Question #339498

A random sample of 11 observations was taken from normal population. The sample mean and

standard deviation are 74.5 and 9 accordingly. Can we infer at 5% significance level that the

population mean is greater than 70?

5. Repeat number 4 with assuming the population standard deviation = 9


1
Expert's answer
2022-05-11T12:09:32-0400

4. The following null and alternative hypotheses need to be tested:

H0:μ70H_0:\mu\le70

Ha:μ>70H_a:\mu>70

This corresponds to a right-tailed test, for which a t-test for one mean, with unknown population standard deviation, using the sample standard deviation, will be used.

Based on the information provided, the significance level is α=0.05,\alpha = 0.05, df=n1=10df=n-1=10 degrees of freedom, and the critical value for a right-tailed test is tc=1.812461.t_c = 1.812461.

The rejection region for this right-tailed test is R={t:t>1.812461}.R = \{t: t> 1.812461\}.

The t-statistic is computed as follows:


t=xˉμs/n=74.5709/111.6583t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{74.5-70}{9/\sqrt{11}}\approx1.6583


Since it is observed that t=1.6583<1.812461=tc,t=1.6583< 1.812461=t_c, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value for right-tailed, df=10df=10 degrees of freedom, t=1.6583t=1.6583 is p=0.064124,p=0.064124, and since p=0.064124>0.05=α,p=0.064124>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 70, at the α=0.05\alpha = 0.05 significance level.


5. The following null and alternative hypotheses need to be tested:

H0:μ70H_0:\mu\le70

Ha:μ>70H_a:\mu>70

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=74.5709/111.6583z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{74.5-70}{9/\sqrt{11}}\approx1.6583


Since it is observed that z=1.6583>1.6449=tc,z=1.6583> 1.6449=t_c, it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for right-tailed is p=P(Z>1.6583)=0.048628,p=P(Z>1.6583)=0.048628, and since p=0.048628<0.05=α,p=0.048628<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 greater than 70, at the α=0.05\alpha = 0.05 significance level.



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