Question

A recent study stated that if a person chewed gum, the average number of sticks of gum he or she chewed daily was 8. Researcher believes the number of sticks is greater. He selected a random sample of 36 gum chewers and found the mean number of sticks of gum chewed per day was 9. The standard deviation of the population is 1. At α = 0.05 level of significance test the claim that the number of sticks of gum a person chews per day is actually greater than 8.

Answer 1

Solution :

Given that,

Population mean = = 8

Sample mean = = 9

Sample standard deviation = s = 1

Sample size = n = 36

Level of significance = = 0.05

This is a two tailed test.

The null and alternative hypothesis is,

Ho: 8

Ha: 8

The test statistics,

t = ( - )/ (s/)

= ( 9 - 8 ) / ( 1 /36)

= 6

Critical t value

t_c = 1.69

Since it is observed that t = 6 >t_c = 1.69, it is then concluded that the null hypothesis is rejected.

p-value = 0

The p-value is p = 0p=0, and since p = 0 < 0.05, it is concluded that the null hypothesis is rejected.

Conclusion:

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population

mean μ is greater than 8, at the 0.05 significance level.