Answer to Question #348726 in Statistics and Probability for Nalin

Question

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

Answer 1

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

$H_0:\mu=5.5$

$H_1:\mu\not=5.5$

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 $\alpha = 0.05,$ and the critical value for a two-tailed test is $z_c = 1.96.$

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

5. The z-statistic is computed as follows:

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