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#### A drug company is considering marketing a new local anesthetic

###### Statistics

A drug company is considering marketing a new local anesthetic. The effective time of the anesthetic the drug company is currently producing has a normal distribution with an average of 7.4 minutes with a standard deviation of 1.2 minutes. The chemistry of the new anesthetic is such that the effective time should be normal with the same standard deviation, but the mean effective time may be lower. If it is lower, the drug company will market the new anesthetic; otherwise, they will continue to produce the older one. A sample of size 36 results in a sample mean of 7.1.  A hypothesis test will be done to help make the decision.

1)Referring to Table 9-4, the appropriate hypotheses are:

a)

b)

c)

d)

2.            Referring to Table 9-4, for a test with a level of significance of 0.10, the critical value would be ________.

3.            Referring to Table 9-4, the value of the test statistic is ________.

4.            Referring to Table 9-4, the p-value of the test is ________.

5.            True or False: Referring to Table 9-4, the null hypothesis will be rejected with a level of significance of 0.10.

6.            True or False: Referring to Table 9-4, if the level of significance had been chosen as 0.05, the null hypothesis would be rejected.

7.            True or False: Referring to Table 9-4, if the level of significance had been chosen as 0.05, the company would market the new anesthetic.

8.            Referring to Table 9-4, what is the power of the test if the average effective time of the anesthetic is 7.0 using a 0.05 level of significance?

9.            Referring to Table 9-4, what is the probability of making a Type II error if the average effective time of the anesthetic is 7.0 using a 0.05 level of significance?

10.          Referring to Table 9-4, what is the power of the test if the average effective time of the anesthetic is 7.5 using a 0.05 level of significance?

11.          Referring to Table 9-4, what is the probability of making a Type II error if the average effective time of the anesthetic is 7.5 using a 0.05 level of significance?

12.          Referring to Table 9-4, what is the power of the test if the average effective time of the anesthetic is 7.0 using a 0.10 level of significance?

13.          Referring to Table 9-4, what is the probability of making a Type II error if the average effective time of the anesthetic is 7.0 using a 0.10 level of significance?

14.          Referring to Table 9-4, what is the power of the test if the average effective time of the anesthetic is 7.5 using a 0.10 level of significance?

15.          Referring to Table 9-4, what is the probability of making a Type II error if the average effective time of the anesthetic is 7.5 using a 0.10 level of significance?

TABLE 9-5

A bank tests the null hypothesis that the mean age of the bank's mortgage holders is less than or equal to 45, versus an alternative that the mean age is greater than 45. They take a sample and calculate a p-value of 0.0202.

16.          True or False: Referring to Table 9-5, the null hypothesis would be rejected at a significance level of   = 0.05.

17.          True or False: Referring to Table 9-5, the null hypothesis would be rejected at a significance level of   = 0.01.

18.          True or False: Referring to Table 9-5, the bank can conclude that the average age is greater than 45 at a significance level of   = 0.01.

19.          Referring to Table 9-5, if the same sample was used to test the opposite one-tailed test, what would be that test's p-value?

a)            0.0202

b)            0.0404

c)            0.9596

d)            0.9798

TABLE 9-6

The quality control engineer for a furniture manufacturer is interested in the mean amount of force necessary to produce cracks in stressed oak furniture. She performs a two-tailed test of the null hypothesis that the mean for the stressed oak furniture is 650. The calculated value of the Z test statistic is a positive number that leads to a p-value of 0.080 for the test.

20.          True or False: Referring to Table 9-6, if the test is performed with a level of significance of 0.10, the null hypothesis would be rejected.

21.          True or False: Referring to Table 9-6, if the test is performed with a level of significance of 0.10, the engineer can conclude that the mean amount of force necessary to produce cracks in stressed oak furniture is 650.

22.          True or False: Referring to Table 9-6, if the test is performed with a level of significance of 0.05, the null hypothesis would be rejected.

23.          True or False: Referring to Table 9-6, if the test is performed with a level of significance of 0.05, the engineer can conclude that the mean amount of force necessary to produce cracks in stressed oak furniture is 650.

24.          True or False: Referring to Table 9-6, suppose the engineer had decided that the alternative hypothesis to test was that the mean was greater than 650. Then if the test is performed with a level of significance of 0.10, the null hypothesis would be rejected.

25.          Referring to Table 9-6, suppose the engineer had decided that the alternative hypothesis to test was that the mean was greater than 650. What would be the p-value of this one-tailed test?

a)            0.040

b)            0.160

c)            0.840

d)            0.960

26.          Referring to Table 9-6, suppose the engineer had decided that the alternative hypothesis to test was that the mean was less than 650. What would be the p-value of this one-tailed test?

a)            0.040

b)            0.160

c)            0.840

d)            0.960

27.          True or False: Referring to Table 9-6, suppose the engineer had decided that the alternative hypothesis to test was that the mean was less than 650. Then if the test is performed with a level of significance of 0.10, the null hypothesis would be rejected.

TABLE 9-7

28.          Referring to Table 9-7, the parameter the company officials is interested in is:

a)            the mean number of viewers who “like the ads a lot”.

b)            the total number of viewers who “like the ads a lot”.

c)            the mean number of company officials who “like the ads a lot”.

d)            the proportion of viewers who “like the ads a lot”.

29.          Referring to Table 9-7, state the alternative hypothesis for this study.

30.          Referring to Table 9-7, what critical value should the company officials use to determine the rejection region?

31.          Referring to Table 9-7, the null hypothesis will be rejected if the test statistics is

a)            greater than 2.3263

b)            less than 2.3263

c)            greater than ?2.3263

d)            less than ?2.3263

32.          True or False: Referring to Table 9-7, the null hypothesis would be rejected.

33.          Referring to Table 9-7, the lowest level of significance at which the null hypothesis can be rejected is ______.

34.          Referring to Table 9-7, the largest level of significance at which the null hypothesis will not be rejected is ______.

35.          True of False: Referring to Table 9-7, the company officials can conclude that there is sufficient evidence to show that the series of television advertisements are less successful than the typical ad using a level of significance of 0.01.

36.          True or False: Referring to Table 9-7, the company officials can conclude that there is sufficient evidence to show that the series of television advertisements are less successful than the typical ad using a level of significance of 0.05.

37.          True or False: Referring to Table 9-7, the value of   is 0.90.

38.          Referring to Table 9-7, what will be the p-value if these data were used to perform a two-tailed test?

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