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.

Answer 2

We can reject the null Hypothesis, i.e.

$H_{1} :\mu >8 \ is \ accepted.$

Here the null hypothesis is,

$H_{0}:\mu=8$

Null Hypothesis: The null hypothesis is a kind of hypothesis which explains the population parameter whose purpose is to test the validity of the given experimental data.

An alternate hypothesis is,

$H_{1}:\mu> 8$ (right-tailed test)

The test statistics is,

$z=\frac{\bar x-{\mu }}{\sqrt{\sigma ^{2}/n}}$

$given \ n=36$

$\bar x{=9 \ \ \sigma =1} }$

$z=\frac{9-8 }{\sqrt{1^{2}/36}}=6$

So, $Z_{cal} =6$

$Z_{tab} \ at\left ( \alpha =0.05 \right )is=1.645$

So, $Z_{cal}> Z_{tab}$

Therefore, we can reject the null Hypothesis i.e., $H_{1} :\mu> 8 \ is \ accepted.$

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