07.01

According to a study, the average man's facial hair grows a half an inch per month. A new drug company believes that men who take its drug daily will have slower facial hair growth. Before putting its product on the market, the company plans on running a test of the effectiveness of the drug. Which null and alternative hypotheses should the company use for this study? (4 points)

H₀: x? = 0.5, H_a: x? ? 0.5

H₀: x? = 0.5, H_a: x? > 0.5

H₀: ? = 0.5, H_a: ? ? 0.5

H₀: ? = 0.5, H_a: ? > 0.5

H₀: ? = 0.5, H_a: ?

Suppose you carry out a significance test of H₀: ? = 8 versus H_a: ? > 8 based on sample size n = 25 and obtain t = 2.15. Find the p-value for this test. What conclusion can you draw at the 5% significance level? Explain. (4 points)

The p-value is 0.02. We reject H₀at the 5% significance level because the p-value 0.02 is less than 0.05.

The p-value is 0.02. We fail to reject H₀at the 5% significance level because the p-value 0.02 is less than 0.05.

The p-value is 0.48. We reject H₀at the 5% significance level because the p-value 0.48 is greater than 0.05.

The p-value is 0.48. We fail to reject H₀at the 5% significance level because the p-value 0.48 is greater than 0.05.

The p-value is 0.52. We fail to reject H₀at the 5% significance level because the p-value 0.52 is greater than 0.05.

The average low-density lipoprotein (LDL) cholesterol level of adult males is 142 milligrams per deciliter. You want to determine whether the true mean male LDL cholesterol level is higher than the average. You decide to pull a random sample of 10 adult males and measure their LDL cholesterol level. You find the sample mean is 145.3 milligrams per deciliter and the standard deviation is 5.39 milligrams per deciliter. What are the test statistic and the p-value? (2 points)

The test statistic is 3.3 and the p-value is 0.0422.

The test statistic is 0.0422 and the p-value is 3.3.

The test statistic is 1.94 and the p-value is 0.0424.

The test statistic is 0.0422 and the p-value is 1.94.

The test statistic is 1.7 and the p-value is 1.94.

A fast-food chain claims one large order of its fries weighs 170 grams. Joe thinks he is getting less than what the restaurant advertises. He weighs the next 12 random orders of fries before he eats them and finds the sample mean is 165.9 grams and the standard deviation is 11.98 grams. What conclusion can be drawn at ? = 0.10? (2 points)

There is not sufficient evidence to prove the fast-food chain advertisement is true.

There is sufficient evidence to prove the fast-food chain advertisement is false.

Joe has sufficient evidence to reject the fast-food chain's claim.

Joe does not have sufficient evidence to reject the fast-food chain's claim.

There is not sufficient data to reach any conclusion.

A study is being conducted to compare vitamin C and zinc to determine which is better at fighting colds. Customers believe vitamin C is better at fighting colds. What are the appropriate hypotheses for this testing scenario? Let ?_cequal the mean of the effectiveness of vitamin C and ?_zequal the mean of the effectiveness of zinc. (2 points)

H₀: ?_c- ?_z= 0

H_a: ?_c- ?_z> 0

H₀: ?_c- ?_z= 0

H_a: ?_c- ?_z

H₀: ?_c- ?_z= 0

H_a: ?_c- ?_z? 0

H₀: x?_c- x?_z= 0

H_a: x?_c- x?_z> 0

H₀: x?_c- x?_z= 0

H_a: x?_c- x?_z? 0

The owners of Shaving Ice are concerned that many of their customers are starting to purchase their icy treats from Cold as Ice because of its new shaved-ice syrup, which has fewer grams of sugar than Shaving Ice. A random sample of 100 strawberry shaved ices from Shaving Ice found a mean of 22 grams of sugar and a standard deviation of 3.2 grams. A random sample of 100 strawberry shaved ices from Cold as Ice found a mean of 18 grams of sugar and a standard deviation of 2.1 grams. Which of the following formulas gives a 99% confidence interval for the difference in mean grams of sugar between a Shaving Ice strawberry shaved ice and a Cold as Ice strawberry shaved ice? (2 points)

4+2.63 3.2 2.1 + V100 100 O 4+2.63 3.22 2.12 + 99 99 O 3.22 2.12 4+2.576 + 99 99 O 3.22 2.12 4+2.63 + 100 100 O 3.22 2.12 4+2.576 + 100 100