Question

Suppose a consumer product researcher wanted to find out whether a highlighter lasted longer than the manufacturer's claim that their highlighters could write continuously for 14 hours. The researcher tested 40 highlighters and recorded the number of continuous hours each highlighter wrote before drying up. Test the hypothesis that the highlighters wrote for more than 14 continuous hours. Following are the summary statistics: =14.5 hours, s =1.2 hours Report the test statistic, p-value, your decision regarding the null hypothesis. At the 5% significance level, state your conclusion about the original claim. Round all values to the nearest thousandth.

1.) ______

A) z = 9.583; p = ; Reject the null hypothesis; there is strong evidence to suggest that the highlighters last longer than 14 hours.

B) t = 2.635; p = 0.006; Fail to reject the null hypothesis; there is not strong evidence to suggest that the highlighters last longer than 14 hours.

C) t = 2.635; p = 0.006; Reject the null hypothesis; there is strong evidence to suggest that the highlighters last longer than 14 hours.

D) z = 9.583; p = ; Fail to reject the null hypothesis; there is not strong evidence to suggest that the highlighters last longer than 14 hours.

Answer 1

C) t = 2.635; p = 0.006; Reject the null hypothesis; there is strong evidence to suggest that the highlighters last longer than 14 hours.

Explanation:

Data given and notation

`\bar X=14.5` represent the sample mean

`s=1.2` represent the sample standard deviation

`n=40` sample size

`\mu_o =14` represent the value that we want to test

`\alpha=0.05` represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

`p_v` represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is higher than 14, the system of hypothesis would be:

Null hypothesis:
`\mu \leq 14`

Alternative hypothesis:
`\mu > 14`

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

`t=(\bar X-\mu_o)/((s)/(√(n)))` (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

`t=(14.5-14)/((1.2)/(√(40)))=2.635`

P-value

The first step is calculate the degrees of freedom, on this case:

`df=n-1=40-1=39`

Since is a one side test the p value would be:

`p_v =P(t_((39))>2.635)=0.006`

Conclusion

If we compare the p value and the significance level given
`\alpha=0.05` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis

The best option would be:

C) t = 2.635; p = 0.006; Reject the null hypothesis; there is strong evidence to suggest that the highlighters last longer than 14 hours.