Question

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: 11; 5; 15; 3; 10; 10; 8; 9. 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% level.

A. Note: If you are using a Student's t-distribution for the problem, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)
What is the test statistic? (If using the z distribution round your answers to two decimal places, and if using the t distribution round your answers to three decimal places.)

B. What is the p-value? (Round your answer to four decimal places.)

C. Construct a 95% confidence interval for the true mean. Label the point estimate and the lower and upper bounds of the confidence interval. (Round your answers to three decimal places.)

Answer 1

Here in this scenario our claim is that the mean number of sick days an employee takes per year is to be about 10.

Based on given sample size of 8 we have to test this claim. To test this claim we need to use t distribution i.e one sample t test because here the population standard deviations is unknown and sample size is less than 30.

So here t distribution is appropriate we cannot use z distribution here.

Further the test is performed at 0.05 level of significance as below,

First we need to compute the sample mean and sample Standerd deviation as below,

The sample size is n = 8n=8. The provided sample data along with the data required to compute the sample mean \bar XXˉ and sample variance s^2s2 are shown in the table below:

	X	X2
	11	121
	5	25
	15	225
	3	9
	10	100
	10	100
	8	64
	9	81
Sum =	71	725

The sample mean \bar XXˉ is computed as follows:

The t critical value is calculated using t table or using Excel at 7 degrees of freedom.

A) The t test Statistic tcal = -0.864.

B) The p value is 0.4161.

C) The 95% Confidence Interval for population mean is (5.797, 11.953).

Thank you.