atSuppose a professor is interested in seeing whether or not they hold their student's attention longer than averageUniversity. Assuming the average number of times that a student checks their phone within a 50minute class is 4.2, determine whether or not the data the professor has collected supports their hypothesis.Below are counts of number of times the professor has seen individual students in their class check their phoneswithin one class.2, 2, 1, 1, 6, 5, 5, 0, 3, 3, 1, 1, 1, 1, 31. Using the data provided, determine the null and alternative hypotheses. Then determine how many tails thisresearch question has2. Using the data provided, use what you have learned in Excel to calculate the standard deviation for thesample and the SEMs for the sample3. Use the calculated SEMs and other values from question 2, calculate your obtained t-statistic for this researchquestion4. Using Appendix B in your text, the degrees of freedom for this research question, and the number of tailsyour hypothesis has to determine the t-critical value5. Based on your calculated t-statistic and t-critical value, engage in hypothesis testing. What is your decisionregarding the research question

The observed number of times, the sampled individual students were seen checking their phones is: 2, 2, 1, 1, 6, 5, 5, 0, 3, 3, 1, 1, 1, 1, 3 Part (1) Considering μ as the average no. of times students check their phones at the population level, so if student shows more attention, students are likely to look up the phone less than average, then the hypothesis is set up as: H0:μ⩾4.2; the average no. of times students check their phone is higher than 4.2. H1:μ<4.2; the average no. of times students check their phones is not higher than 4.2. Considering the true parameter is being claimed to be higher than a claimed value, the critical region is defined on the higher end of the distribution. So this is a one-tailed (a right-tailed) test. Part (2) The data is assigned in excel as follows: The sample standard deviation is obtained using the excel function "STDEV.S" as follows: So sample standard deviation, s=1.7995 Considering the standard error of mean for the sample, i.e., SEMs=sample standard deviation ssample size n, the SEMs is calculated as follows (here, n=15): And SEMs=0.4646. Part (3) Using the sample observation in the question, the sample mean is: x¯=∑i=1nxin xi is the each individual observation Using the excel function "AVERAGE()" the sample mean is calculated as: So, x¯=2.3333. So the t-test statistic is calculated as: t=x¯-μ0SEMs μ0=claimed value of true mean=4.2=2.3333-4.20.4646≈-4.0179 Part (4) The required degrees of freedom is: df=n-1=15-1=14. The no. of tails for the hypothesis is 1, i.e., a one-tailed test. Considering the standard significance level of α=5%=0.05, the t-critical value for 14 df is considered as follows: Moreover, using the excel function "T.INV(probability,deg_freedom)", the t-critical value (right-tailed) for a 5% significance level with14 df is obtained as (provided that probability=1-α=1-5%=0.95): So the t-critical value is tcritical=-1.761 (Since the test is left tailed). Part (5) The decision rule is if tcritical < tcalculated, H0 is rejected. Since, tcalculated =-4.0179<tcritical =-1.761, H0 is rejected. So the average no. of times students check their phones is inferred to be not higher than 4.2, i.e., the teacher may conclude that the attention of students is not longer than the claimed time of 4.2.