It is reported that the average woman's hair grows a quarter of an inch per month. A new vitamin company believes that women who take its vitamin daily will have increased hair growth. Before putting its product on the market, the company plans on running a test on the effectiveness of the vitamins. Which null and alternative hypotheses should the company use for this study? (2 points)

	H0: x = 0.25, Ha: x 0.25
	H0: x = 0.25, Ha: x > 0.25
	H0: = 0.25, Ha: 0.25
	H0: = 0.25, Ha: > 0.25
	H0: = 0.25, Ha: < 0.25

2.

(07.01) You are conducting a significance test of H₀: = 5 against H_a: > 5. After checking the conditions are met from a simple random sample of 30 observations, you obtain t = 2.35. Based on this result, describe the p-value. (2 points)

	The p-value falls between 0.15 and 0.2.
	The p-value falls between 0.025 and 0.05.
	The p-value falls between 0.01 and 0.02.
	The p-value falls between 0.005 and 0.01.
	The p-value is less than 0.005.

3.

(07.01) The weight of baby seals is believed to be Normally distributed, with a mean of 69.5 pounds. The average weight of a random sample of 25 baby seals is found to be 69.75 pounds, with a standard deviation of 4.2 pounds. What is the standard error of the mean? (2 points)

	0.84
	0.30
	0.25
	0.17
	We are unable to determine the standard error of the mean because we do not know .

4.

(07.01) Suppose you carry out a significance test of H₀: = 7 versus H_a: 7 based on sample size n = 28 and obtain t = 1.45. Find the p-value for this test. What conclusion can you draw at the 5% significance level? Explain. (2 points)

	The p-value is 0.0793. We reject H0 at the 5% significance level because the p-value 0.0793 is greater than 0.05.
	The p-value is 0.0793. We fail to reject H0 at the 5% significance level because the p-value 0.0793 is greater than 0.05.
	The p-value is 0.4207. We fail to reject H0 at the 5% significance level because the p-value 0.4207 is greater than 0.05.
	The p-value is 0.1586. We fail to reject H0 at the 5% significance level because the p-value 0.1586 is greater than 0.05.
	The p-value is 0.1586. We reject H0 at the 5% significance level because the p-value 0.1586 is greater than 0.05.

5.

(07.01) Bottles of grape soda are assumed to contain 300 milliliters of soda. There is some variation from bottle to bottle because the filling machine is not perfectly precise. Usually, the distribution of the contents is approximately Normal. An inspector measures the contents of nine randomly selected bottles from one day of production. The results are 298.7, 299.4, 301.9, 301.3, 300.9, 300.6, 301.7, 300.4, and 301.5 milliliters. Do these data provide convincing evidence at = 0.05 that the mean amount of soda in all the bottles filled that day differs from the target value of 300 milliliters? (2 points)

	Because the p-value of 0.0414 is greater than the significance level of 0.05, we fail to reject the null hypothesis. We conclude the data provide insufficient evidence that the mean amount of soda in all the bottles filled that day differs from the target value of 300 milliliters.
	Because the p-value of 0.0414 is greater than the significance level of 0.05, we reject the null hypothesis. We conclude the data provide convincing evidence that the mean amount of soda in all the bottles filled that day differs from the target value of 300 milliliters.
	Because the p-value of 0.0828 is greater than the significance level of 0.05, we reject the null hypothesis. We conclude the data provide convincing evidence that the mean amount of soda in all the bottles filled that day differs from the target value of 300 milliliters.
	Because the p-value of 0.0828 is greater than the significance level of 0.05, we fail to reject the null hypothesis. We conclude the data provide insufficient evidence that the mean amount of soda in all the bottles filled that day differs from the target value of 300 milliliters.
	Because the p-value of 1.9819 is greater than the significance level of 0.05, we fail to reject the null hypothesis. We conclude the data provide insufficient evidence that the mean amount of soda in all the bottles filled that day differs from the target value of 300 milliliters.