# Hypothesis Testing, Levels of Significance and P-Values

26. According to a dietary study, a high sodium intake may be related to ulcers, stomach cancer, and migraine headaches. The human requirement for salt is only 220 milligrams per day, which is surpassed in most single servings of ready-to-eat cereals. If a random sample of 20 similar servings of certain cereal has a mean sodium content of 244 milligrams and a standard deviation of 24.5 milligrams, does this suggest at the 0.05 level of significance that the average sodium content for a single serving of such cereal is greater than 220 milligrams? Assume the distribution of sodium contents to be normal.

28. According to Chemical Engineering an important property of fiber is its water absorbency. The average percent absorbency of 25 randomly selected pieces of cotton fiber was found to be 20 with a standard deviation of 1.5. A random sample of 25 pieces of acetate yielded an average percent of 12 with a standard deviation of 1.25. Is there strong evidence that the population mean percent absorbency for cotton fiber is significantly higher than the mean for acetate? Assume that the percent absorbency is approximately normally distributed and that the population variances in percent absorbency for the two fibers are the same. Use a significance level of 0.05.

30. A random sample of size n1 = 25, taken from a normal population with a standard deviation ơ1 = 5.2, has a mean x1 = 81. A second random sample of size n2 = 36, taken from a different normal population with a standard deviation ơ2 = 3.4, has a mean x2 = 76. Test the hypothesis that μ1 = μ2 against the alternative μ1 does not equal μ2. Quote a P-value in your conclusion.

38. A UCLA researcher claims that the average life span of mice can be extended by as much as 8 months when the calories in their food are reduced by approximately 40% from the time they are weaned. The restricted diets are enriched to normal levels by vitamins and protein. Suppose that a random sample of 10 mice are fed a normal diet and live an average life span of 32.1 months with a standard deviation of 3.2 months, while a random sample of 15 mice are fed the restricted diet and live an average life span of 37.6 months with a standard deviation of 2.8 months. Test the hypothesis at the 0.05 level of significance that the average life span of mice on this restricted diet is increased by 8 months against the alternative that the increase is less than 8 months. Assume the distributions of life spans for the regular and restricted diets are approximately normal with equal variances.

40. In a study conducted at the Virginia Polytechnic Institute and State University, the plasma ascorbic acid levels of pregnant women were compared for smokers versus nonsmokers. Thirty0two women in the last 3 months of pregnancy, free of major health disorders, and ranging in age from 15 to 32 years, were selected for the study. Prior to the collection of 20ml of blood, the participants were told to avoid breakfast, forgo their vitamin supplements, and avoid foods high in ascorbic acid content. From the blood samples, the following plasma ascorbic acid values of each subject were determined in milligrams per 100 milliliters:

Plasma Ascorbic Acid Values

Nonsmokers Smokers

0.97 1.16 0.48

0.72 0.86 0.71

1.00 0.85 0.98

0.81 0.58 0.68

0.62 0.57 1.18

1.32 0.64 1.36

1.24 0.98 0.78

0.99 1.09 1.64

0.90 0.92

0.74 0.78

0.88 1.24

0.94 1.18

Is there sufficient evidence to conclude that there is a difference between plasma ascorbic acid levels of smokers and nonsmokers? Assume that the two sets of data came from normal populations with unequal variances. Use a P-value.

48. If the distribution of life spans in Exercise 21 is approximately normal, how large a sample is required in order that the probability of committing a type II error be 0.1 when the true mean is 35.9 months? Assume that ơ = 5.8 months.

Exercise 21: In a research report, it is claimed that mice with an average life span of 32 months will live to be about 40 months old when 40% of the calories in their food are replaced by vitamins and protein. Is there any reason to believe that μ < 40 if 64 mice that are placed on this diet have an average life of 38 months with a standard deviation of 5.8 months? Use a P-value in your conclusion.

Answer: The hypotheses are H0: μ = 40 months and H1: μ < 40 months

Now, z = (38-40) / 5.8√64 = -2.76, and P-value = P(Z < -2.76) = 0.0029

Decision: reject H0.

50. How large should the samples be in Exercise 31 if the power of our test is to be 0.95 when the true difference between thread types A and B is 8 kilograms?

Exercise 31: A manufacturer claims that the average tensile strength of thread A exceeds the average tensile strength of thread B by at least 12 kilograms. To test his claim, 50 pieces of each type of thread are tested under similar conditions. Type A thread had an average tensile strength of 86.7 kilograms with a standard deviation of 6.28 kilograms, while type B thread had an average tensile strength of 77.8 kilograms with a standard deviation of 5.61 kilograms. Test the manufacturer's claim using a 0.05 level of significance.

Answer: The hypotheses are H0: μA - μB = 12 kilograms and H1: μA - μB > 12 kilograms

α = 0.1. Degrees of freedom = 7.38, hence we use 7 degrees of freedom with the critical region t > 2.998.

Computation: t = (110-97.4)-10 / √78.800/5 + 913.333/7 = 0.22

Decision: Fail to reject H0.

54. Nine subjects were used in an experiment to determine if an atmosphere involving exposure to carbon monoxide has an impact on breathing capability. The subjects were exposed to breathing chambers, one of which contained a high concentration of CO. Several breathing measures were made for each subject for each chamber. The subjects were exposed to the breathing chambers in random sequence. The data give the breathing frequency in number of breaths taken per minute. Make a one-sided test of the hypothesis that mean breathing frequency is the same for the two environments. Use α = 0.05. Assume that breathing frequency is approximately normal.

Subject With CO Without CO

1 30 30

2 45 40

3 26 25

4 25 23

5 34 30

6 51 49

7 46 41

8 32 35

9 30 28

#### Solution Summary

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