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




1
Expert's answer
2022-06-07T15:33:33-0400

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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