Question #341208

A random sample of ten measurements were obtained from a normally distributed population with mean u=6.5. The sample values are X-4.2 and s 2.




a. Test the null hypothesis that the mean of the population against the alternative hypothesis, μ = 6.5. Use a = 0.05.




b. Test the null hypothesis that the mean of the population against the alternative hypothesis, u 6.5. Use a = 0.05

1
Expert's answer
2022-05-16T15:55:51-0400

a.

The following null and alternative hypotheses need to be tested:

H0:μ=6.5H_0:\mu=6.5

H1:μ>6.5H_1:\mu>6.5

This corresponds to a right-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

Based on the information provided, the significance level is α=0.05,\alpha = 0.05, and the critical value for a right-tailed test is zc=1.6449.z_c = 1.6449.

The rejection region for this right-tailed test is R={z:z>1.6449}.R = \{z:z> 1.6449\}.

The z-statistic is computed as follows:


z=xˉμσ/n=4.26.52/103.6366z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{4.2-6.5}{2/\sqrt{10}}\approx-3.6366

Since it is observed that z=3.6366<1.6449=zc,z=-3.6366<1.6449=z_c, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value is p=P(z>3.6366)=0.999862,p=P(z>-3.6366)=0.999862, and since p=0.999862>0.05=α,p=0.999862>0.05=\alpha, it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean μ\mu

is greater than 90, at the α=0.05\alpha = 0.05 significance level.


b.

The following null and alternative hypotheses need to be tested:

H0:μ=6.5H_0:\mu=6.5

H1:μ6.5H_1:\mu\not=6.5

This corresponds to a two-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

Based on the information provided, the significance level is α=0.05,\alpha = 0.05, and the critical value for a two-tailed test is zc=1.96.z_c = 1.96.

The rejection region for this two-tailed test is R={z:z>1.96}.R = \{z: |z| > 1.96\}.

The z-statistic is computed as follows:


z=xˉμσ/n=4.26.52/103.6366z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{4.2-6.5}{2/\sqrt{10}}\approx-3.6366

Since it is observed that z=3.6366<1.96=zc,z =- 3.6366 <1.96=z_c, it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is p=2P(z<3.6366)=0.000276,p=2P(z<-3.6366)=0.000276, and since p=0.000276<0.05=α,p=0.000276<0.05=\alpha, it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean μ\mu

is different than 6.5, at the α=0.05\alpha = 0.05 significance level.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS