Answer:
There is not sufficient evidence to support the claim that the technique performs differently than the traditional method.
Explanation:
The null hypothesis is:
![H_(0) = 95](https://img.qammunity.org/2022/formulas/mathematics/college/kf6e8v7llkl1g1w297gbvcr14t24z79s14.png)
The alternate hypotesis is:
![H_(1) \\eq 95](https://img.qammunity.org/2022/formulas/mathematics/college/ru9aoa6jsn6wmmgfj8qlebt0faq2aks6oj.png)
The test statistic is:
![z = (X - \mu)/((\sigma)/(√(n)))](https://img.qammunity.org/2022/formulas/mathematics/college/59im90558cjdobm60unnw2lrn6ewzh3ena.png)
In which X is the sample mean,
is the value tested at the null hypothesis,
is the standard deviation and n is the size of the sample.
A researcher used the technique with 260 students and observed that they had a mean of 94 hours. Assume the standard deviation is known to be 6.
This means, respectively, that
![n = 260, X = 94, \sigma = 6](https://img.qammunity.org/2022/formulas/mathematics/college/x8qq9b9r9y4o97cxvm4akhe4tjmolhq71c.png)
The test-statistic is:
![z = (X - \mu)/((\sigma)/(√(n)))](https://img.qammunity.org/2022/formulas/mathematics/college/59im90558cjdobm60unnw2lrn6ewzh3ena.png)
![z = (94 - 95)/((6)/(√(260)))](https://img.qammunity.org/2022/formulas/mathematics/college/fopenbgrkzhzklaeux0josy88rp3tktknr.png)
![z = -2.69](https://img.qammunity.org/2022/formulas/mathematics/college/ze4dbrnyji4lrf4ajnhy1su2x8ki9rwdui.png)
The pvalue is:
2(P(Z < -2.69))
P(Z < -2.69) is the pvalue of Z when X = -2.69, which looking at the z-table, is 0.0036
2*(0.0036) = 0.0072
0.0072 < 0.01, which means that the null hypothesis is accepted, that is, there is not sufficient evidence to support the claim that the technique performs differently than the traditional method.