In: Statistics and Probability
17)The cost of a daily newspaper varies from city to city. However, the variation among prices remains steady with a population standard deviation of $0.20. A study was done to test the claim that the mean cost of a daily newspaper is $1.00. Eleven costs yield a mean cost of $0.94 with a standard deviation of $0.18. Do the data support the claim at the 1% 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, however.)
Part (e)
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.)
z=
Part (g)
Sketch a picture of this situation. Label and scale the horizontal
axis and shade the region(s) corresponding to the
p-value.
we use t-test here because sample size is n=11. which is small so we prefer t-test here.
The provided sample mean is and the sample standard deviation is , and the sample size is .
(1) Null and Alternative Hypotheses
The following null and alternative hypotheses need to be tested:
This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used.
(2) Rejection Region
Based on the information provided, the significance level is , and the critical value for a two-tailed test is
The rejection region for this two-tailed test is .
(3) Test Statistics
The t-statistic is computed as follows:
(4) Decision about the null hypothesis
Since it is observed that , it is then concluded that the null hypothesis is not rejected.
Using the P-value approach: The p-value is p= 0.2948, and since , it is concluded that the null hypothesis is not rejected.
(5) Conclusion
It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean is different than 1, at the 0.01 significance level.
Confidence Interval
The 99% confidence interval is .
Graphically