Question

A large company that produces allergy medications claims that Americans lose an average of 40 hours of work to problems related to seasonal allergies. A consumer advocacy group believes that company's claim exhibits self-interest bias. In other words, the advocacy group believes that the actual mean number of missed hours is less than claimed. The advocacy group would like to obtain statistical evidence about this issue and takes a random sample of 100 American workers. They find that these 100 people lost and average of 38 hours due to seasonal allergy symptoms. The standard deviation is 9.5 hours.

Required:
a. What are the correct null and alternative hypotheses in this situation?
b. Based on your selection from part (b), what is the value of the standardized test statistic for this significance test?
c. Based on your selection from part (b). what is the P-value for this significance test?

Answer 1

a) See step by step explanation

b) z(s) = -2,1

c) P-value = 0,0179

Explanation:

Company opinion

Normal distribution

Population mean μ = 40 h

Population standard deviation unknown

Sample size n = 100

Sample mean μ = 38 h

Sample standard deviation s = 9,5

a) Test Hypothesis

Null Hypothesis H₀ μ = μ₀

Alternative Hypothesis Hₐ μ < μ₀

CI we will choose 95% then significance value α = 5 % α = 0,05

The sign (-) in the alternative hypothesis mean we will develop a one tail-test to the left, as n = 100 s a normal z-table case

z(c) for α = 0,05 z(c) = - 1,74

b) z(c) = - 1,74

c) P-value ???

z(s) = ( 38 - 40 ) / 9,5 / √100

z(s) = -2*10 / 9,5

z(s) = - 2,105

From z-Table P-value = 0,0179

P-value < 0,05

Then as P value is smaller than α, we Reject H₀, and accet the claim that company claim exhibits self-interest bias