Question

A sample​ mean, sample​ size, and population standard deviation are provided below. Use the​ one-mean z-test to perform the required hypothesis test at the 10​% significance level.
x = 20​, n = 36​, sigma = 9​, H0​: mu = 25​, H a​mu : < 25.

Answer 1

Answer:

$z=\frac{20-25}{\frac{9}{\sqrt{36}}}=-3.33$  

The p value for this case is given by:

$p_v =P(z$  

For this case the p value is a very low value compared to the significance level of 0.1 so then we can reject the null hypothesis and we can conclude that the true mean is significantly less than 25 at 10% of significance.

Step-by-step explanation:

Information given

$\bar X=20$ represent the sample mean  

$\sigma=9$ represent the population deviation

$n=36$ sample size  

$\mu_o =25$ represent the value to verify

$\alpha=0.1$ represent the significance level

tzwould represent the statistic

$p_v$ represent the p value

System of hypothesis

We want to test the hypothesis that the true mean is lower than 25  and the system of hypothesis are:  

Null hypothesis:$\mu \geq 25$  

Alternative hypothesis:$\mu < 25$  

The statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)  

Replacing the info given:

$z=\frac{20-25}{\frac{9}{\sqrt{36}}}=-3.33$  

The p value for this case is given by:

$p_v =P(z$  

For this case the p value is a very low value compared to the significance level of 0.1 so then we can reject the null hypothesis and we can conclude that the true mean is significantly less than 25 at 10% of significance.