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## Math 115

1) In 2002, the mean age of an inmate on death row was 40.7 years. A
sociologist wants to test the claim that the mean age of a death-row
inmate has changed since then. She randomly selects 32 death-row
inmates and finds that their mean age is 38.9 with a sample standard
deviation of 9.6 years. Test the claim at the α = 0.05 level of significance.
Claim: µ ≠ 40.7
H0: µ = 40.7

H1:
µ ≠ 40.7
Critical Value(s) = ± 2.040
Test Statistic =
38.9− 40.7

Fail to reject!
Conclusion:
There is not sufficient sample evidence to support the claim that
the mean age is different from 40.7.
2) Nexium is a drug that can be used to reduce the acid produced by the
body and heal damage to the esophagus due to acid reflux. Suppose the
manufacturer claims that more than 94% of patients taking Nexium are
healed within 8 weeks. In clinical trials 213 of 224 patients were healed
after 8 weeks. Test the manufacturers claim at the α = 0.01 level of
significance.
Claim: P > 0.94
H0: P ≤ 0.94

H1:
P > 0.94
Critical value(s) = 2.33

Test Statistic =

Fail to Reject!
Conclusion:
There is not sufficient sample evidence to support the claim that
more than 94% of patients are healed within 8 weeks.
3) A researcher claims that the mean height of women today is greater the
mean height of women in 1974 which was 63.7 inches. She obtains a simple
random sample of 45 women and finds the sample mean to be 63.9 inches.
Assume that the population standard deviation is 3.5 inches. Test the
researcher’s claim using a level of significance of α = 0.05.
Claim: µ > 63.7

H0:
µ ≤ 63.7

H1:
µ > 63.7

Critical value(s) =
1.645
Test Statistic =

Fail to reject!
Conclusion:
There is not sufficient sample evidence to support the claim that
the mean height today is greater than in 1994.
4) A drug company manufactures a 200-mg pain reliever. Specifications
demand that the standard deviation of the amount of the active
ingredient must not exceed 5 mg. You select a random sample of 30
tablets from a certain batch and find that the sample standard deviation
is 7.3 mg. Assume the amount of the active ingredient is normally
distributed. Test the claim that the standard deviation of the amount of
the active ingredient is greater than 5 mg using α = 0.05.
Claim: σ > 5

H0:
σ ≤ 5
H1:
σ > 5

Critical value(s) =
42.557
Test Statistic =

Reject Ho !
Conclusion:
The sample data support the claim that σ > 5.
Using this data set determine:
i) the equation of the linear regression line (y = a + bx)
a = 6.55

y = 6.55 – 0.714x
ii) the correlation coefficient, r.
r = - 0.948
iii) Using α = 0.05, is there a linear correlation between x and y?
Critical values = ± 0.811
Claim: ρ ≠ 0, so H0 : ρ = 0 & H1: ρ ≠ 0.
Test statistic is r = -0.948 which falls in the critical region so we
reject H0.
Yes, there is correlation between x & y.

Source: http://www.math.utk.edu/~kbonee/115/exam3/115-Ex3s-F07.pdf

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