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 overbarequals18​, nequals33​, sigmaequals5​, Upper H 0​: mu equals 21​, Upper H Subscript a​: mu less than 21

Answer 1

Answer:

$z=\frac{18-21}{\frac{5}{\sqrt{33}}}=-3.447$  

$p_v =P(z<-3.447)=0.00028$  

We can see that $p_v<\alpha=0.1$ so we can conclude that we have enough evidence to reject the null hypothesis, and the true mean is significantly less than 21 at 10% of signficance.  

Step-by-step explanation:

Data given and notation  

$\bar X=18$ represent the sample mean

$\sigma=5$ represent the population standard deviation

$n=33$ sample size  

$\mu_o =21$ represent the value that we want to test  

$\alpha=0.1$ represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

$p_v$ represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is less than 21 or no, the system of hypothesis would be:  

Null hypothesis:$\mu \geq 21$  

Alternative hypothesis:$\mu < 21$  

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

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

Calculate the statistic  

We can replace in formula (1) the info given like this:  

$z=\frac{18-21}{\frac{5}{\sqrt{33}}}=-3.447$  

P-value  

Since is a left side test the p value would be:  

$p_v =P(z<-3.447)=0.00028$  

Conclusion  

We can see that $p_v<\alpha=0.1$ so we can conclude that we have enough evidence to reject the null hypothesis, and the true mean is significantly less than 21 at 10% of signficance.  

Answer 2

Using the z-distribution, it is found that since the test statistic is less than the critical value for the left-tailed test, there is enough evidence to conclude that the population mean is less than 21.

The null hypothesis is:

$H_0: \mu = 21$

The alternative hypothesis is:

$H_1: \mu < 21$

We have the standard deviation for the population, thus, the z-distribution is used. The test statistic is given by:

$z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

The parameters are:

• $\overline{x}$ is the sample mean.
• $\mu$ is the value tested at the null hypothesis.
• $\sigma$ is the standard deviation of the population.
• n is the sample size.

For this problem, the values of the parameters are: $\overline{x} = 18, \mu = 21, \sigma = 5, n = 33$

Hence, the value of the test statistic is:

$z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{18 - 21}{\frac{5}{\sqrt{33}}}$

$z = -3.45$

The critical value for a left-tailed test, as we are testing if the mean is less than a value, with a 0.1 significance level is $z^{\ast} = -1.28$

Since the test statistic is less than the critical value for the left-tailed test, there is enough evidence to conclude that the population mean is less than 21.

A similar problem is given at https://brainly.com/question/15704150