Questions:

All these questions are on the Normal Distribution (z)

1/ A trucking firm suspects the claim that the average lifetime of certain tires is
at least 28,000 miles. To check the claim, the firm tests 40 such tires on its trucks
and gets a mean lifetime of 27,563 miles with a standard deviation of 1,348 miles.
Perform the test at the 0.01 level of significance.

(click here for solution) (5)

2/ A law student wants to test the claim that convicted embezzlers spend an average
of 12.3 months in jail. He tests a random sample of 35 such cases and gets an
average of 11.5 months with s = 3.8 months. Test at the 0.05 level of significance.

(click here for solution) (5)

3/ A diesel engine manufacturer claims that his engines have a mean thermal efficiency
of at least 32.3%. Tests performed on 40 of his engines yielded a mean thermal efficiency
of 31.8% with a standard deviation of 2.2%. Test his claim at the 0.05 level of significance.
Find the tail probability or p-value associated with these test results.

(click here for solution) (6)

4/ A food study reports that 1 pound of goodfish yields an average of 2.41 ounces
of FPC (fish-protein concentrate) used to enrich food. Is this average supported by
a study of 30 samples of goodfish which show an average of 2.44 ounces of FPC/lb. and a standard deviation of 0.07 ounces if we use the:

a) 0.05 level of significance;

b) 0.01 level of significance?

c) Find the tail probability corresponding to the 2 values of z in a) and b) to
decide whether Ho can be rejected at the 0.02 level of significance.

(click here for solution) (12)

5/ In a study to test if there is a difference between the average heights of adult
females born in 2 different countries, random samples yielded these results:

	sample size	sample mean	sample st. dev.
sample 1	n1 = 120	x1 = 62.7	s1 = 2.5
sample 2	n2 = 150	x2 = 61.8	s2 = 2.62

Test at the 0.05 level the null hypothesis that the 2 population means are equal.

(click here for solution) (5)

6/ In a study of the relationship between family size and intelligence, 40 "only children"
had an average IQ of 101.5 with a standard deviation of 6.7 and 50 "first borns"
in two child families had an average IQ of 105.9 with a standard deviation of 5.8.
At the 0.05 level of significance, test if the difference of the 2 means is significant.

(click here for solution) (5)

Solutions:

1)

Lower Tail Test on a Mean
Ho: l m 28,000 miles	Ha: l < 28,000 miles
Level of significance: a = 1 %	Criteria: if z < -2.33, reject Ho
Statistic:
Conclusion: we cannot reject the null hypothesis Ho.

2)

2-Tail Test on a Mean
Ho: l = 12.3 months	Ha: l ! 12.3 months
Level of significance: a = 5 %	Criteria: if | z | > 1.96, reject Ho
Statistic:
Conclusion: we cannot reject the null hypothesis Ho.

3)

Lower Tail Test on a Mean
Ho: l m 32.3 % efficiency rating	Ha: l < 32.3 % efficiency rating
Level of significance: a = 5 %	Criteria: if z < -1.645, reject Ho
Statistic:
Conclusion: we cannot reject the null hypothesis Ho. the p-value = the probability of being in the tail created by the statistic z = -1.44 the p-value = .5000 - .4251 = .0749 which is > 5%, that's why we don't reject Ho.

4)

2-Tail Test on a Mean
Ho: l = 2.41 ounces	Ha: l ! 2.41 ounces
Level of significance: a) a = 5 %, b) a = 1 %	Criteria: a) if | z | > 1.96, reject Ho b) if | z | > 2.575, reject Ho
Statistic:
Conclusion: a) we must reject the null hypothesis Ho. b) we cannot reject Ho; p-value = 2(.5000 - .4906) = .0188 since .0188 < .02, we could reject Ho at the 2% level of significance.

5)

Hypothesis Test on Difference of Means
Ho: l1 = l2	Ha: l1 ! l2
Level of significance: a = 5 %	Criteria: if | z | > 1.96, reject Ho
Statistic:
Conclusion: we must reject the null hypothesis Ho.

6)

Hypothesis Test on Difference of Means
Ho: l1 = l2	Ha: l1 ! l2
Level of significance: a = 5 %	Criteria: if | z | > 1.96, reject Ho
Statistic:
Conclusion: we must reject the null hypothesis Ho.

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