stats 2
GUYAyESip5
stats_2.docx1. As part of the study on ongoing fright symptoms due to exposure to horror movies at a young age, the following table was presented to describe the lasting impact these movies have had during bedtime and waking life:
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Waking symptoms |
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Bedtime symptoms |
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Yes |
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No |
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Yes |
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35 |
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33 |
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No |
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32 |
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19 |
(a) What percent of the students have lasting waking-life symptoms? (Round your answer to two decimal places.) % (b) What percent of the students have both waking-life and bedtime symptoms? (Round your answer to two decimal places.) % (c) Test whether there is an association between waking-life and bedtime symptoms. State the null and alternative hypotheses. (Use α = 0.01.) Null Hypothesis:
H0: There is no relationship between waking and bedtime symptoms.H0: Waking symptoms cause bedtime symptoms. H0: There is a relationship between waking and bedtime symptoms.H0: Bedtime symptoms cause waking symptoms.
Alternative Hypothesis:
Ha: Bedtime symptoms cause waking symptoms.Ha: Waking symptoms cause bedtime symptoms. Ha: There is a relationship between waking and bedtime symptoms.Ha: There is no relationship between waking and bedtime symptoms.
State the χ2 statistic and the P-value. (Round your answers for χ2 and the P-value to three decimal places.)
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χ2 |
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df |
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P |
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Conclusion:
We have enough evidence to conclude that there is a relationship.We do not have enough evidence to conclude that there is a relationship.
2. Here are the row and column totals for a two-way table with two rows and two columns.
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a |
b |
200 |
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c |
d |
200 |
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200 |
200 |
400 |
Find two different sets of counts a, b, c, and d for the body of the table. This demonstrates that the relationship between two variables cannot be obtained solely from the two marginal distributions of the variables. (Start with the given value of a.)
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a |
b |
c |
d |
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Set 1 |
30 |
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Set 2 |
115 |
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3. A study examined patterns and characteristics of volunteer service for young people from high school through early adulthood. Here are some data that can be used to compare males and females on participation in unpaid volunteer service or community service and motivation for participation.
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Participants |
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Motivation |
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Gender |
Strictly Voluntary |
Court-ordered |
Other |
Non-participants |
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Men |
31.1% |
3.8% |
7.2% |
57.9% |
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Women |
42.3% |
1.5% |
7.8% |
48.4% |
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Note that the percents in each row sum to 100%.
(a) Graphically compare the volunteer-service profiles for men and women. Describe any differences that are striking.
Men dominate every category.There are no striking differences. Women dominate every category.Women have a noticeably higher percentage in the strictly voluntary category.
(b) Find the proportion of men who volunteer. Do the same for women. Compute the relative risk of being a volunteer for females versus males. (Round your answer for relative risk to two decimal places.)
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Men |
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Women |
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Relative risk |
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Write a clear sentence contrasting females and males using relative risk as your numerical summary.
Men and women participate equally.A higher percentage of women were participants. A higher percentage of men were participants.none of the above
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Magazine readership |
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Model dress |
Women |
Men |
General interest |
Total |
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Not sexual |
360 |
522 |
242 |
1124 |
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Sexual |
201 |
94 |
90 |
385 |
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Total |
561 |
616 |
332 |
1509 |
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(a) Summarize the data numerically and graphically. (Compute the conditional distribution of model dress for each audience. Round your answers to three decimal places.)
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Women |
Men |
General |
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Not sexual |
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Sexual |
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(b) Perform the significance test that compares the model dress for the three categories of magazine readership. Summarize the results of your test and give your conclusion. (Use α = 0.01. Round your value for χ2 to two decimal places, and round your P-value to four decimal places.)
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χ2 = |
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P-value = |
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Conclusion
Reject the null hypothesis. There is not significant evidence of an association between target audience and model dress.Fail to reject the null hypothesis. There is not significant evidence of an association between target audience and model dress. Reject the null hypothesis. There is significant evidence of an association between target audience and model dress.Fail to reject the null hypothesis. There is significant evidence of an association between target audience and model dress.
(c) All of the ads were taken from the March, July, and November issues of six magazines in one year. Discuss this fact from the viewpoint of the validity of the significance test and the interpretation of the results.
This is an SRS. This gives us no reason to believe our conclusions are suspect.This is not an SRS. This gives us no reason to believe our conclusions are suspect. This is an SRS. This gives us reason to believe our conclusions might be suspect.This is not an SRS. This gives us reason to believe our conclusions might be suspect.
5. A research project based on a study of older adults examined the relationship between physical activity and pet ownership. The data collected included information concerning pet owner characteristics and the type of pet owned. Here are data giving the relationship between pet ownership status and gender.
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Pet ownership status |
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Gender |
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Non-pet owners |
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Dog owners |
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Cat owners |
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Female |
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1026 |
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161 |
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85 |
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Male |
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917 |
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169 |
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81 |
Analyze the data. (Round your χ2 to three decimal places and your P-value to four decimal places.)
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χ2 |
= |
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df |
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P-value |
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Summarize your results and conclusions.
The relationship between gender and pet ownership is significant.The relationship between gender and pet ownership is not significant.
6. The index of biotic integrity (IBI) is a measure of the water quality in streams. IBI and land-use measures for a collection of streams in the Ozark Highland ecoregion of Arkansas were collected as part of a study. The data data226.dat gives the data for IBI and the area of the watershed in square kilometers for streams in the original sample with area less than or equal to 70km2.
(a) Use numerical and graphical methods to describe the variable IBI. Do the same for area. (Round your answers for x to two decimal places and your answers for s to three decimal places.)
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IBI: |
x = |
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s = |
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area: |
x = |
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s = |
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(b) Run the simple linear regression and summarize the results. (Let x = area and y = IBI. Round your slope, intercept, and r to three decimal places. Round F to two decimal places and your P-value to four decimal places.)
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y |
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+ x |
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F |
= |
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P |
= |
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r |
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Is area of watershed a good predictor for IBI at the 5% significance level?
Yes, the area of watershed is a good predictor for IBI.No, there is insufficient evidence to say the area of watershed is a good predictor for IBI.
Interpret the intercept.
The increase in IBI when the watershed area is increased by one unitThe value of IBI when watershed area is 0 The increase in the watershed area when IBI is increased by one unitThe area of watershed when IBI value is 0
Interpret the slope.
The increase in the watershed area when IBI is increased by one unitThe area of watershed when IBI value is 0 The increase in IBI when the watershed area is increased by one unitThe value of IBI when watershed area is 0
Interpret the P-value.
If the P-value if less than 0.05 then we can conclude that the intercept is 0If the P-value is less than 0.05 then we can conclude that the watershed area is a bad predictor for IBI. If the P-value is greater than 0.05 then we can conclude that the watershed area is a good predictor for IBI.If the P-value is less than 0.05 then we can conclude that the watershed area is a good predictor for IBI.
7. The Leaning Tower of Pisa is an architectural wonder. Engineers concerned about the tower's stability have done extensive studies of its increasing tilt. Measurements of the lean of the tower over time provide much useful information. The following table gives measurements for the years 1975 to 1987. The variable "lean" represents the difference between where a point on the tower would be if the tower were straight and where it actually is. The data are coded as tenths of a millimeter in excess of 2.9 meters, so that the 1975 lean, which was 2.9642 meters, appears in the table as 642. Only the last two digits of the year were entered into the computer. (data128.dat)
(a) Plot the data. Consider whether or not the trend in lean over time appears to be linear. (Do this on paper. Your instructor may ask you to turn in this graph.) (b) What is the equation of the least-squares line? (Round your answers to two decimal places.) y = + x What percent of the variation in lean is explained by this line? (Round your answer to one decimal place.) % (c) Give a 99% confidence interval for the average rate of change (tenths of a millimeter per year) of the lean. (Round your answers to two decimal places.) ( , )
8. Find a 95% confidence interval for the slope in each of the following settings. (Round your answers to three decimal places.)
(a) n = 23, = 1.2 + 11.10x, and
SEb1 = 73.10
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(b) n = 23, = 12.0 + 3.10x, and
SEb1 = 73.10
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(c) n = 100, = 1.2 + 11.10x, and
SEb1 = 73.10
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9. Can a pretest on mathematics skills predict success in a statistics course? The 82 students in an introductory statistics class took a pretest at the beginning of the semester. The least-squares regression line for predicting the score y on the final exam from the pretest score x was = 9.4 + 0.75x. The standard error of b1 was 0.41.
(a) Test the null hypothesis that there is no linear relationship between the pretest score and the score on the final exam against the two-sided alternative. (Round your test statistic to three decimal places and your P-value to four decimal places.)
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t = |
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df = |
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P = |
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Conclusion
We reject H0 at the 5% significance level.We do not reject H0 at the 5% significance level.
(b) Would you reject this null hypothesis versus the one-sided alternative that the slope is positive? Explain your answer. P = Conclusion
We could reject H0 at the 5% significance level.We could not reject H0 at the 5% significance level.
10. How are returns on common stocks in overseas markets related to returns in U.S. markets? Measure U.S. returns by the annual rate of return on the Standard & Poor's 500-Stock Index and overseas returns by the annual rate of return on the Morgan Stanley EAFE (Europe, Australasia, Far East) index. Both are recorded in percents. Regress the EAFE returns on the S&P 500 returns for the 30 years 1971 to 2000. Here is part of the Minitab output for this regression. The regression equation is EAFE = 4.76 + 0.663 S&P.
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"Analysis of Variance" |
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Source |
DF |
SS |
MS |
F |
P |
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Regression |
1 |
3445.9 |
3445.9 |
9.5 |
0.005 |
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Residual Error |
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Total |
29 |
13598.3 |
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What are the values of the regression standard error s and the squared correlation r 2 ? (Enter your answers to four decimal places.)
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s |
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r 2 |
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ubmit Answer
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ave Progress
