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

Answer:

The average number of sticks of gum a person chewed daily was greater than 8.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 8

Sample mean, $\bar{x}$ = 9

Sample size, n = 36

Alpha, α = 0.05

Population standard deviation, σ = 1

First, we design the null and the alternate hypothesis

$H_{0}: \mu \leq 8\text{ gums}\\H_A: \mu > 8\text{ gums}$

We use one-tailed z test to perform this hypothesis.

Formula:

$z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }$

Putting all the values, we have

$z_{stat} = \displaystyle\frac{9 - 8}{\frac{1}{\sqrt{36}} } = 6$

Now, $z_{critical} \text{ at 0.05 level of significance } = 1.64$

Since,  

$z_{stat} > z_{critical}$

We fail to accept the null hypothesis and reject the null hypothesis. We accept the alternate hypothesis.

Thus, the average number of sticks of gum a person chewed daily was greater than 8.