Statistics AP/GT - UH

Math 3339

Homework 6 (Chapters 8 & 10)

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1. *Water samples are taken from water used for cooling as it is being discharged from a power plant into a river. It has been determined that as long as the mean temperature of the discharged water is at most 150 F, there will be no negative effects on the river's ecosystem. To investigate whether the plant is in compliance with regulations that prohibit a mean discharge-water temperature above 150 , 50 water sample will be tank at randomly selected times, and the temperature of each sample recorded.

a. Determine an appropriate null and alternative hypothesis for this test. b. In the context of this situation, describe type I and type II errors.

a) H0: ? = 150, Ha: ? > 150 b) Type I, if it is determined that the mean discharge-water temperature is above 150 when it is not.

Type II, if it is determined that the mean discharge-water temperature is not above 150, when it really is greater than 150.

2. Section 8.4.2, Problem 1

For each of the following scenarios state whether H0 should be rejected or not. State any assumptions that you make beyond the information that is given.

(a) H0 : ? = 4, H1 : ? 4, n = 15, = 3.4, S = 1.5, = .05. (b) H0 : ? = 21, H1 : ? < 21, n = 75, = 20.12, S = 2.1, = .10.

(c) H0 : ? = 10, H1 : ? 10, n = 36, p-value = 0.061.

a) Assumption is that we have a normal distribution since the sample size is less than 30.

Test

statistic

=

t

=

3.4-4 1.5/15

=

-1.54919

p-value = P(T -1.54919 or T 1.54919) = 2*pt(-1.54919,14) = 0.1436

Since the p-value is greater than we FAIL TO REJECT H we FAIL TO REJECT H0.

b) Since the sample size is greater than 75 we can apply the central limit theorem and say that we have a Normal distribution

Test

statistic

=

t

=

20.12-21 2.1/75

=

-3.629

p-value = P(T -3.629) = pt(-3.629,74) = 0.00026

Since the p-value is less than we REJECT H0

c) If we assume = 0.05, we would FAIL TO REJECT H0.

Problems came from Devore, Jay and Berk, Kenneth, Modern Mathematical Statistics with Applications, Thomson Brooks/Cole, 2007.

3. Section 8.4.2, Problem 2. Use the "test.vs.grade" data and test the null hypothesis that the mean test score for the population is 70 against the alternative that it is greater than 70. Find a p-value and state your conclusion if = 0.05. Repeat for the null hypothesis ? = 75.

H0: ? = 70 and Ha: ? > 70

> t.test(test_vs_grade$Test,mu = 70, alternative = "greater")

One Sample t-test

data: test_vs_grade$Test

t = 4.786, df = 158, p-value = 1.939e-06

alternative hypothesis: true mean is greater than 70

95 percent confidence interval:

74.35786

Inf

sample estimates:

mean of x

76.66038

p-value 0, RH0, there is very strong evidence that the mean test score for the population is significantly greater than 70.

H0: ? = 75 and Ha: ? > 75

> t.test(test_vs_grade$Test,mu = 75, alternative = "greater")

One Sample t-test

data: test_vs_grade$Test

t = 1.1931, df = 158, p-value = 0.1173

alternative hypothesis: true mean is greater than 75

95 percent confidence interval:

74.35786

Inf

sample estimates:

mean of x

76.66038

p-value = 0.1173, FRH0, there is no evidence that the mean test score for the population is significantly greater than 75.

Problems came from Devore, Jay and Berk, Kenneth, Modern Mathematical Statistics with Applications, Thomson Brooks/Cole, 2007.

4. A sample of 12 radon detectors of a certain type was selected, and each was exposed to 100 pCI/L of radon. The resulting readings were as follows: 105.6, 90.9, 91.2, 96.9, 96.5, 91.3, 100.1, 105.0, 99.6, 107.7, 103.3, 92.4

Does this data suggest that the population mean reading under these conditions differs from 100? Set up an appropriate hypothesis test to answer this question. H0: ? = 100, Ha: ? 100

> radon=c(105.6, 90.9, 91.2, 96.9, 96.5, 91.3, 100.1, 105.0, 99.6, 107.7, 103.3, 92.4) > t.test(radon,mu=100)

One Sample t-test data: radon t = -0.92138, df = 11, p-value = 0.3766 alternative hypothesis: true mean is not equal to 100 95 percent confidence interval:

94.49322 102.25678 sample estimates: mean of x

98.375

p-value = 0.3766, FRH0, there is no evidence that the mean readings differs from 100.

Problems came from Devore, Jay and Berk, Kenneth, Modern Mathematical Statistics with Applications, Thomson Brooks/Cole, 2007.

5. *For healthy individuals the level of prothrombin in the blood is approximately normally distributed with

mean 20 mg/100 mL and standard deviation 4 mg/100 mL. Low levels indicate low clotting ability. In studying

the effect of gallstones on prothrombin, the level of each patient in a sample is measured to see if there is a

deficiency. Let ? be the true average level of prothrombin for gallstone patients.

a. What are the appropriate null and alternative hypotheses? b. Let denote the sample average level of prothrombin in a sample of n = 20 randomly selected gallstone

patients. Consider the test procedure with test statistic and rejection region 17.92. What is the probability distribution of the test statistic when H0 is true (i.e. determine center, spread, and shape of )? What is the probability of a type I error for this test procedure? c. What is the probability distribution of the test statistic, when ? = 16.7? Using the test procedure of part (b), what is the probability that gallstone patients will be judged not deficient in prothrombin, when in fact

? = 16.7 (a type II error)?

d. How would you change the test procedure of part (b) to obtain a test with significance level 0.05? What

impact would this change have on the error probability of part (c)?

a. H0: ? = 20 mg/100 mL Ha: ? < 20 mg/100 mL

b.

If

H0

is

true

then

the

mean

(center)

is

20

mg/100

mL

and

the

standard

deviation

of

is

4 20

=

0.8944,

since we know the population standard deviation we can use the Normal distribution. Type 1 error is

rejecting the null hypothesis when in fact it is true. So the probability of a type 1 error is the

probability that the sample mean is in the rejection region when the true mean is really 20.

P(Type 1 error) = P( 17.92) = pnorm(17.92,20,0.8944) = 0.01

c. What changes is the mean at 16.7, the standard deviation stays the same at 0.8944 and it is still a Normal distribution. Type 2 error is failing to reject the null hypothesis when in fact it is false. So in this example, the probability of type 2 error is the probability that the sample mean is not in the rejection region when the true population mean is 16.7. P(Type 2 error) = P( > 17.92) = 1 ? pnorm(17.92,16.7,0.8944) = 0.0863

d. If we change to = 0.05, the rejection region becomes qnorm(0.05,20,0.8944) = 18.52884. That is if

we get any sample mean less than 18.52884 we would reject the null hypothesis. So the probability of a type 2 error becomes P( > 18.52884) = 1-pnorm(18.52884,16.7,0.8944) = 0.0204 which is less than in part (c).

Problems came from Devore, Jay and Berk, Kenneth, Modern Mathematical Statistics with Applications, Thomson Brooks/Cole, 2007.

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