An SRS of 100 postal employees found that the average time these employees had worked for the postal service was = 7 years, with standard deviation s = 2 years. Assume the distribution of the time the population of employees have worked for the postal service is approximately normal, with mean μ. Are these data evidence that μ has changed from the value of 7.5 years of 20 years ago? To determine this, we test the hypotheses

H0: μ = 7.5, Ha: μ ≠ 7.5

using the one-sample t test.


1.
The appropriate degrees of freedom for this test are
A.
9.
B.
10.
C.
99.
D.
100.


An SRS of 100 postal employees found that the average time these employees had worked for the postal service was = 7 years, with standard deviation s = 2 years. Assume the distribution of the time the population of employees have worked for the postal service is approximately normal, with mean μ. Are these data evidence that μ has changed from the value of 7.5 years of 20 years ago? To determine this, we test the hypotheses

H0: μ = 7.5, Ha: μ ≠ 7.5

using the one-sample t test.


2.
The P-value for the one-sample t test is
A.
larger than 0.10.
B.
between 0.10 and 0.05.
C.
between 0.05 and 0.01.
D.
below 0.01.


An SRS of 100 postal employees found that the average time these employees had worked for the postal service was = 7 years, with standard deviation s = 2 years. Assume the distribution of the time the population of employees have worked for the postal service is approximately normal, with mean μ. Are these data evidence that μ has changed from the value of 7.5 years of 20 years ago? To determine this, we test the hypotheses

H0: μ = 7.5, Ha: μ ≠ 7.5

using the one-sample t test.


3.
Suppose the mean and standard deviation obtained were based on a sample of 25 postal workers rather than 100. The P-value would be
A.
larger.
B.
smaller.
C.
unchanged because the difference between and the hypothesized value μ = 7.5 is unchanged.
D.
unchanged because the variability measured by the standard deviation stays the same.


An SRS of 100 postal employees found that the average time these employees had worked for the postal service was = 7 years, with standard deviation s = 2 years. Assume the distribution of the time the population of employees have worked for the postal service is approximately normal, with mean μ. Are these data evidence that μ has changed from the value of 7.5 years of 20 years ago? To determine this, we test the hypotheses

H0: μ = 7.5, Ha: μ ≠ 7.5

using the one-sample t test.


4.
A 95% confidence interval for the population mean time μ that postal service employees have spent with the postal service is
A.
7 ± 2.
B.
7 ± 1.984.
C.
7 ± 0.4.
D.
7 ± 0.2.


5.
The water diet requires one to drink two cups of water every half hour from when one gets up until one goes to bed, but otherwise allows one to eat whatever one likes. Four adult volunteers agree to test the diet. They are weighed prior to beginning the diet and after six weeks on the diet. The weights (in pounds) are
Person
1
2
3
4
Weight before the diet
180
125
240
150
Weight after six weeks
170
130
215
152

For the population of all adults, assume that the weight loss after six weeks on the diet (weight before beginning the diet – weight after six weeks on the diet) is normally distributed with mean μ. To determine if the diet leads to weight loss, we test the hypotheses

H0: μ = 0, Ha: μ > 0.

Based on these data we conclude
A.
we would not reject H0 at significance level 0.10.
B.
we would reject H0 at significance level 0.10 but not at 0.05.
C.
we would reject H0 at significance level 0.05 but not at 0.01.
D.
we would reject H0 at significance level 0.01.


6.
Which of the following is an example of a matched pairs design?
A.
A teacher compares the pre-test and post-test scores of students.
B.
A teacher compares the scores of students using a computer-based method of instruction with the scores of other students using a traditional method of instruction.
C.
A teacher compares the scores of students in her class on a standardized test with the national average score.
D.
A teacher calculates the average of scores of students on a pair of tests and wishes to see if this average is larger than 80%.


A food company is developing a new breakfast drink and their market analysts are currently working on preliminary taste testing studies. To help with their marketing strategy, they were first interested in whether preference for the new product was related to a person's gender. There were 100 male and 100 female volunteers available for the taste test. Both the males and females tasted the product and rated the flavor on a scale of 1 to 10, 1 being “very unpleasant” and 10 being “very pleasant.” The mean rating for males was = 6.4, with a standard deviation s1 = 1.5. The mean rating for females was = 7.0, with a standard deviation s2 = 2.0. Let μ1 and μ2 represent the mean ratings we would observe for the populations of males and females, respectively, and assume our samples can be regarded as samples from these populations.


7.
A 90% confidence interval for μ1 μ2 is (use the conservative value for the degrees of freedom)
A.
–0.6 ± 0.25.
B.
–0.6 ± 0.32.
C.
–0.6 ± 0.42.
D.
–0.6 ± 0.5.


A food company is developing a new breakfast drink and their market analysts are currently working on preliminary taste testing studies. To help with their marketing strategy, they were first interested in whether preference for the new product was related to a person's gender. There were 100 male and 100 female volunteers available for the taste test. Both the males and females tasted the product and rated the flavor on a scale of 1 to 10, 1 being “very unpleasant” and 10 being “very pleasant.” The mean rating for males was = 6.4, with a standard deviation s1 = 1.5. The mean rating for females was = 7.0, with a standard deviation s2 = 2.0. Let μ1 and μ2 represent the mean ratings we would observe for the populations of males and females, respectively, and assume our samples can be regarded as samples from these populations.


8.
Suppose the researcher had wished to test the hypotheses

H0: μ1 = μ2, Ha: μ1μ2.

The numerical value of the two-sample t statistic is
A.
2.00.
B.
2.40.
C.
3.25.
D.
9.60.


A food company is developing a new breakfast drink and their market analysts are currently working on preliminary taste testing studies. To help with their marketing strategy, they were first interested in whether preference for the new product was related to a person's gender. There were 100 male and 100 female volunteers available for the taste test. Both the males and females tasted the product and rated the flavor on a scale of 1 to 10, 1 being “very unpleasant” and 10 being “very pleasant.” The mean rating for males was = 6.4, with a standard deviation s1 = 1.5. The mean rating for females was = 7.0, with a standard deviation s2 = 2.0. Let μ1 and μ2 represent the mean ratings we would observe for the populations of males and females, respectively, and assume our samples can be regarded as samples from these populations.


9.
Which of the following would lead us to believe that the t-procedures were not safe to use here?
A.
The sample medians and means for the two groups were slightly different.
B.
The distributions of the data were moderately skewed.
C.
The data are integers between 1 and 10 and so cannot be normal.
D.
Only the most severe departures from normality would lead us to believe the t- procedures were not safe to use.


10.
A radio talk show host with a large audience is interested in the proportion p of adults in his listening area who think the drinking age should be lowered to 18. To find this out he poses the following question to his listeners. “Do you think that the drinking age should be reduced to 18, in light of the fact that 18 year olds are eligible for military service?” He asks listeners to phone in and vote “yes” if they agree the drinking age should be lowered and “no” if not. Of the 100 people who phoned in, 70 answered “yes.” The Wilson estimate, , of the proportion who think the drinking age should be reduced is
A.
70.
B.
0.70.
C.
0.69.
D.
not able to be determined from the information given.


The college newspaper of a large midwestern university periodically conducts a survey of students on campus to determine the attitude on campus concerning issues of interest. Pictures of the students interviewed along with a quote of their response are printed in the paper. Students are interviewed by a reporter “roaming” the campus who selects students to interview “haphazardly.” On a particular day the reporter interviews eight students and asks them if they feel there is adequate student parking on campus. Five of the students say no.


11.
The sample proportion who respond “no” is
A.
0.375.
B.
0.625.
C.
0.667.
D.
0.700.


The college newspaper of a large midwestern university periodically conducts a survey of students on campus to determine the attitude on campus concerning issues of interest. Pictures of the students interviewed along with a quote of their response are printed in the paper. Students are interviewed by a reporter “roaming” the campus who selects students to interview “haphazardly.” On a particular day the reporter interviews eight students and asks them if they feel there is adequate student parking on campus. Five of the students say no.


12.
Which of the following assumptions for inference about a proportion using a confidence interval are violated in this example?
A.
The data are an SRS from the population of interest.
B.
The population is at least 10 times as large as the sample.
C.
We are interested in inference about a proportion.
D.
There appear to be no violations.


After the football team once again lost a game to the college's arch rival, the alumni association conducted a survey to see if alumni were in favor of firing the coach. An SRS of 100 alumni from the population of all living alumni was taken. 64 of the alumni in the sample were in favor of firing the coach. Let p represent the proportion of all living alumni who favor firing the coach.


13.
A 95% confidence interval for p is
A.
0.63 ± 0.047.
B.
0.63 ± 0.078.
C.
0.63 ± 0.093.
D.
0.63 ± 0.128.


After the football team once again lost a game to the college's arch rival, the alumni association conducted a survey to see if alumni were in favor of firing the coach. An SRS of 100 alumni from the population of all living alumni was taken. 64 of the alumni in the sample were in favor of firing the coach. Let p represent the proportion of all living alumni who favor firing the coach.


14.
Suppose you wished to see if the majority of alumni are in favor of firing the coach. To do this you test the hypotheses

H0: p = 0.50, Ha: p > 0.50.

The P-value of your test is
A.
between .10 and .05.
B.
between .05 and .01.
C.
between .01 and .001.
D.
below .001.


An old saying in golf is “you drive for show and you putt for dough.” The point is that good putting is more important than long driving for shooting low scores and hence winning money. To see if this is the case, data on the top 69 money winners on the PGA tour in 1993 are examined. The average number of putts per hole for each player is used to predict their total winnings using the simple linear regression model

1993 winningsi = β0 + β1(average number of putts per hole)i + εi

where the deviations εi are assumed to be independent and normally distributed with mean 0 and standard deviation σ. This model was fit to the data using the method of least squares. The following results were obtained from statistical software.

R2 = 0.081
s = 281,777
Variable
Parameter Estimate
Std. Err. of Parameter Est.
Constant
7,897,179
3,023,782
Avg. Putts
–4,139,198
1,698,371


15.
The explanatory variable in this study is
A.
1993 winnings.
B.
average number of putts per hole.
C.
the slope, β1.
D.
–4,139,198.


An old saying in golf is “you drive for show and you putt for dough.” The point is that good putting is more important than long driving for shooting low scores and hence winning money. To see if this is the case, data on the top 69 money winners on the PGA tour in 1993 are examined. The average number of putts per hole for each player is used to predict their total winnings using the simple linear regression model

1993 winningsi = β0 + β1(average number of putts per hole)i + εi

where the deviations εi are assumed to be independent and normally distributed with mean 0 and standard deviation σ. This model was fit to the data using the method of least squares. The following results were obtained from statistical software.

R2 = 0.081
s = 281,777
Variable
Parameter Estimate
Std. Err. of Parameter Est.
Constant
7,897,179
3,023,782
Avg. Putts
–4,139,198
1,698,371


16.
The quantity s = 281,777 is an estimate of the standard deviation σ of the deviations in the simple linear regression model. The degrees of freedom for s are
A.
69.
B.
68.
C.
67.
D.
281,777.


An old saying in golf is “you drive for show and you putt for dough.” The point is that good putting is more important than long driving for shooting low scores and hence winning money. To see if this is the case, data on the top 69 money winners on the PGA tour in 1993 are examined. The average number of putts per hole for each player is used to predict their total winnings using the simple linear regression model

1993 winningsi = β0 + β1(average number of putts per hole)i + εi

where the deviations εi are assumed to be independent and normally distributed with mean 0 and standard deviation σ. This model was fit to the data using the method of least squares. The following results were obtained from statistical software.

R2 = 0.081
s = 281,777
Variable
Parameter Estimate
Std. Err. of Parameter Est.
Constant
7,897,179
3,023,782
Avg. Putts
–4,139,198
1,698,371


17.
The intercept of the least-squares regression line is
A.
7,897,179.
B.
–4,139,198.
C.
3,023,782.
D.
1,698,371.


An old saying in golf is “you drive for show and you putt for dough.” The point is that good putting is more important than long driving for shooting low scores and hence winning money. To see if this is the case, data on the top 69 money winners on the PGA tour in 1993 are examined. The average number of putts per hole for each player is used to predict their total winnings using the simple linear regression model

1993 winningsi = β0 + β1(average number of putts per hole)i + εi

where the deviations εi are assumed to be independent and normally distributed with mean 0 and standard deviation σ. This model was fit to the data using the method of least squares. The following results were obtained from statistical software.

R2 = 0.081
s = 281,777
Variable
Parameter Estimate
Std. Err. of Parameter Est.
Constant
7,897,179
3,023,782
Avg. Putts
–4,139,198
1,698,371


18.
Suppose the researchers test the hypotheses

H0: β1 = 0, Ha: β1 < 0

The value of the t statistic for this test is
A.
2.61.
B.
2.44.
C.
0.081.
D.
–2.44.


A random sample of 79 companies from the Forbes 500 list (which actually consists of nearly 800 companies) was selected, and the relationship between sales (in hundreds of thousands of dollars) and profits (in hundreds of thousands of dollars) was investigated by regression. The following simple linear regression model was used

Profitsi = β0 + β1(Sales)i + εi

where the deviations εi were assumed to be independent and normally distributed, with mean 0 and standard deviation σ. This model was fit to the data using the method of least squares. The following ANOVA table was obtained from statistical software.
SOURCE
DF
SUM OF SQUARES
Model
1
32,809,212
Error
 
16,734,234
 


19.
The degrees of freedom for the residual SS, the error sum of squares, is
A.
1.
B.
2.
C.
77.
D.
78.


A random sample of 79 companies from the Forbes 500 list (which actually consists of nearly 800 companies) was selected, and the relationship between sales (in hundreds of thousands of dollars) and profits (in hundreds of thousands of dollars) was investigated by regression. The following simple linear regression model was used

Profitsi = β0 + β1(Sales)i + εi

where the deviations εi were assumed to be independent and normally distributed, with mean 0 and standard deviation σ. This model was fit to the data using the method of least squares. The following ANOVA table was obtained from statistical software.
SOURCE
DF
SUM OF SQUARES
Model
1
32,809,212
Error
 
16,734,234
 


20.
Total SS, the total sum of squares, has value
A.
16,074,978.
B.
16,734,234.
C.
32,809,212.
D.
49,543,448.


Has the number of home runs hit by major league teams been changing over time? For the 41 years from 1960 to 2000, the average number of home runs hit per game per team for each season was computed in order to assess any change over time. Initially, simple linear regression was used to study the trend in home runs hit over the period 1960 to 2000, by using year to predict the average number of home runs per game per team in that year. However, it was pointed out that after the 1976 season the manufacturer of major league baseballs was changed from Spaulding to Rawlings. Because the change in the baseball used might affect the number of home runs (for example, if Rawlings produces a livelier ball, this would likely lead to more home runs), it was decided to include an additional variable, namely



A multiple regression analysis was performed using the model

μAvg. home runs per game per team = β0 + β1(Year) + β2(Manufacturer).

The following results were obtained.
Analysis of Variance
Source
df
Sum of Squares
Model
2
 
0.744384
Error
38
 
2.539330
 
Parameter Estimates
Variable
df
Parameter Estimate
Standard Error
Intercept
1
–27.91180
12.8900
Year
1
   0.014977
0.00650
Manufacturer
1
–0.107816
0.11573
 


21.
Using the regression equation, the predicted average number of home runs per game per team in 1978 would be
A.
1.463.
B.
1.713.
C.
2.431.
D.
29.624.


Has the number of home runs hit by major league teams been changing over time? For the 41 years from 1960 to 2000, the average number of home runs hit per game per team for each season was computed in order to assess any change over time. Initially, simple linear regression was used to study the trend in home runs hit over the period 1960 to 2000, by using year to predict the average number of home runs per game per team in that year. However, it was pointed out that after the 1976 season the manufacturer of major league baseballs was changed from Spaulding to Rawlings. Because the change in the baseball used might affect the number of home runs (for example, if Rawlings produces a livelier ball, this would likely lead to more home runs), it was decided to include an additional variable, namely



A multiple regression analysis was performed using the model

μAvg. home runs per game per team = β0 + β1(Year) + β2(Manufacturer).

The following results were obtained.
Analysis of Variance
Source
df
Sum of Squares
Model
2
 
0.744384
Error
38
 
2.539330
 
Parameter Estimates
Variable
df
Parameter Estimate
Standard Error
Intercept
1
–27.91180
12.8900
Year
1
   0.014977
0.00650
Manufacturer
1
–0.107816
0.11573
 


22.
The value of the regression standard error is
A.
0.067.
B.
0.259.
C.
0.372.
D.
1.593.


Has the number of home runs hit by major league teams been changing over time? For the 41 years from 1960 to 2000, the average number of home runs hit per game per team for each season was computed in order to assess any change over time. Initially, simple linear regression was used to study the trend in home runs hit over the period 1960 to 2000, by using year to predict the average number of home runs per game per team in that year. However, it was pointed out that after the 1976 season the manufacturer of major league baseballs was changed from Spaulding to Rawlings. Because the change in the baseball used might affect the number of home runs (for example, if Rawlings produces a livelier ball, this would likely lead to more home runs), it was decided to include an additional variable, namely



A multiple regression analysis was performed using the model

μAvg. home runs per game per team = β0 + β1(Year) + β2(Manufacturer).

The following results were obtained.
Analysis of Variance
Source
df
Sum of Squares
Model
2
 
0.744384
Error
38
 
2.539330
 
Parameter Estimates
Variable
df
Parameter Estimate
Standard Error
Intercept
1
–27.91180
12.8900
Year
1
   0.014977
0.00650
Manufacturer
1
–0.107816
0.11573
 


23.
The value of R2 is
A.
0.067.
B.
0.227.
C.
0.259.
D.
0.744.


Has the number of home runs hit by major league teams been changing over time? For the 41 years from 1960 to 2000, the average number of home runs hit per game per team for each season was computed in order to assess any change over time. Initially, simple linear regression was used to study the trend in home runs hit over the period 1960 to 2000, by using year to predict the average number of home runs per game per team in that year. However, it was pointed out that after the 1976 season the manufacturer of major league baseballs was changed from Spaulding to Rawlings. Because the change in the baseball used might affect the number of home runs (for example, if Rawlings produces a livelier ball, this would likely lead to more home runs), it was decided to include an additional variable, namely



A multiple regression analysis was performed using the model

μAvg. home runs per game per team = β0 + β1(Year) + β2(Manufacturer).

The following results were obtained.
Analysis of Variance
Source
df
Sum of Squares
Model
2
 
0.744384
Error
38
 
2.539330
 
Parameter Estimates
Variable
df
Parameter Estimate
Standard Error
Intercept
1
–27.91180
12.8900
Year
1
   0.014977
0.00650
Manufacturer
1
–0.107816
0.11573
 


24.
A 90% confidence interval for β1, the coefficient of Year, based on these results is
A.
0.014977 ± 0.0065.
B.
0.014977 ± 0.0110.
C.
0.014977 ± 0.0128.
D.
0.0065 ± 0.014977.



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