Question #342154

Show your solution using the critical value method.


a previous study on the effect of blended learning module on senior high school student's achievement revealed that mean score pi is 85 with standard deviation of 3. juliana, a senior high school statistic teacher, surveyed a class of 30 students test the hypothesis at a 0.10 level of significance


1
Expert's answer
2022-05-18T09:48:51-0400

The following null and alternative hypotheses need to be tested:

H0:μ=85H_0:\mu=85

Ha:μ85H_a:\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=0.10,\alpha = 0.10, df=n1=29df=n-1=29 degrees of freedom, and the critical value for a two-tailed test is tc=1.699127.t_c =1.699127.

The rejection region for this two-tailed test is R={t:t>1.699127}.R = \{t: |t|>1.699127\}.

The t-statistic is computed as follows:


t=Xˉμs/n=83854302.7386t=\dfrac{\bar{X}-\mu}{s/\sqrt{n}}=\dfrac{83-85}{4\sqrt{30}}\approx-2.7386

Since it is observed that t=2.7386>1.699127=tc,|t|= 2.7386 >1.699127= t_c, it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for two-tailed df=29df=29 degrees of freedom, t=2.7386t=-2.7386 is p=0.00617,p =0.00617, and since p=0.00617<0.10=α,p=0.00617<0.10=\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=0.10\alpha = 0.10 significance level.


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