Question

The effectiveness of a new bug repellent is tested on 1515 subjects for a 10 hour period. (Assume normally distributed population.) Based on the number and location of the bug bites, the percentage of surface area exposed protected from bites was calculated for each of the subjects. The results were as follows:

?⎯⎯⎯=94x¯=94,  ?=9 s=9

The new repellent is considered effective if it provides a percent repellency of at least 9191. Using ?=0.05α=0.05, construct a hypothesis test with null hypothesis ?=91μ=91 and alternative hypothesis ?>91μ>91 to determine whether the mean repellency of the new bug repellent is greater than 9191 by computing the following:

A) the degree of freedom

B) the critical t Value

C) the test statistics

Answer 1

Null hypothesis

vs

Alternative hypothesis

A) the degree of freedom = n - 1 = 15 - 1 = 14

B) the critical t Value = 1.761(by using t table)

C) t test statistic formula is

  

=1.29

t test statistic =1.29 > t critical value = 1.761

Therefore, we reject H0 at

We have sufficient evidence at to say that, the mean repellency of the new bug repellent is greater than 91