Question #305006

Last year the employees of the city health department donated an average of $ 8 to the rescue squad. Test the hypothesis at the 0.01 level of significance that the average contribution this year is still $ 8 if a random sample of 35 employees showed an average donation of $ 8.90 with a standard deviation of $ 1.75.

1
Expert's answer
2022-03-03T09:59:42-0500

The following null and alternative hypotheses need to be tested:

H0:μ=8H_0:\mu=8

H1:μ8H_1:\mu\not=8

This corresponds to a two-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=0.01,\alpha = 0.01, df=n1=34df=n-1=34 degrees of freedom, and the critical value for a two-tailed test is tc=2.728394.t_c=2.728394.

The rejection region for this two-tailed test is R={t:t>2.728394}.R = \{t: |t| > 2.728394\}.

The t-statistic is computed as follows:


t=xˉμs/n=8.9081.75/353.042555t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{8.90-8}{1.75/\sqrt{35}}\approx3.042555

Since it is observed that t=3.042555>2.728394=tc,|t| = 3.042555 > 2.728394=t_c , it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for two-tailed, df=34df=34 degrees of freedom, t=3.042555,t=3.042555, is p=0.0045,p=0.0045, and since p=0.0045<0.01=α,p=0.0045<0.01=\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 8, at the α=0.01\alpha = 0.01 significance level.



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