Question

A random sample of 76 teenagers were asked how many movies they see in a year. It came out that the mean number of movies-per-year is 51.2 and the standard deviation is 15.4. Use a .05 significance level to test the claim that the mean number of movies-per-year is 52. Test claim using the critical value method .

Answer 1

The sample mean is Xˉ=51.2, the sample standard deviation is s=15.4, and the sample size is n=76. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: μ =52 Ha: μ ≠52 This corresponds to a Two-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used. (2a) Critical Value Based on the information provided, the significance level is α=0.05, and the degree of freedom is n-1=76-1=75. Therefore the critical value for this Two-tailed test is tc​=1.9921. This can be found by either using excel or the t distribution table. (2b) Rejection Region The rejection region for this Two-tailed test is |t|>1.9921 i.e. t>1.9921 or t<-1.9921 (3)Test Statistics The t-statistic is computed as follows: (4) The p-value The p-value is the probability of obtaining sample results as extreme or more extreme than the sample results obtained, under the assumption that the null hypothesis is true. In this case, the p-value is 0.6519 (5) The Decision about the null hypothesis (a) Using the critical method Since it is observed that |t|=0.4529 < tc​=1.9921, it is then concluded that the null hypothesis is Not rejected. (b) Using p-value method Using the P-value approach: The p-value is p=0.6519, and since p=0.6519>0.05, it is concluded that the null hypothesis is Not rejected. (6) Conclusion It is concluded that the null hypothesis Ho is Not rejected. Therefore, there is Not enough evidence to claim that the population mean μ is different than 52, at the 0.05 significance level.

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