Question

When the 127 workers aged 18-25 in the 2008 General Social Survey were asked how many hours they worked in the previous week, the mean was 37.47 with a standard deviation of 13.63. Test that the mean hours worked per week for this sample is different from a population mean of 40 hours worked per week at the 10% significance level.

a) What is the appropriate test to conduct.

-One-sample t-test for means

One-sample t-test for proportions

One-sample z-test for proportions

One-sample z-test for means

b) Choose all the basic assumptions that are satisfied from the following statements.

-There is a large enough sample size (n>30).

-These data are normally distributed.

-These data were obtained from a random sample of voters.

c) Based on the significance level at which you are testing, what is (are) the critical value(s) for the test?

d) Calculate the appropriate test statistic, showing all your work. Enter the standard error you calculated in the box below (without units).

e) Calculate the appropriate test statistic, showing your work. Enter the test statistic you calculated in the box below (without units).

f) Calculate the corresponding p-value from the appropriate table.

g) What are the null and alternative hypotheses?

h) What conclusions can you draw from the hypothesis test? Be sure to comment on evidence from both the test statistic and p-value

i) Construct a 95% confidence interval for the estimate of the mean difference. Enter the lower bound of the interval you calculated in the box below (without units).

j) Construct a 95% confidence interval for the estimate of the mean difference. Enter the upper bound of the interval you calculated in the box below (without units).

k) How would you interpret the confidence interval?

l) What connections can you draw between the confidence interval and the hypothesis test?

Answer 1

a)

-One-sample t-test for means

b)

-There is a large enough sample size (n>30).

-These data are normally distributed.

c)

Level of Significance ,    α =    0.10
Sample Size ,   n =    127
degree of freedom=   DF=n-1=   126

critical t value, t* =    ±   1.6570   [Excel formula =t.inv(α/no. of tails,df) ]      

d)

sample std dev ,    s =    13.6300          
Sample Size ,   n =    127          
Sample Mean,    x̅ =   37.4700          
                  
degree of freedom=   DF=n-1=   126          
                  
Standard Error , SE = s/√n =   13.6300   / √    127   =   1.2095

e)

t-test statistic= (x̅ - µ )/SE = (   37.470   -   40   ) /    1.2095   =   -2.09

f)

p-Value   =   0.0385

g)

Ho :   µ =   40
Ha :   µ ╪   40

h)

decsion : |test stat | > |critical value| , reject Ho

Conclusion: There is enough evidence to say that the mean hours worked per week for this sample is different from a population mean of 40 hours worked per week at the 10% significance level.

Decision:   p-value<α, Reject null hypothesis   
Conclusion: There is enough evidence to say that the mean hours worked per week for this sample is different from a population mean of 40 hours worked per week at the 10% significance level.

i)

Level of Significance ,    α =    0.05          
degree of freedom=   DF=n-1=   126          
't value='   tα/2=   1.9790   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   13.6300   / √   127   =   1.209467
margin of error , E=t*SE =   1.9790   *   1.20947   =   2.393499
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    37.47   -   2.393499   =   35.076501

j)

Interval Upper Limit = x̅ + E =    37.47   -   2.393499   =   39.863499

k)

95%   confidence interval is (   35.08   < µ <   39.86   )

40 does not lie in the CI, so reject Ho

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