Question #333854

An academic organization claimed that Grade 11 students study time is at most 200 minutes per day, on average. Another survey was conducted to find whether the claim is true. The group took a random sample of 30 students and found a mean study time of 240 minutes with standard deviation of 100 minutes.



Answer the following:



Parameter:


Claim:


Claim ( in symbol ):


Ho: Ho:


Ha: Ha:


What is the significance level or a?


Is it two-tailed or one-tailed test?

1
Expert's answer
2022-05-02T01:41:23-0400

Parameter: mean

Claim: Grade 11 students study time is at most 200 minutes per day, on average.

The following null and alternative hypotheses need to be tested:

H0:μ200H_0:\mu\le200

Ha:μ>200H_a:\mu>200

The significance level is α=0.05.\alpha = 0.05.

One-tailed test.

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.

The significance level is α=0.05,\alpha = 0.05, df=n1=301=29df=n-1=30-1=29 degrees of freedom, and the critical value for a right-tailed test is tc=1.699127.t_c = 1.699127.

The rejection region for this right-tailed test isR={t:t>1.699127}.R = \{t: t > 1.699127\}.

The t-statistic is computed as follows:


t=xˉμs/n=240200100/302.1909t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{240-200}{100/\sqrt{30}}\approx2.1909

Since it is observed that t=2.1909>1.699127=tc,t = 2.1909>1.699127= t_c , it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for right-tailed, df=29df=29 degrees of freedom, t=2.1909t=2.1909 is p=0.018322,p=0.018322, and since p=0.018322<0.05=α,p=0.018322<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 200, at the α=0.05\alpha = 0.05 significance level.



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