Question

The cost of a daily newspaper varies from city to city. However, the variation among prices remains steady with a standard deviation of $0.20. A study was done to test the claim that the mean cost of a daily newspaper is $1.00. Ten costs yield a mean cost of $0.97 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.)

1.Sketch a picture of this situation. Label and scale the horizontal axis and shade the region(s) corresponding to the p-value.

2. 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 lower and upper bounds to two decimal places.)

Answer 1

Q1) As we are testing here whether the mean is 1, therefore the null and the alternative hypothesis here are given as:

The test statistic here is computed as:

For n - 1 = 9 degrees of freedom, we get the p-value from the t distribution tables as:
p = 2P( t₉ < -0.5270) = 2*0.3055 = 0.6110

As the p-value here is 0.6110 > 0.01 which is the level of significance here, therefore the test is not significant here and we cannot reject the null hypothesis here. Therefore we dont have sufficient evidence here that mean cost is $1.

b) For 9 degrees of freedom, we have from t distribution tables here:
P( -2.262 < t₉ < 2.262) = 0.95

Therefore the confidence interval here is obtained as:

This is the required 95% confidence interval here.