Question #344869

Hi:u=85 Ha:u=85




The sample mean is 83,the sample size is 39,and the standard deviation is 5. Use a=0.05

1
Expert's answer
2022-05-26T11:06:52-0400

The following null and alternative hypotheses need to be tested:

H0:μ=85H_0:\mu=85

H1:μ85H_1:\mu\not=85

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=0.05,\alpha = 0.05, df=n1=38df=n-1=38 and the critical value for a two-tailed test is tc=2.024394.t_c =2.024394.

The rejection region for this two-tailed test is R={t:t>2.024394}.R = \{t:|t|>2.024394\}.

The t-statistic is computed as follows:



t=xˉμs/n=83855/39=2.498t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{83-85}{5/\sqrt{39}}=-2.498

Since it is observed that t=2.498>2.024394=tc,|t|=2.498>2.024394=t_c, it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for two-tailed, df=38df=38 degrees of freedom, t=2.498t=-2.498 is p=0.016935,p= 0.016935, and since p=0.016935<0.05=α,p= 0.016935<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 85, at the α=0.05\alpha = 0.05 significance level.


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