STAT 500 Practice Problem Solutions Hypothesis Testing

STAT 500 Practice Problem Solutions – Hypothesis Testing 1. The benign mucosal cyst is the most common lesion of a pair of sinuses in the upper jawbone. In a random sample of 800 males, 35 persons were observed to have a benign mucosal cyst. a. Would it be appropriate to use a normal approximation in conducting a statistical test of the null hypothesis Ho: p ≥ .096 (the highest incidence in previous studies among males)? Explain. po = 0.096 and n = 800 Since npo = 800(0.096) = 76.9 and n(1- po) = 800(0.904) = 723.1 are ate least 5, we can use the normal approximation, Z, method.

b. Conduct a statistical test of the research hypothesis Ha: p < 0.096 by computing the p-value manually and drawing a conclusion using the p-value approach at a 1% Type I error rate. Ho: p ≥ 0.096 (or one can write Ho: p = 0.096) and Ha: p < 0.096 �= This is a one-sided left-tailed test. 𝒑𝒑 Test Statistic: Z * =

pˆ − p0 = p0 (1 − po ) n

𝟑𝟑𝟑𝟑

𝟖𝟖𝟖𝟖𝟖𝟖

= 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎

0.044 − 0.096 = −5.02 0.096(1 − 0.096) 800

p-value = P(Z < Z*) = P(Z < -5.02). From standard normal table, we can see that the pvalue would be close to 0.000 With our p-value being less than α = 0.01, we reject the null hypothesis. We have statistical evidence to conclude the incidences of benign mucosal cysts among males is significantly less that 0.096

c. What is the rejection region for this test? With a left-tailed test and α = 0.01, the critical value would be -Zα = -Z0.01 = -2.33 This makes the rejection region Z* ≤ -2.33

d. Use Minitab to verify your results. 1

Test and CI for One Proportion Test of p = 0.096 vs p < 0.096 Sample 1

X 35

N 800

Sample p 0.043750

95% Upper Bound 0.055645

Z-Value -5.02

P-Value 0.000

Using the normal approximation.

2. Some mushrooms were found in a forest. You do not know much about whether those are poisonous. There are two hypotheses: • •

The mushrooms are poisonous and cannot be eaten The mushrooms are not poisonous and can be eaten

How will you set up the hypotheses? Give a brief explanation. Ho: The mushrooms are poisonous and cannot be eaten. Ha: The mushrooms are not poisonous and can be eaten Explanation: We should set the hypotheses in a way so that the Type I error is more serious. Here we put “the mushrooms are poisonous and cannot be eaten” in Ho since a false rejection of Ho is a more serious error than failing to reject Ho. That is, the error to say the mushrooms are poisonous when they are not is a more serious error than to say the mushrooms are not poisonous when they are.

3. A dealer in recycled paper places empty trailers at various sites. The trailers are gradually filled by individuals who bring in old newspapers and magazines, and are picked up on several schedules. One such schedule involves pickup every second week. This schedule is desirable if the average amount of recycled paper is more than 1,600 cubic feet per 2-week period. The dealer’s records for eighteen 2-week periods show the following volumes (in cubic feet) at a particular site (recycled_paper.txt) where x¯ = 1,718.3 and s = 137.8.

a. By hand, compute a 95% confidence interval of μ and provide an interpretation of your interval.

2

With sample size of 18 the DF are 18 – 1 = 17. For a 95% confidence interval the α is 0.05, thus from the t-table under α/2 = 0.025 we get a t-value of 2.11

x ±t*

s 137.8 = 1718.3 ± 2.11 * = 1718.3 ± 2.11 * 32.5 = 1718.3 ± 68.6 n 18

1649.7 ≤ µ ≤ 1786.9 We are 95% confident that the average amount of recycled paper picked up every two weeks is from 1649.7 cubic feet to 1786.9 cubic feet.

b. Is there strong evidence that μ is greater than 1,600? Conduct the test by hand using the pvalue approach with a 5% level of significance. The test calls for a one-sided, right-tailed test: Ho: µ ≤ 1600 versus Ha: µ > 1600 (note that Ho could be written as µ =1600). With sample size of only 18 the data would have to be approximately normal for the tmethod to be appropriately used. We check that with Minitab later. Test Statistic: t* =

X − µ 0 1718.3 − 1600 118.3 = = = 3.64 S 137.8 32.5 n 18

p-value = P(t > t*) = p(t > 3.64) at 17 degrees of freedom. From the t-table in the row for 17 degrees of freedom we find that 3.64 falls between 3.646 and 3.965 which correspond to right-tail probabilities of 0.001 and 0.0005, respectively. As a result, the p-value for our test statistic of 3.64 falls between 0.0005 and 0.001. Since this range of our p-value is less than our α value of 0.05, we reject the null hypothesis. We have statistical evidence to conclude that the average amount of recycled paper picked up every two weeks exceeds 1600 cubic feet. c. Use Minitab to verify your results in parts a and b, and to check normality of the data. Confidence Interval:

3

One-Sample T N 18

Mean 1718.3

StDev 137.8

SE Mean 32.5

95% CI (1649.8, 1786.8)

Hypothesis Test: (note that Minitab p-value of 0.001 falls within our p-value range). One-Sample T Test of μ = 1600 vs > 1600 N 18

Mean 1718.3

StDev 137.8

SE Mean 32.5

95% Lower Bound 1661.8

T 3.64

P 0.001

Normality Check: All points within confidence bounds so okay to assume normality. Probability Plot of ft3 Normal - 95% CI

99

Mean 1718 StDev 137.8 N 18 AD 0.169 P-Value 0.921

95 90

Percent

80 70 60 50 40 30 20 10 5

1

1200

1400

1600

1800

2000

2200

ft3

4. The undergraduate GPA of 18 students from a large MBA class of 800 students is selected. The data are given as (mba_student_gpa.txt). Use the data in the file above and Minitab to test the research hypothesis that the average undergraduate GPA of the MBA class differs from 3.5. Use the p-value approach to perform the 4

test at a default level of significance. Remember to check normality of the data. Also, use the ttable to find the p-value range based on the test statistic produced in Minitab. Recall that the default level of significance is α of 0.05. The hypothesis test calls for a twosided, two-tailed test. Ho: µ = 3.5 versus Ha: µ ≠ 3.5 One-Sample T: gpa Test of μ = 3.5 vs ≠ 3.5 Variable gpa

N 18

Mean 3.5389

StDev 0.3421

SE Mean 0.0806

95% CI (3.3688, 3.7090)

T 0.48

P 0.636

T-table p-value range: With the alternative being a two-tailed test, the p-value would be found by: 2 x P(t ≥ |t*|) of 2 x P(t ≥ |0.48|) With sample size of 18 the degrees of freedom are 17. Looking across that t-table in that row of 17 degrees of freedom, we see that the test statistic falls between 0.257 and 0.689 which correspond to right-tail probabilities of 0.4 and 0.25, respectively. The range of the p-value would be twice these values (remember two-sided so we double these right-tail probabilities from the table). Therefore, the p-value range based on the t-table would be from 0.5 to 0.8 which does include the output p-value of 0.636 The software will correctly calculate the p-value based on what alternative you provide. Check Normality: All points within confidence bounds so okay to assume normality.

5

Probability Plot of gpa Normal - 95% CI

99

Mean 3.539 StDev 0.3421 N 18 AD 0.559 P-Value 0.127

95 90

Percent

80 70 60 50 40 30 20 10 5

1

2.5

3.0

3.5

4.0

4.5

gpa

6