Question #349087

A sample of 160 people has a mean age of 27 with a population standard deviation of 5. Test the hypothesis that the population mean is 26.7 at =0.05

1
Expert's answer
2022-06-09T14:47:35-0400

The following null and alternative hypotheses need to be tested:

H0:μ=26.7H_0:\mu=26.7

H1:μ26.7H_1:\mu\not=26.7

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=2726.75/160=0.759z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{27-26.7}{5/\sqrt{160}}=0.759


6. Since it is observed that z=0.759<1.96=zc,|z|=0.759<1.96=z_c, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value for two-tailed is p=2P(Z>0.759)=0.447853,p=2P(Z>0.759)=0.447853, and since p=0.447853>0.05=α,p=0.447853>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 different than 26.7, at the α=0.05\alpha = 0.05 significance level.


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