Question

A random sample of 12 second-year university students enrolled in a business statistics course was drawn. At the course's completion, each student was asked how many hours he or she spent doing homework in statistics. The data are listed below. 34 37 41 34 34 33 41 37 37 28 36 39 It is known that the population standard deviation is 8. The instructor has recommended that students devote 3 hours per week for the duration of the 12-week semester, for a total of 36 hours. Test to determine whether there is evidence at the 0.08 significance level that the average student spent less than the recommended amount of time. Fill in the requested information below. A. The value of the standardized test statistic: Note: For the next part, your answer should use interval notation. An answer of the form (-?, a) is expressed (-infty, a), an answer of the form (b, ?) is expressed (b, infty), and an answer of the form (-?, a) U (b, ?) is expressed (-infty, a)U(b, infty). B. The rejection region for the standardized test statistic: C. The p-value is D. Your decision for the hypothesis test: A. Reject H1. B. Reject H0. C. Do Not Reject H0. D. Do Not Reject H1. (1 point) 45 people are randomly selected and the accuracy of their wristwatches is checked, with positive errors representing watches that are ahead of the correct time and negative errors representing watches that are behind the correct time. The 45 values have a mean of 92sec and a standard deviation of 174sec. Use a 0.01 significance level to test the claim that the population of all watches has a mean of 0sec. The test statistic is The P-Value is The final conclusion is A. There is sufficient evidence to warrant rejection of the claim that the mean is equal to 0 B. There is not sufficient evidence to warrant rejection of the claim that the mean is equal to 0 (1 point) It is necessary for an automobile producer to estimate the number of miles per gallon achieved by its cars. Suppose that the sample mean for a random sample of 140 cars is 29.5 miles and assume the standard deviation is 2.5 miles. Now suppose the car producer wants to test the hypothesis that ?, the mean number of miles per gallon, is 26.5 against the alternative hypothesis that it is not 26.5. Conduct a test using ? = .05 by giving the following: (a) positive critical z score (b) negative critical z score (c) test statistic The final conclusion is A. There is not sufficient evidence to reject the null hypothesis that ? = 26.5. B. We can reject the null hypothesis that ? = 26.5 and accept that ? ? 26.5. (1 point) Golf-course designers have become concerned that old courses are becoming obsolete since new technology has given golfers the ability to hit the ball so far. Designers, therefore, have proposed that new golf courses need to be built expecting that the average golfer can hit the ball more than 245 yards on average. Suppose a random sample of 101 golfers be chosen so that their mean driving distance is 250.2 yards, with a standard deviation of 48.3. Conduct a hypothesis test where H0 : ? = 245 and H1 : ? > 245 by computing the following: (a) test statistic (b) p-value p = (c) If this was a two-tailed test, then the p-value is

A random sample of 12 second-year university students enrolled in a business statistics course was drawn. At the course's completion, each student was asked how many hours he or she spent doing homework in statistics. The data are listed below.

34 37 41 34 34 33
41 37 37 28 36 39

It is known that the population standard deviation is 8. The instructor has recommended that students devote 3 hours per week for the duration of the 12-week semester, for a total of 36 hours. Test to determine whether there is evidence at the 0.08 significance level that the average student spent less than the recommended amount of time. Fill in the requested information below.

A. The value of the standardized test statistic:

Note: For the next part, your answer should use interval notation. An answer of the form (-?, a) is expressed (-infty, a), an answer of the form (b, ?) is expressed (b, infty), and an answer of the form (-?, a) U (b, ?) is expressed (-infty, a)U(b, infty).

B. The rejection region for the standardized test statistic:

C. The p-value is

D. Your decision for the hypothesis test:

A. Reject H1.
B. Reject H0.
C. Do Not Reject H0.
D. Do Not Reject H1.

(1 point) 45 people are randomly selected and the accuracy of their wristwatches is checked, with positive errors representing watches that are ahead of the correct time and negative errors representing watches that are behind the correct time. The 45 values have a mean of 92sec and a standard deviation of 174sec. Use a 0.01 significance level to test the claim that the population of all watches has a mean of 0sec.

The test statistic is

The P-Value is

The final conclusion is

A. There is sufficient evidence to warrant rejection of the claim that the mean is equal to 0
B. There is not sufficient evidence to warrant rejection of the claim that the mean is equal to 0

(1 point) It is necessary for an automobile producer to estimate the number of miles per gallon achieved by its cars. Suppose that the sample mean for a random sample of 140 cars is 29.5 miles and assume the standard deviation is 2.5 miles. Now suppose the car producer wants to test the hypothesis that ?, the mean number of miles per gallon, is 26.5 against the alternative hypothesis that it is not 26.5. Conduct a test using ? = .05 by giving the following:

(a) positive critical z score
(b) negative critical z score
(c) test statistic

The final conclusion is

A. There is not sufficient evidence to reject the null hypothesis that ? = 26.5.
B. We can reject the null hypothesis that ? = 26.5 and accept that ? ? 26.5.

(1 point) Golf-course designers have become concerned that old courses are becoming obsolete since new technology has given golfers the ability to hit the ball so far. Designers, therefore, have proposed that new golf courses need to be built expecting that the average golfer can hit the ball more than 245 yards on average. Suppose a random sample of 101 golfers be chosen so that their mean driving distance is 250.2 yards, with a standard deviation of 48.3.

Conduct a hypothesis test where H0 : ? = 245 and H1 : ? > 245 by computing the following:
(a) test statistic
(b) p-value p =
(c) If this was a two-tailed test, then the p-value is

A random sample of 12 second-year university students enrolled in a business statistics course was drawn. At the course's completion, each student was asked how many hours he or she spent doing homework in statistics. The data are listed below.

34 37 41 34 34 33
41 37 37 28 36 39

It is known that the population standard deviation is 8. The instructor has recommended that students devote 3 hours per week for the duration of the 12-week semester, for a total of 36 hours. Test to determine whether there is evidence at the 0.08 significance level that the average student spent less than the recommended amount of time. Fill in the requested information below.

A. The value of the standardized test statistic:

Note: For the next part, your answer should use interval notation. An answer of the form (-?, a) is expressed (-infty, a), an answer of the form (b, ?) is expressed (b, infty), and an answer of the form (-?, a) U (b, ?) is expressed (-infty, a)U(b, infty).

B. The rejection region for the standardized test statistic:

C. The p-value is

D. Your decision for the hypothesis test:

A. Reject H1.
B. Reject H0.
C. Do Not Reject H0.
D. Do Not Reject H1.

(1 point) 45 people are randomly selected and the accuracy of their wristwatches is checked, with positive errors representing watches that are ahead of the correct time and negative errors representing watches that are behind the correct time. The 45 values have a mean of 92sec and a standard deviation of 174sec. Use a 0.01 significance level to test the claim that the population of all watches has a mean of 0sec.

The test statistic is

The P-Value is

The final conclusion is

A. There is sufficient evidence to warrant rejection of the claim that the mean is equal to 0
B. There is not sufficient evidence to warrant rejection of the claim that the mean is equal to 0

(1 point) It is necessary for an automobile producer to estimate the number of miles per gallon achieved by its cars. Suppose that the sample mean for a random sample of 140 cars is 29.5 miles and assume the standard deviation is 2.5 miles. Now suppose the car producer wants to test the hypothesis that ?, the mean number of miles per gallon, is 26.5 against the alternative hypothesis that it is not 26.5. Conduct a test using ? = .05 by giving the following:

(a) positive critical z score
(b) negative critical z score
(c) test statistic

The final conclusion is

A. There is not sufficient evidence to reject the null hypothesis that ? = 26.5.
B. We can reject the null hypothesis that ? = 26.5 and accept that ? ? 26.5.

(1 point) Golf-course designers have become concerned that old courses are becoming obsolete since new technology has given golfers the ability to hit the ball so far. Designers, therefore, have proposed that new golf courses need to be built expecting that the average golfer can hit the ball more than 245 yards on average. Suppose a random sample of 101 golfers be chosen so that their mean driving distance is 250.2 yards, with a standard deviation of 48.3.

Conduct a hypothesis test where H0 : ? = 245 and H1 : ? > 245 by computing the following:
(a) test statistic
(b) p-value p =
(c) If this was a two-tailed test, then the p-value is

Added by Rosa R.

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