Respuesta :
Answer:
There is not sufficient evidence to support the claim that the technique performs differently than the traditional method.
Step-by-step explanation:
The null hypothesis is:
[tex]H_{0} = 95[/tex]
The alternate hypotesis is:
[tex]H_{1} \neq 95[/tex]
The test statistic is:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis, [tex]\sigma[/tex] 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 [tex]n = 260, X = 94, \sigma = 6[/tex]
The test-statistic is:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]z = \frac{94 - 95}{\frac{6}{\sqrt{260}}}[/tex]
[tex]z = -2.69[/tex]
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.
