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Sagot :
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]$\beta=P\left(Z < 1.645-\frac{2 \sqrt{10}}{1.25}\right)=P(Z < -3.409)=0.000326$[/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{\left(z_\alpha+z_s\right)^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]$\left(-\infty, \bar{x}+z_\alpha \frac{\sigma}{\sqrt{n}}\right)$[/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.
What is null hypothesis?
In a scientific setting, a hypothesis (plural: hypotheses) is a testable claim about the relationship between two or more variables or a suggested explanation for a phenomenon that has been observed.
Part a
We have the following data given:
n=10 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{-\mu}{\sqrt{n}}}$[/tex]
If we replace the values given we got:
[tex]$z=\frac{40.5-40}{\frac{1.24}{\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$[/tex] we have enough evidence to FAIL to reject the null hypothesis. So then there is not evidence to support the claim that the mean is greater than 40 .
Part b
The p value on this case is given by:
[tex]p_v=P(Z > 1.265)=1-P(Z < 1.265)=1-0.897=0.1029[/tex]
Part c
For this case the probability of type II error is defined as the probability of incorrectly retaining the null hypothesis and is defined like this:
[tex]\beta=P\left(Z < z_\alpha-\frac{(x-\mu) \sqrt{n}}{\sigma}\right)[/tex]
Where [tex]$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$[/tex], so then we can replace like this:
[tex]\beta=P\left(Z < 1.645-\frac{2 \sqrt{10}}{1.25}\right)=P(Z < -3.409)=0.000326$[/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{\left(z_\alpha+z_\beta\right)^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]\left(-\infty, \bar{x}+z_\alpha \frac{\sigma}{\sqrt{n}}\right)[/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.
To learn more about null hypothesis visit:https://brainly.com/question/28920252
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