A recent newspaper article reported that the average leisure time for males is less than 20 hours per week. A random sample of 30 men was selected and found to have a mean leisure time of 16.9 hours per week with a standard deviation of 6 hours per week. Use a 5 percent level of significance to test the validity of this claim.

1)State the data type and parameter of interest.

Data type:

Parameter of interest:

2)State the Null and Alternative hypotheses

H_o:

H_A:

3)State the test statistic, its distribution and the level of significance. State the required condition

Test statistic:

Distribution of the standardized test statistic:

Level of significance:

Required conditions:

4) State the decision rule (in terms of the p-value):

5) Obtain the p-value (from the appropriate output) and compare with the decision rule:

6) Derive your conclusion regarding the claim in the newspaper article report.

Transcribed Image Text:

t-Test of a Mean z-Test of a Mean Sample mean CSample standard deviatid Sample size Hypothe sized mean Alpha Sample mean Population standard deviation Sample size Hypothesized mean Alpha z Stat P(Z<=z) one-tail z Critical one-tail P(Z<=z) two-tail z Critical two-tail t Stat P(T<=t) one-tail t Critical one-tail 16.9 6 16.9 -2.83 -2.83 6 0.0042 0.0023 30 1.6449 0.0047 30 1.6991 20 P(T<-t) two-tail t Critical two-tail 20 0.0084 0.05 2.0452 0.05 1.9600 t-Test of a Mean z-Test of a Mean Sample mean CSample standard deviatid Sample size Hypothe sized mean Alpha Sample mean Population standard deviation Sample size Hypothesized mean Alpha z Stat P(Z<=z) one-tail z Critical one-tail P(Z<=z) two-tail z Critical two-tail t Stat P(T<=t) one-tail t Critical one-tail 16.9 6 16.9 -2.83 -2.83 6 0.0042 0.0023 30 1.6449 0.0047 30 1.6991 20 P(T<-t) two-tail t Critical two-tail 20 0.0084 0.05 2.0452 0.05 1.9600

Answer rating: 100% (QA)

1 Data Type Continuous Parameter of interest mean leisure time 2 ... View the full answer