The mean number of sick days an employee takes per year is believed to be about 10. Members of a personnel department do not believe this figure. They randomly survey 8 employees. The number of sick days they took for the past year are as follows: 10; 6; 13; 3; 10; 10; 7; 10. Let X = the number of sick days they took for the past year. Should the personnel team believe that the mean number is about 10? Conduct a hypothesis test at the 5% level1). What is the P value? Explain what the p-value means for this problem.2). Construct a 95% confidence interval for the true mean. Sketch the graph of the situation. Label the point estimate and the lower and upper bounds of the confidence interval. (Round your answers to three decimal places.) A Hypothesis Tests - 1 sample xy! The mean number of sick da XSearch results for ' The mea Xb Answered: A random survey XToday's Christmas Radio I XW Confidence Intervals - MATH X +webassign.net/web/Student/Assignment-Responses/submit?dep=24266164&tags=autosave#Q8Part (f)What is the p-value? (Round your answer to four decimal places.).2190Explain what the p-value means for this problem.O If Ho is true, then there is a chance equal to the p-value that the average number of sick days for employees is at least as different from 10 as the mean of the sample is different from 10.If Ho is false, then there is a chance equal to the p-value that the average number of sick days for employees is at least as different from 10 as the mean of the sample is different from 10.O If Ho is false, then there is a chance equal to the p-value the average number of sick days for employees is not at least as different from 10 as the mean of the sample is different from 10.If Ho is true, then there is a chance equal to the p-value the average number of sick days for employees is not at least as different from 10 as the mean of the sample is different from 10.Incorrect. The interpretation of a p-value is based on the assumption that Ho is true.Part (g)Part (h)Indicate the correct decision ("reject" or "do not reject" the null hypothesis), the reason for it, and write an appropriate conclusion.(i) Alpha (Enter an exact number as an integer, fraction, or decimal.)a = |0.05(ii) Decision:reject the null hypothesisrejectnull hypothesis(iii) Reason for decision:Since a > p-value, we do not reject the null hypothesis.Since a < p-value, we reject the null hypothesis.Since a > p-value, we reject the null hypothesis.Since a < p-value, we do not reject the null hypothesis. A Hypothesis Tests - 1 sample Xy! The mean number of sick da XSearch results for ' The mea Xb Answered: A random survey XToday's Christmas Radio 1 X W Confidence Intervals - MATH x +webassign.net/web/Student/Assignment-Responses/submit?dep=24266164&tags=autosave#Q8(iii) Reason for decision:Since a > p-value, we do not reject the null hypothesis.Since a < p-value, we reject the null hypothesis.Since a > p-value, we reject the null hypothesis.Since a < p-value, we do not reject the null hypothesis.(iv) Conclusion:There is sufficient evidence to conclude that the average number of sick days used per year by an employee is not equal to 10 days.There is not sufficient evidence to conclude that the average number of sick days used per year by an employee is not equal to 10 days.Part (i)Construct a 95% confidence interval for the true mean. Sketch the graph of the situation. Label the point estimate and the lower and upper bounds of the confidence interval. (Round your answers to three decimal places.)95% C.I.2.166.4710.8Additional MaterialsOBook

Name: The mean number of sick days an employee takes per year is believed to
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Description: The mean number of sick days an employee takes per year is believed to be about 10. Members of a personnel department do not believe this figure. They randomly survey 8 employees. The number of sick days they took for the past year are as follows: 10

It is given that the mean number of sick days an employee takes per year is : 10 However, the member of a personnel department do not believe this figure. The number of sick days they took for the past year are as follows: 10; 6; 13; 3; 10; 10; 7; 10. Let X = the number of sick days they took for the past year The test of hypothesis is given by : where ; mean and standard deviation is obtained using excel command : and Also, B3: B10 are taking the values from 10 to 10 i.e. ( 10,6,13,3,10,10,7,10) Therefore, x¯ = 8.625s.d (s) = 3.114 1. The test - statistic formula is given by : Thus the test statistic , t = -1.2489. To find P-value we need to know three things: 1) Test statistic , t = -1.2489. 2) Degrees of freedom , df = n-1 = 8-1 = 7. 3) Since we have not equal sign in our alternative hypotheses so it is two tailed test. It can be easily found using MS-Excel. The function to find critical value in Excel is as follows; =tdist ( test statistic , deg_freedom , tails). Therefore , to get a p- value please type =tdist( 1.2489 , 7 , 2 ) in an empty cell and press ENTER. Th p-value value would be 0.2518. Note : while entering test statistic value in excel please ignore the minus sign. The P-value decision rule : If P-value ≤ significance level , then we reject null and supports alternative . If P-value > significance level , then we fail to reject null that is we support null and reject alternative. In our example P-value =0.2518 and significance level is given as 0.05. Here P-value is greater than 0.05 so we fail to reject the null hypotheses and it simply means that we are in favor of accepting null hypotheses. Decision : Fail to reject the null hypotheses. Conclusion : There is not sufficient evidence to conclude that the average number of sick day used per year by an employee is not equal to 10 days. 2.Here it is asked to construct 95% confidence interval for true mean. The formula for 99% confidence interval is as follows; Here we know all values except t'. We can easily find this value by using MS-EXCEL. To find this we need to know the two things; Level of confidence , c = 0.95 Degrees of freedom = n -1 = 8-1 = 7 The excel function to find critical value , t' is = TINV( 1 - c, degrees of freedom ) Thus , the critical value , t' = 2.365. We can illustrate this by curve as followsThe confidence interval is as follows; Confidence interval =(8.625 ± 2.365 3.1148)= ( 8.625 ± 2.604)=( 6.021 , 11.229) Thus , the 95% confidence interval for mean is ( lower bound = 6.021 , upper bound = 11.229).