In IBM SPSS, where in the analyze menu can you find Wilcoxon test?

A. Correlate

B. Mixed models

C. Compare means

D. Nonparamatic tests

A new thermostat has been engineered for the frozen food cases in large supermarkets. Both the old and new thermostats hold temperatures at an average of 25°F. However, it is hoped that the new thermostat might be more dependable in the sense that it will hold temperatures closer to 25°F. One frozen food case was equipped with the new thermostat, and a random sample of 26 temperature readings gave a sample variance of 4.7. Another similar frozen food case was equipped with the old thermostat, and a random sample of 14 temperature readings gave a sample variance of 12.2. Test the claim that the population variance of the old thermostat temperature readings is larger than that for the new thermostat. Use a 5% level of significance. How could your test conclusion relate to the question regarding the dependability of the temperature readings? (Let population 1 refer to data from the old thermostat.)

(a) What is the level of significance?


State the null and alternate hypotheses.

H₀: σ₁² = σ₂²; H₁: σ₁² ≠ σ₂²H₀: σ₁² = σ₂²; H₁: σ₁² > σ₂²    H₀: σ₁² > σ₂²; H₁: σ₁² = σ₂²H₀: σ₁² = σ₂²; H₁: σ₁² < σ₂²

(b) Find the value of the sample F statistic. (Round your answer to two decimal places.)


What are the degrees of freedom?

What assumptions are you making about the original distribution?

The populations follow independent chi-square distributions. We have random samples from each population.The populations follow dependent normal distributions. We have random samples from each population.    The populations follow independent normal distributions.The populations follow independent normal distributions. We have random samples from each population.

(c) Find or estimate the P-value of the sample test statistic. (Round your answer to four decimal places.)

p-value > 0.1000.050 < p-value < 0.100    0.025 < p-value < 0.0500.010 < p-value < 0.0250.001 < p-value < 0.010p-value < 0.001

(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis?

At the α = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant.At the α = 0.05 level, we reject the null hypothesis and conclude the data are not statistically significant.    At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are statistically significant.At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.

(e) Interpret your conclusion in the context of the application.

Reject the null hypothesis, there is sufficient evidence that the population variance is larger in the old thermostat temperature readings.Fail to reject the null hypothesis, there is sufficient evidence that the population variance is larger in the old thermostat temperature readings.    Reject the null hypothesis, there is insufficient evidence that the population variance is larger in the old thermostat temperature readings.Fail to reject the null hypothesis, there is insufficient evidence that the population variance is larger in the old thermostat temperature readings.

Rothamsted Experimental Station (England) has studied wheat production since 1852. Each year, many small plots of equal size but different soil/fertilizer conditions are planted with wheat. At the end of the growing season, the yield (in pounds) of the wheat on the plot is measured. For a random sample of years, one plot gave the following annual wheat production (in pounds).

Use a calculator to verify that, for this plot, the sample variance is s² ≈ 0.310.

Another random sample of years for a second plot gave the following annual wheat production (in pounds).

Use a calculator to verify that the sample variance for this plot is s² ≈ 0.091.

Test the claim that the population variance of annual wheat production for the first plot is larger than that for the second plot. Use a 1% level of significance.

(a) What is the level of significance?

State the null and alternate hypotheses.

H_o: σ₁² = σ₂²; H₁: σ₁² > σ₂²H_o: σ₁² > σ₂²; H₁: σ₁² = σ₂²    H_o: σ₂² = σ₁²; H₁: σ₂² > σ₁²H_o: σ₁² = σ₂²; H₁: σ₁² ≠ σ₂²

(b) Find the value of the sample F statistic. (Use 2 decimal places.)


What are the degrees of freedom?

What assumptions are you making about the original distribution?

The populations follow independent chi-square distributions. We have random samples from each population.The populations follow independent normal distributions. We have random samples from each population.    The populations follow dependent normal distributions. We have random samples from each population.The populations follow independent normal distributions.

(c) Find or estimate the P-value of the sample test statistic. (Use 4 decimal places.)

p-value > 0.1000.050 < p-value < 0.100    0.025 < p-value < 0.0500.010 < p-value < 0.0250.001 < p-value < 0.010p-value < 0.001

(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis?

At the α = 0.01 level, we reject the null hypothesis and conclude the data are not statistically significant.At the α = 0.01 level, we reject the null hypothesis and conclude the data are statistically significant.    At the α = 0.01 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.At the α = 0.01 level, we fail to reject the null hypothesis and conclude the data are statistically significant.

(e) Interpret your conclusion in the context of the application.

Fail to reject the null hypothesis, there is sufficient evidence that the variance in annual wheat production is greater in the first plot.Reject the null hypothesis, there is insufficient evidence that the variance in annual wheat production is greater in the first plot.    Reject the null hypothesis, there is sufficient evidence that the variance in annual wheat production is greater in the first plot.Fail to reject the null hypothesis, there is insufficient evidence that the variance in annual wheat production is greater in the first plot.

You don't need to be rich to buy a few shares in a mutual fund. The question is, how reliable are mutual funds as investments? This depends on the type of fund you buy. The following data are based on information taken from a mutual fund guide available in most libraries.

A random sample of percentage annual returns for mutual funds holding stocks in aggressive-growth small companies is shown below.

Use a calculator to verify that s² ≈ 347.723 for the sample of aggressive-growth small company funds.

Another random sample of percentage annual returns for mutual funds holding value (i.e., market underpriced) stocks in large companies is shown below.

Use a calculator to verify that s² ≈ 137.336 for value stocks in large companies.

Test the claim that the population variance for mutual funds holding aggressive-growth in small companies is larger than the population variance for mutual funds holding value stocks in large companies. Use a 5% level of significance. How could your test conclusion relate to the question of reliability of returns for each type of mutual fund?

(a) What is the level of significance?

State the null and alternate hypotheses.

H_o: σ₁² = σ₂²; H₁: σ₁² > σ₂²H_o: σ₁² > σ₂²; H₁: σ₁² = σ₂²    H_o: σ₂² = σ₁²; H₁: σ₂² > σ₁²H_o: σ₁² = σ₂²; H₁: σ₁² ≠ σ₂²

(b) Find the value of the sample F statistic. (Use 2 decimal places.)


What are the degrees of freedom?

What assumptions are you making about the original distribution?

The populations follow dependent normal distributions. We have random samples from each population.The populations follow independent normal distributions.    The populations follow independent normal distributions. We have random samples from each population.The populations follow independent chi-square distributions. We have random samples from each population.

(c) Find or estimate the P-value of the sample test statistic. (Use 4 decimal places.)

p-value > 0.1000.050 < p-value < 0.100    0.025 < p-value < 0.0500.010 < p-value < 0.0250.001 < p-value < 0.010p-value < 0.001

(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis?

At the α = 0.05 level, we reject the null hypothesis and conclude the data are not statistically significant.At the α = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant.    At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are statistically significant.

(e) Interpret your conclusion in the context of the application.

Fail to reject the null hypothesis, there is sufficient evidence that the variance in percentage annual returns for mutual funds is greater in the aggressive-growth in small companies.Reject the null hypothesis, there is insufficient evidence that the variance in percentage annual returns for mutual funds is greater in the aggressive-growth in small companies.    Reject the null hypothesis, there is sufficient evidence that the variance in percentage annual returns for mutual funds is greater in the aggressive-growth in small companies.Fail to reject the null hypothesis, there is insufficient evidence that the variance in percentage annual returns for mutual funds is greater in the aggressive-growth in small companies.

A new kind of typhoid shot is being developed by a medical research team. The old typhoid shot was known to protect the population for a mean time of 36 months, with a standard deviation of 3 months. To test the time variability of the new shot, a random sample of 21 people were given the new shot. Regular blood tests showed that the sample standard deviation of protection times was 2.1 months. Using a 0.05 level of significance, test the claim that the new typhoid shot has a smaller variance of protection times.

(a) What is the level of significance?


State the null and alternate hypotheses.

H_o: σ² = 9; H₁: σ² < 9H_o: σ² = 9; H₁: σ² ≠ 9    H_o: σ² < 9; H₁: σ² = 9H_o: σ² = 9; H₁: σ² > 9

(b) Find the value of the chi-square statistic for the sample. (Round your answer to two decimal places.)


What are the degrees of freedom?


What assumptions are you making about the original distribution?

We assume a exponential population distribution.We assume a uniform population distribution.    We assume a binomial population distribution.We assume a normal population distribution.

(c) Find or estimate the P-value of the sample test statistic.

P-value > 0.1000.050 < P-value < 0.100    0.025 < P-value < 0.0500.010 < P-value < 0.0250.005 < P-value < 0.010P-value < 0.005

(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis?

Since the P-value > α, we fail to reject the null hypothesis.Since the P-value > α, we reject the null hypothesis.    Since the P-value ≤ α, we reject the null hypothesis.Since the P-value ≤ α, we fail to reject the null hypothesis.

(e) Interpret your conclusion in the context of the application.

At the 5% level of significance, there is insufficient evidence to conclude that the new typhoid shot has a smaller variance of protection times.At the 5% level of significance, there is sufficient evidence to conclude that the new typhoid shot has a smaller variance of protection times.    

(f) Find a 90% confidence interval for the population standard deviation. (Round your answers to two decimal places.)

Interpret the results in the context of the application.

We are 90% confident that σ lies above this interval.We are 90% confident that σ lies within this interval.    We are 90% confident that σ lies outside this interval.We are 90% confident that σ lies below this interval.

The type of household for the U.S. population and for a random sample of 411 households from a community in Montana are shown below.

Use a 5% level of significance to test the claim that the distribution of U.S. households fits the Dove Creek distribution.

(a) What is the level of significance?


State the null and alternate hypotheses.

H₀: The distributions are the same.
H₁: The distributions are different.H₀: The distributions are the same.
H₁: The distributions are the same.    H₀: The distributions are different.
H₁: The distributions are the same.H₀: The distributions are different.
H₁: The distributions are different.

(b) Find the value of the chi-square statistic for the sample. (Round the expected frequencies to two decimal places. Round the test statistic to three decimal places.)


Are all the expected frequencies greater than 5?

YesNo    

What sampling distribution will you use?

binomialStudent's t    normalchi-squareuniform

What are the degrees of freedom?


(c) Find or estimate the P-value of the sample test statistic. (Round your answer to three decimal places.)


(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis that the population fits the specified distribution of categories?

Since the P-value > α, we fail to reject the null hypothesis.Since the P-value > α, we reject the null hypothesis.    Since the P-value ≤ α, we reject the null hypothesis.Since the P-value ≤ α, we fail to reject the null hypothesis.

(e) Interpret your conclusion in the context of the application.

At the 5% level of significance, the evidence is sufficient to conclude that the community household distribution does not fit the general U.S. household distribution.At the 5% level of significance, the evidence is insufficient to conclude that the community household distribution does not fit the general U.S. household distribution.

1. Enter into SPSS

2. Name the variables and indicate the levels of measurement as noted in the variable view screen.

3. Run descriptive statistics on each of the variables - mean, median, mode, standard deviation and create a histogram with a normal curve

4. Submit your output for this assignment

Jim Mead is a veterinarian who visits a Vermont farm to examine prize bulls. In order to examine a bull, Jim first gives the animal a tranquilizer shot. The effect of the shot is supposed to last an average of 65 minutes, and it usually does. However, Jim sometimes gets chased out of the pasture by a bull that recovers too soon, and other times he becomes worried about prize bulls that take too long to recover. By reading journals, Jim has found that the tranquilizer should have a mean duration time of 65 minutes, with a standard deviation of 15 minutes. A random sample of 10 of Jim's bulls had a mean tranquilized duration time of close to 65 minutes but a standard deviation of 26 minutes. At the 1% level of significance, is Jim justified in the claim that the variance is larger than that stated in his journal? Find a 95% confidence interval for the population standard deviation.

(a) What is the level of significance?


State the null and alternate hypotheses.

H_o: σ² = 225; H₁: σ² ≠ 225H_o: σ² = 225; H₁: σ² > 225    H_o: σ² = 225; H₁: σ² < 225H_o: σ² > 225; H₁: σ² = 225

(b) Find the value of the chi-square statistic for the sample. (Round your answer to two decimal places.)


What are the degrees of freedom?


What assumptions are you making about the original distribution?

We assume a normal population distribution.We assume a binomial population distribution.    We assume a uniform population distribution.We assume a exponential population distribution.

(c) Find or estimate the P-value of the sample test statistic.

P-value > 0.1000.050 < P-value < 0.100    0.025 < P-value < 0.0500.010 < P-value < 0.0250.005 < P-value < 0.010P-value < 0.005

(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis?

Since the P-value > α, we fail to reject the null hypothesis.Since the P-value > α, we reject the null hypothesis.    Since the P-value ≤ α, we reject the null hypothesis.Since the P-value ≤ α, we fail to reject the null hypothesis.

(e) Interpret your conclusion in the context of the application.

At the 1% level of significance, there is insufficient evidence to conclude that the variance of the duration times of the tranquilizer is larger than stated in the journal.At the 1% level of significance, there is sufficient evidence to conclude that the variance of the duration times of the tranquilizer is larger than stated in the journal.    

(f) Find the requested confidence interval for the population standard deviation. (Round your answers to two decimal place.)

Interpret the results in the context of the application.

We are 95% confident that σ lies above this interval.We are 95% confident that σ lies outside this interval.    We are 95% confident that σ lies within this interval.We are 95% confident that σ lies below this interval.

A transect is an archaeological study area that is 1/5 mile wide and 1 mile long. A site in a transect is the location of a significant archaeological find. Let x represent the number of sites per transect. In a section of Chaco Canyon, a large number of transects showed that x has a population variance σ² = 42.3. In a different section of Chaco Canyon, a random sample of 26 transects gave a sample variance s² = 50.5 for the number of sites per transect. Use a 5% level of significance to test the claim that the variance in the new section is greater than 42.3. Find a 95% confidence interval for the population variance.

(a) What is the level of significance?


State the null and alternate hypotheses.

H_o: σ² = 42.3; H₁: σ² < 42.3H_o: σ² = 42.3; H₁: σ² > 42.3    H_o: σ² > 42.3; H₁: σ² = 42.3H_o: σ² = 42.3; H₁: σ² ≠ 42.3

(b) Find the value of the chi-square statistic for the sample. (Round your answer to two decimal places.)


What are the degrees of freedom?


What assumptions are you making about the original distribution?

We assume a normal population distribution.We assume a exponential population distribution.    We assume a uniform population distribution.We assume a binomial population distribution.

(c) Find or estimate the P-value of the sample test statistic.

P-value > 0.1000.050 < P-value < 0.100    0.025 < P-value < 0.0500.010 < P-value < 0.0250.005 < P-value < 0.010P-value < 0.005

(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis?

Since the P-value > α, we fail to reject the null hypothesis.Since the P-value > α, we reject the null hypothesis.    Since the P-value ≤ α, we reject the null hypothesis.Since the P-value ≤ α, we fail to reject the null hypothesis.

(e) Interpret your conclusion in the context of the application.

At the 5% level of significance, there is insufficient evidence to conclude conclude that the variance is greater in the new section.At the 5% level of significance, there is sufficient evidence to conclude conclude that the variance is greater in the new section.    

(f) Find the requested confidence interval for the population variance. (Round your answers to two decimal places.)

Interpret the results in the context of the application.

We are 95% confident that σ² lies outside this interval.We are 95% confident that σ² lies above this interval.    We are 95% confident that σ² lies below this interval.We are 95% confident that σ² lies within this interval.

Truthfulness in online profiles. Many teens have posted profiles on a social-
networking website. A sample survey in 2007 asked a random sample of teens

with online profiles if they included false information in their profiles. Of 170
younger teens (aged 12 to 14), 117 said yes. Of 317 older teens (aged 15 to 17),
152 said yes.
16

(a) Do these samples satisfy the guidelines for the large-sample confidence
interval?
(b) Give a 95% confidence interval for the difference between the proportions
of younger and older teens who include false information in their online
profiles.

Pauly’s Pizza claims that the mean time it takes for them to deliver a pizza to dorms at Nat’s college is 31 minutes. After a long wait one night, Nat decides to test this claim. He randomly selects 15 dormitory residents and asks them to record the time it takes for Pauly’s to deliver the next time they order pizza. Here are the results (in minutes). The sample mean is x=33.8 and the sample standard deviation is s= 7.72 .

31 38 39 25 26 45 42 32 23 38 42 21 40 37 28

a. We want to use this information to construct a 90% confidence interval to estimate the true mean delivery time. State the parameter our confidence interval will estimate (in context).

b. Identify the conditions that must be met to use this procedure and explain how you know that each one has been satisfied. For the Nearly Normal Condition, include a picture of the histogram of the sample data. Make sure to include labels. Hint: Start the distribution from 20 and use a bin width of 4.

c. Find the appropriate critical value () and the standard error of the sample mean (). SHOW YOUR WORK! Round the standard error to two decimal places.

d. Use the formula shown in your notes to get the 90% confidence interval by hand. SHOW YOUR WORK! Round to one decimal places.

e. Interpret the confidence interval constructed in part (d) in the context of the problem.

f. Interpret the confidence level in the context of the problem.

g. Suppose you wanted to estimate the mean delivery time to Nat’s college with 90% confidence to have a margin of error no more than 5 minutes. Calculate how large a sample you would need. Assume min. SHOW YOUR WORK! Remember to round your final answer UP to the nearest whole number.

h. Recall that Pauly’s Pizza claimed the average time it would take to deliver to Nat’s college is 31 minutes. Does your 95% confidence interval support this claim?

i. What is the name of the significance test that can we perform to test the claim made?

j. What hypotheses would we use if we wished to conduct a two-sided test?

k. Calculate the t-score using the formula shown in class. Round to two decimal places. SHOW YOUR WORK! l. Use your t-score (with corresponding degrees of freedom) to estimate the p-value with our t-table. m. It turns out the exact p-value is 0.182. Interpret the p-value in context.

n. What decision would you draw based on the size of the p-value?

o. Are our confidence interval and significance test results in agreement?

Is the number of calories in a beer related to the number of carbohydrates​ and/or the percentage of alcohol in the​ beer? The accompanying table has data for 35 beers. The values for three variables are​ included: the number of calories per 12​ ounces, the alcohol​ percentage, and the number of carbohydrates​ (in grams) per 12 ounces. Complete parts a through d.

Let

Upper X 1X1

represent alcohol percentage and let

Upper X 2X2

represent the number of carbohydrates.

ModifyingAbove Upper Y with caret equals nothing plus left parenthesis nothing right parenthesis Upper X 1 plus left parenthesis nothing right parenthesis Upper X 2

The group of interest contains everything that isn’t the unwanted group. For this example, the spades, clubs, and diamonds would be the group of interest. However, the hearts would not be in the grouping because they do not want a heart in the hand being dealt.

Show using R for i and j

1. Find the probability in question (hand does not contain a heart).

10. Find the probability of having no more than 4 hearts in a hand.

The table below shows the number of cars sold last month by seven employees at Concord Motors and their number of years of sales experience.

Management would like to use simple regression analysis to estimate monthly car sales using the number of years of sales experience. The 95% confidence interval for the regression slope is ________.

A. (-1.842, 3.158)

B. (-0.481, 1.797)

C. (0.058, 1.258)

D. (0.408, 0.908)