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; 6; 13; 3; 11; 9; 8; 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% level.

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.)

1. State the distribution to use for the test. (Enter your answer in the form z or t_df where df is the degrees of freedom.)

2. 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.)

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

Explain what the p-value means for this problem. If

H₀ 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

H₀ 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.    If

H₀ 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

H₀ 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.

4. 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.)

Answer 1

1)

t distribution with df = 7

.......

2)

Ho :   µ =   10                  
Ha :   µ ╪   10       (Two tail test)          
                          
Level of Significance ,    α =    0.05                  
sample std dev ,    s = √(Σ(X- x̅ )²/(n-1) ) =   3.1820                  
Sample Size ,   n =    8                  
Sample Mean,    x̅ = ΣX/n =    8.8750                  
                          
degree of freedom=   DF=n-1=   7                  
                          
Standard Error , SE = s/√n =   3.1820   / √    8   =   1.1250      
t-test statistic= (x̅ - µ )/SE = (   8.875   -   10   ) /    1.1250   =   -1.000

3)

p-Value   =   0.3506

H0 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.

.................

Level of Significance ,    α =    0.05          
degree of freedom=   DF=n-1=   7          
't value='   tα/2=   2.3646   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   3.1820   / √   8   =   1.125000
margin of error , E=t*SE =   2.3646   *   1.12500   =   2.660202
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    8.88   -   2.660202   =   6.214798
Interval Upper Limit = x̅ + E =    8.88   -   2.660202   =   11.535202
95%   confidence interval is (   6.215   < µ <   11.535   )

.............

THANKS

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