PART I: Examine the given statement, then express the null hypothesis H0 and alternative hypothesis H1 in symbolic form. Be sure to use the correct symbol (Âľ,p,) for the indicated parameter.

1.The mean IQ of statistics students is at least 110.

H0:

H1:

2.The percentage of men who watch golf on TV is not 70%, as is claimed by the Madison Advertising Company.

H0:

H1:

3.Salaries among women business analysts have a standard deviation greater than $3,000.

H0:

H1:

4. Find the critical z values in each of the following exercises. Assume that the normal distribution applies.

a.Two-tailed test with = 0.01

b.Right-tailed test with = 0.05

c. Left-tailed test with = 0.10

5. Use the 2 table to find the critical values of 2 based on the information given.

a. H1: 0.237
n = 15, df = 14
= 0.05

b. H1:
n = 23
= 0.025

c.H1: > 0.237
n = 12
= 0.10

PART II: Perform each of the following hypothesis tests.

1.In a Gallup survey, 1087 randomly selected adults were asked “Do you have occasion to use alcoholic beverages such as liquor, wine, or beer, or are you a total abstainer?” Sixty-two percent of the subjects said that they used alcoholic beverages. Consider a hypothesis test that uses a 0.05 significance level to test the claim that the majority (more than 50%) of adults use alcoholic beverages.

Claim: The majority (more than 50%) of adults use alcoholic beverages.

H0:

H1:

=

Critical Value(s):

Test Statistic:

Decision:

Conclusion:

2.In previous tests, baseballs were dropped 24 ft onto a concrete surface, and they bounced an average of 92.84 in. In a test of a sample of 40 new balls, the bounce heights had a mean of 92.67 in. and a standard deviation of 1.79 in. Use a 0.05 significance level to determine whether there is sufficient evidence to support the claim that the new balls have bounce heights with a mean different from 92.84 in.

Claim: The new balls have bounce heights with a mean different from 92.84 in

H0:
H1:

=

Critical Value(s):

Test Statistic:

Decision:

Conclusion:

3.With individual lines at its various windows, the Jefferson Valley Bank found that the standard deviation for normally distributed waiting times on Friday afternoons was 6.2 min. The bank experimented with a single main waiting line and found that for a simple random sample of 25 customers, the waiting times have a standard deviation of 3.8 min. Use a 0.05 significance level to test the claim that a single line causes lower variation among the waiting times.

Claim: a single line causes lower variation among the waiting times.

H0:
H1:

=

Critical Value(s):

Test Statistic:

Decision:

Conclusion:

4.Mental measurements of young children are often made by giving them blocks and telling them to build a tower as tall as possible. One experiment of block building was repeated a month later, with the time (in seconds) listed in the accompanying table. Is there sufficient evidence to support the claim that there is a difference between the two times? Use a 0.01 significance level.

CHILDABCDEFGHIJKLMN
FIRST 3019192329524220123914811731
SECOND30 614 81415142217 811301417
Difference 013 515153728-2-531 351 314

_
d =

Claim: There is a difference between the two times.

H0:

H1:

=

n =

df =

Critical Value: t =

Test Statistic: t =

Decision:

Conclusion:

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Wechsler IQ scores typically follow a distribution with mean 100 and standard deviation 15. Using the Full Scale IQ scores for all the test subjects, first test the hypothesis that the standard deviation in IQ is 15 points. Based on the outcome of that test, conduct the appropriate tests that of the claim that the mean is 100 points. Use a 0.05 level of significance.

Claim: The standard deviation in IQ is 15 points.

H0:

H1:
= 0.05

Critical Value(s):

Test Statistic:

Decision:

Conclusion:

Claim: The mean IQ is 100.

H0:

H1:

= 0.05

Critical Value(s):

Test Statistic:

Decision:

Conclusion: