Respuesta :
Answer:
a) [tex] z= \frac{40.5-40}{\frac{1.25}{\sqrt{10}}}=1.265[/tex]
Now we can calculate the p value since we have a right tailed test the p values is given by:
[tex] p_v = P(Z>1.265) = 1-P(Z<1.265) =1-0.897=0.1029[/tex]
And since the [tex] p_v >\alpha[/tex] we have enough evidence to FAIL to reject the null hypothesis. So then there is not evidence to support the claim that the mean is greater than 40.
b) [tex] p_v = P(Z>1.265) = 1-P(Z<1.265) =1-0.897=0.1029[/tex]
c) [tex] \beta = P(Z< 1.645 - \frac{2\sqrt{10}}{1.25}) = P(Z<-3.409)= 0.000326[/tex]
d) We want to ensure that the probability of error type II not exceeds 0.1, and for this case we can use the following formula:
[tex] n = \frac{(z_{\alpha} +z_{\beta})^2 \sigma^2}{(x-\mu)^2}[/tex]
The true mean for this case is [tex] \mu = 44[/tex] and we want [tex] \beta<0.1[/tex] so then [tex] z_{1-0.1} =z_{0.9}=1.29[/tex] represent the value on the normal standard distribution that accumulates 0.1 of the area on the right tail. And we can replace like this:
[tex] n = \frac{(1.65+1.29)^2 1.25^2}{(44-40)^2} =0.844 \approx 1[/tex]
e) For this case we can calculate a one sided confidence interval given by:
[tex] (-\infty , \bar x +z_{\alpha} \frac{\sigma}{\sqrt{n}})[/tex]
And if we replace we got:
[tex] 40.5 +1.65 \frac{1.25}{\sqrt{10}}=41.152[/tex]
And the confidence interval would be [tex] (-\infty,41.152)[/tex]
And since 40 is on the confidence interval we don't have enough evidence to reject the null hypothesis on this case.
Step-by-step explanation:
Part a
We have the following data given:
[tex] n =10 [/tex] represent the sample size
[tex] \bar x = 40.5[/tex] represent the sample mean
[tex]\sigma =1.25[/tex] represent the population deviation.
We want to test the following hypothesis:
Null: [tex] \mu \leq 40[/tex]
Alternative:[tex] \mu >40[/tex]
The significance level provided was [tex]\alpha =0.05[/tex]
The statistic for this case since we have the population deviation is given by:
[tex] z= \frac{\bar x -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
If we replace the values given we got:
[tex] z= \frac{40.5-40}{\frac{1.25}{\sqrt{10}}}=1.265[/tex]
Now we can calculate the p value since we have a right tailed test the p values is given by:
[tex] p_v = P(Z>1.265) = 1-P(Z<1.265) =1-0.897=0.1029[/tex]
And since the [tex] p_v >\alpha[/tex] we have enough evidence to FAIL to reject the null hypothesis. So then there is not evidence to support the claim that the mean is greater than 40.
Part b
The p value on this case is given by:
[tex] p_v = P(Z>1.265) = 1-P(Z<1.265) =1-0.897=0.1029[/tex]
Part c
For this case the probability of type II error is defined as the probability of incorrectly retaining the null hypothesis and is defined like this:
[tex] \beta = P(Z< z_{\alpha} - \frac{(x-\mu)\sqrt{n}}{\sigma}})[/tex]
Where [tex] z_{\alpha}=1.645[/tex] represent the critical value for the test that accumulates 0.05 of the area on the right tail of the normal standard distribution.
The true mean on this case is assumed [tex]\mu = 42[/tex], so then we can replace like this:
[tex] \beta = P(Z< 1.645 - \frac{2\sqrt{10}}{1.25}) = P(Z<-3.409)= 0.000326[/tex]
Part d
We want to ensure that the probability of error type II not exceeds 0.1, and for this case we can use the following formula:
[tex] n = \frac{(z_{\alpha} +z_{\beta})^2 \sigma^2}{(x-\mu)^2}[/tex]
The ture mean for this case is [tex] \mu = 44[/tex] and we want [tex] \beta<0.1[/tex] so then [tex] z_{1-0.1} =z_{0.9}=1.29[/tex] represent the value on the normal standard distribution that accumulates 0.1 of the area on the right tail. And we can replace like this:
[tex] n = \frac{(1.65+1.29)^2 1.25^2}{(44-40)^2} =0.844 \approx 1[/tex]
Part e
For this case we can calculate a one sided confidence interval given by:
[tex] (-\infty , \bar x +z_{\alpha} \frac{\sigma}{\sqrt{n}})[/tex]
And if we replace we got:
[tex] 40.5 +1.65 \frac{1.25}{\sqrt{10}}=41.152[/tex]
And the confidence interval would be [tex] (-\infty,41.152)[/tex]
And since 40 is on the confidence interval we don't have enough evidence to reject the null hypothesis on this case.
