T_A -T_B<0)[/tex]
[tex]P(\frac{(T_A -T_B)-(-60)}{\sqrt{9750}}< \frac{60}{\sqrt{9750}})[/tex]
[tex]P(Z<0.608) =0.728[/tex]
A certain pen has been designed so that true average writing lifetime under controlled conditions (involving the use of awriting machine) is at least 10 hours. A random sample of 18 pens is selected, the writing lifetime of each is determined,and a normal probability plot of the resulting data supports the use of a one-sample t test.a. What hypotheses should be tested if the investigators believe a priori that the design specification has been satisfied?b. What conclusion is appropriate if the hypotheses of part (a) are tested,t = —23, and a = .05?c. What conclusion is appropriate if the hypotheses of part (a) are tested, t = —1.8, and u = .01?d. What should be concluded if the hypotheses of part (a) are tested and t = —3.6 ?
Answer:
a) Null hypothesis:[tex]\mu \geq 10[/tex]
Alternative hypothesis:[tex]\mu < 10[/tex]
b) [tex]p_v =P(t_{17}<-2.3)=0.017[/tex]
And on this case since the p value is lower than the significance level we have enough evidence to reject the null hypothesis.
c) [tex]p_v =P(t_{17}<-1.8)=0.0448[/tex]
And on this case since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis.
d) [tex]p_v =P(t_{17}<-3.6)=0.0011[/tex]
And on this case since the p value is lower than the significance level we have enough evidence to reject the null hypothesis.
Step-by-step explanation:
1) Data given and notation
[tex]\bar X[/tex] represent the sample mean
[tex]s[/tex] represent the sample deviation
[tex]n=18[/tex] sample size
[tex]\mu_o =10[/tex] represent the value that we want to test
[tex]\alpha[/tex] represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the true mean is at least 10 hours, the system of hypothesis would be:
Part a
Null hypothesis:[tex]\mu \geq 10[/tex]
Alternative hypothesis:[tex]\mu < 10[/tex]
If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
Part b
For this case we have t=-2.3 , [tex]\alpha=0.05[/tex]
First we need to find the degrees of freedom [tex]df=n-1=18-1=17[/tex]
Now since we are conducting a left tailed test the p value is given by:
[tex]p_v =P(t_{17}<-2.3)=0.017[/tex]
And on this case since the p value is lower than the significance level we have enough evidence to reject the null hypothesis.
Part c
For this case we have t=-1.8 , [tex]\alpha=0.01[/tex]
First we need to find the degrees of freedom [tex]df=n-1=18-1=17[/tex]
Now since we are conducting a left tailed test the p value is given by:
[tex]p_v =P(t_{17}<-1.8)=0.0448[/tex]
And on this case since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis.
Part d
For this case we have t=-3.6 , [tex]\alpha=0.05[/tex]
First we need to find the degrees of freedom [tex]df=n-1=18-1=17[/tex]
Now since we are conducting a left tailed test the p value is given by:
[tex]p_v =P(t_{17}<-3.6)=0.0011[/tex]
And on this case since the p value is lower than the significance level we have enough evidence to reject the null hypothesis.
The student is learning about conducting a one-sample t-test in a hypothesis testing problem related to the lifetime of a pen. The null and alternative hypothesis are defined, and conclusions are drawn based on various test statistics and significance levels.
Explanation:This situation relates to hypothesis testing in statistics, in particular the usage of a one-sample t-test.
The null hypothesis (H0) and alternative hypothesis (Ha) to be tested would be:
H0: μ ≥ 10 hours (The average lifetime of the pen is at least 10 hours)Ha: μ < 10 hours (The average lifetime of the pen is less than 10 hours)The t value represents the test statistic for conducting the hypothesis test, and the 'α' and 'u' are the significance levels.
With t = -23 and α = 0.05, this falls within the rejection region, as the t value is less than the critical value for a one-tailed test at significance level α = 0.05. Hence, we reject the null hypothesis. There is evidence to suggest that the pen's lifetime is less than 10 hours.
Similarly, with t = -1.8 and u = 0.01, the null hypothesis would be retained, as this value is within the acceptance region.The conclusion would be that there is not sufficient evidence to suggest that the pen's lifetime is less than 10 hours.
Finally, with t = -3.6, the null hypothesis is rejected, suggesting that the pen's lifetime is less than 10 hours. However, this is only applicable if the significance level is above the calculated p-value for t = -3.6.
Learn more about Hypothesis Testing here:https://brainly.com/question/34171008
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The spinner for a board game has eight colors the arrow can land on. What are the degrees of freedom for a goodness-of-fit test of the fairness of this spinner?
A. 9
B. 8
C. 7
D. 6
E. We can't determine the degrees of freedom without knowing the sample size.
Answer: Option 'c' is correct.
Step-by-step explanation:
Since we have given that
Number of colors that the arrow can land on = 8
We need to find the degrees of freedom for a goodness of fit test of the fairness of this spinner.
As we know that degree of freedom is 1 less than the size of sample.
So, Mathematically, it is expressed as below:
So, Degree of freedom would be
[tex]v=n-1\\\\v=8-1\\\\v=7[/tex]
Hence, Option 'c' is correct.
equation form for
slope =3
point (2, 17)
Answer:
The equation of the line with slope is 3 and point (2,17) is [tex]y=3x+11[/tex]
Step-by-step explanation:
Given slope is 3 ie., m=3 and point (2,17)
To find the equation of the line with slope and a point:
Using slope point formula
[tex]y-y_{1}=m(x-x_{1})[/tex]
Let the point (2,17) be [tex](x_{1},y_{1})[/tex]
Substitute the values of m=3 and point (2,17)
[tex]y-y_{1}=m(x-x_{1})[/tex]
[tex]y-17=3(x-2)[/tex]
[tex]y-17=3x-6[/tex]
[tex]y=3x-6+17[/tex]
[tex]y=3x+11[/tex]
Therefore the equation of the line is [tex]y=3x+11[/tex]
the air in a small room 12ft by 8ft by 8ft is 3% carbon monoxide. Starting at t=0, fresh air containing no carbon monoxide is blown into the room at a rate of 100ft cubed/min. If air in the room flows out through aa vent at the same rate, when will the air in the room be 0.01% carbon monoxide?
Answer:
The air in the room at 0.01% carbon monoxide at 43.8 min
Step-by-step explanation:
Let be the volume of CO in the room at time t, be v(t) and the total volume of the room be V. The volume percent of CO in the room at a given time is then given by:
[tex]p(t) = \frac{100\times v(t)}{V}[/tex]
Volume percent is the measure of concentration used in this problem. The "Amount" of CO in the room is then measured in terms of the volume of CO in the room.
Let the rate at which fresh air enters the room be f, which is the same as the rate at which air exits the room. We assume that the air in the room mixes instantaneously with the air entering the room, so that the concentration of CO is uniform throughout the room.
As you wrote, the rate at which the volume of CO in the room changes with time is given by
[tex]\frac{dvt}{dt} = 0 \times f -\frac{f}{v} \times v(t) = -\frac{f}{v} \times v(t)[/tex]
This is a simple first-order equation:
[tex]\frac{dv}{v} = -\frac{f}{v} dt[/tex]
[tex]ln(v) - ln(c) = -\frac{f}{v} \times t[/tex]
where ln(c) is the constant of integration.
ln [tex]\frac{v}{c} = -\frac{f}{v} \times t[/tex]
[tex]v(t) = c \times e^{(-f*\frac{t}{V})}[/tex]
In terms of volume percent,
[tex]p(t) = \frac{100*v(t)}{V}= (\frac{C}{V})*exp(\frac{-f \times t}{v})[/tex]
where C = 100*\frac{c}{V} is just another way of writing the constant.
Plugging in the values for the constants, we get:
[tex]p(t) = (\frac{C}{768 cu.ft.})* exp(\frac{-t}{7.68 min})[/tex]
Now use the initial condition (p(0) = 3% at t = 0) to solve for C:
3% = C
[tex]p(t) = (3\%)\times exp(\frac{t}{7.68 min})[/tex]
To find the time when the air in the room reaches a certain value, it is easier to rewrite this solution as:
[tex]\frac{p(t)}{3\%} = exp(\frac{-t}{7.68 min})[/tex]
[tex]t(p) = -(7.68 min)ln(\frac{p}{3\%} )[/tex]
[tex]= (7.68 min)*ln(\frac{3\%}{p})[/tex]
The question asks when p(t) = 0.01%. Plugging this into the above equation, we get:
[tex]t(0.01\%) = (7.68 min)*ln(\frac{3}{0.01}) = 43.8 min[/tex]
The average annual salary for employees in a store is $50,000. It is given that the population standard deviation is $5,000.
Suppose that a random sample of 70 employees will be selected from the population.What is the value of the standard error of the average annual salary?
Round your answer to the nearest integer.
Answer: Standard error of the average annual salary SE = $597.6
Step-by-step explanation:
Given;
Standard deviation = r = $5,000
Number of samples = n = 70
Mean = $50,000
To derive the standard error of mean SE. It is given as
SE = r/√n
SE = $5,000/√70
SE = $5,000/8.3666
SE = $597.6