Question

1.

According to the U.S. Energy Information Administration, the average household expenditure for natural gas was $679 in the winter of 2013. Suppose that a random sample of 50 customers shows a mean expenditure of $712 and assume that the population standard deviation is $80. Test whether the population mean expenditure is greater than $679 using level of significance α = 0.05. State the p-value and conclusion.

a) p-value = 0.0035; reject H0: there is evidence at level of significance α = 0.05 that the mean expenditure is greater than $679. b) p-value = 0.9982; reject H0: there is evidence at level of significance α = 0.05 that the mean expenditure is greater than $679. c) p-value = 0.0018; do not reject H0: there is insufficient evidence at level of significance α = 0.05 that the mean expenditure is greater than $679. d) p-value = 0.0018; mean expenditure is greater than $679.

2.

Which of the following alternative hypotheses best fits the following scenario? The Bureau of Justice Statistics reports that in 2000, prison sentences for drug offenses had a mean length of 47 months. After an extended campaign to discourage such offenders by increasing the penalty for such crimes, a random sample is taken and finds a mean stay of 54 months. Has there been a significant increase in the length of stay?

a) Ha: μ < 47 b) Ha: μ ≠ 47 c) Ha: μ > 47

3.	In a recent hypothesis test, the null hypothesis was NOT rejected. Which of the following types of error could have been made, and what is the probability of such an error?
	a) Type I error with probability of α b) Type II error with probability of β c) Type I error with probability of β d) Type II error with probability of α

4.

In 2006 a study found that 1% of American Internet users who were married or in a long-term relationship met on a blind date or through a dating service. A survey taken this year found 12 such couples in a random sample of 500 such Internet users. Has there been an increase in the proportion of " successful" matches via blind dates and dating services?

a) Yes, the sample gives P(Z > 3.15) = 0.0008, which provides extremely strong evidence of an increase in the proportion of " successful" such matches. b) No, the sample gives P(Z > –3.15) = 0.9992, which does not provide sufficient evidence to claim an increase in the proportion of " successful" such matches. c) No, the sample gives P(Z > 0.0063) = 0.4997, which does not provide sufficient evidence to reject the null hypothesis. d) Yes, the sample gives P(Z > 2.05) = 0.02, which provides solid evidence of an increase in the proportion of " successful" such matches.

Answer 1

1) Here we have to test that

where

n = 50

Sample mean =

Population standard deviation =

Here population standard deviation is known so we use z test.

Test statistic:

z = 2.92               (Round to 2 decimal)

Test statistic = 2.92

alpha = 0.05

Test is one tailed (right tailed) test.

P value = P(z > 2.92)

             = 1 - P(z < 2.92)

             = 1 - 0.9982                    (From statistical table of z values)

            = 0.0018

P value = 0.0018

Here p value < alpha

So we reject H0.

p-value = 0.0018; mean expenditure is greater than $679.