Question #348726

There is a claim that on daily average, students spend 5.5 hours on social media with a standard deviation of 1.25. A researcher wants to test the claim at 5% level of significance.



1. Formulate the null and alternative hypothesis



2. Determine the test statistic to be used



3. Find the corresponding zvalue



4. Identify the rejection region




Expert's answer

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

H0:μ=5.5H_0:\mu=5.5

H1:μ5.5H_1:\mu\not=5.5


2. This corresponds to a two-tailed test, for which a z-test for one mean, with known population standard deviation will be used.


3. Based on the information provided, the significance level is α=0.05,\alpha = 0.05, and the critical value for a two-tailed test is zc=1.96.z_c = 1.96.


4. The rejection region for this two-tailed test is R={z:z>1.96}.R = \{z:|z|>1.96\}.


5. The z-statistic is computed as follows:


z=xˉμσ/nz=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}

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