Homework Help 1234

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STAT 200 QUIZ 3 Solutions Section 6380 Fall 2013

• The quiz covers Chapters 7 and 8.

1. (15 points) Mimi was the 5th seed in 2012 UMUC Tennis Open that took place in August. In this tournament, she won 75 of her 100 serving games. Based on UMUC Sports Network, she wins 80% of the serving games in her 5-year tennis career.

(a) (3 pts) Find a 95% confidence interval estimate of the proportion of serving games Mimi won. (Show work)

The confidence interval estimate is about a proportion, so we will use Formula 7-1 on page 333 of the textbook for the margin of error E.

α = 1 – 0.95 = 0.05, hence α/2 = 0.025. The corresponding critical value z α/2 = z 0.025 =1.96.

Sample proportion 75ˆ 0.75

100 p = = , and

ˆ ˆ1 1 0.75 0.25q p= − = − = .

Margin of error / 2 ˆ ˆ (0.75)(0.25)

(1.96) (1.96)(0.043301) 0.08487 0.0849 100

pq E z

n α= = = = ≈

Therefore, the 95% confidence interval estimate is

ˆ ˆ 0.75 0.0849 0.75 0.0849 0.6651 0.8349 0.665 0.835 (rounded to three decimal places)

p E p p E p

p p

− < < + − < < + < < < <

(b) (2 pts) Based on the confidence interval estimate you got in part (a), is this tournament result consistent with her career record of 80%? Why or why not? Please explain your conclusion.

Since the 80% = 0.80 falls within confidence interval estimate (0.665, 0.835), the tournament result is consistent with her career record of 80%.

(c) (2 pts) A sport reporter commented that Mimi’s performance in the tournament is worse than usual. You decide to test if the reporter’s claim is valid by using hypothesis testing that you just learned from STAT 200 class. What are your null and alternative hypotheses?

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The reporter’s claim in symbolic form is p > 0.80, and the opposite of the claim is p ≤ 0.80. The original claim does not contain equality, hence it becomes the alternative hypothesis. We can therefore express H0 and H1 as follows:

0

1

: 0.80 : 0.80 (reporter's claim)

H p H p

= <

(d) (2 pts) What is the test statistic? (Show work)

Use the “test statistic for testing a claim about a proportion” formula on page 413 of the textbook,

ˆ 0.75 0.80 0.05 1.25

0.04(0.80)(0.20) 100

p p z

pq n

− − − = = = = −

(e) (2 pts) What is the P-value? (Show work)

This is a left-tailed test, so P-value is the area to left of test statistic z.

Refer to Table A-2 of the textbook, P(z < -1.25) = 0.1056. Therefore,

P-value = P(z <-1.25 0.75) = 0.1056.

(f) (2 pts) What is the critical value? (Show work)

This is a left-tailed test, so the area of the critical region is an area of α = 0.05.

Referring to Table A-2, we find the critical value of z = -1.645 is at the boundary of the critical region.

Z = -1.25

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(g) (2 pts) What is your conclusion of the testing at 0.05 significance level? Why?

P-value Method

Because the P-value of 0.1056 is greater than the significance level of α=0.05, we fail to reject the null hypothesis. Hence, we conclude that there is not sufficient evidence to support the reporter’s claim that Mimi’s performance in this tournament is worse than usual.

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Or Traditional Method

Because the test statistic does not fall within the critical region, we fail to reject the null hypothesis. Hence, we conclude that there is not sufficient evidence to support the reporter’s claim that Mimi’s performance in this tournament is worse than usual.

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2. (6 points) A simple random sample of 120 SAT scores has a mean of 1540. Assume that SAT scores have a population standard deviation of 300.

(a) (4 pts) Construct a 95% confidence interval estimate of the mean SAT score. (Show work)

The confidence interval estimate is about a population mean with σ known and sample size greater than 30, so we will use the formula on page 346 of the textbook for the margin of error E.

α = 1 – 0.95 = 0.05, hence α/2 = 0.025. The corresponding critical value z α/2 = z 0.025 =1.96.

Margin of error / 2 300

(1.96)( ) (1.96)(27.38613) 53.67681 53.677 120

E z n

α σ

= = = = ≈

Therefore, the 95% confidence interval estimate is

α=0.05

z = -1.645 Critical Value z = -1.25

Test Statistic

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1540 53.677 1540 53.677 1486.323 1593.677

x E x Eµ µ

µ

− < < + − < < +

< <

(b) (2 pts) Is a 99% confidence interval estimate of the mean SAT score wider than the 95% confidence interval estimate you got from part (a)? Why? [You don’t have to construct the 99% confidence interval]

99% confidence interval estimate is wider than the 95% confidence interval since z α/2 = z 0.005 =2.575 for the 99% confidence interval estimate is greater than z α/2 = z 0.025 =1.96 for the 95% confidence interval.

3. (6 points) Consider the hypothesis test given by

0 1

: 670 : 670.

H H

µ µ = ≠

In a random sample of 70 subjects, the sample mean is found to be 678.2.x = The population standard deviation is known to be 27.σ =

(a) (4 pts) Determine the P-value for this test. (Show work)

We want to test a claim about population mean with σ known and sample size greater than 30, so we will use the “Test Statistic for Testing a Claim About a Mean (with σ known)” formula on page 425 of the textbook.

678.2 670 8.2 2.54

27 3.227 70

xx z

n

µ σ − −

= = = =

Since this is a two-tailed test, P-value is twice the area of the extreme region bounded by the test statistic z.

z = 2.54 Test Statistic

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Referring to Table A-2, P-value = (2) P(z > 2.54) = (2) [ 1 – P(z < 2.54) ] = (2) [ 1 – 0.9945] = (2) (0.0055) = 0.011 (b) (2 pts) Is there sufficient evidence to justify the rejection of 0H at the 0.02α =

level? Explain.

Yes. Because the P-value of 0.011 is less than the significance level of α=0.02, we reject the null hypothesis.

4. (7 pts) The playing times of songs are normally distributed. Listed below are the playing times (in seconds) of 10 songs from a random sample. Use a 0.05 significance level to test the claim that the songs are from a population with a standard deviation less than 1 minute. 448 231 246 246 227 213 239 258 255 257 (a) (1 pts) What are your null hypothesis and alternative hypothesis?

The original claim in symbolic form is σ < 60, and the opposite of the claim is σ ≥ 60. The original claim does not contain equality, hence it becomes the alternative hypothesis. We can therefore express H0 and H1 as follows:

0

1

: 60 seconds : 60 seconds (original claim)

H H

σ σ

= <

(b) (4 pts) What is the test statistic? (Show work)

We want to test a claim about population standard deviation and we know it’s a normal distribution, so we will use the “Test Statistic for Testing a Claim About σ or σ2” formula on page 424 of the textbook.

First we have to use Formula 3-4 or 3-5 on page 97 to find the sample variance s2:

x (x - sample mean)2

448 34596 231 961

246 256 246 256 227 1225 213 2401 239 529 258 16 255 49 257 25

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∑ 2620 40314

sample mean = 2620/10 = 262

Hence, 2

2 ( ) 40314 40314 4479.33 1 10 1 9

x x s

n −

= = = = − −

∑

Test statistic 2

2 2 2

( 1) (10 1)(4479.33) 40314 11.198

60 3600 n s

χ σ − −

= = = =

(c) (2 pts) What is your conclusion? Why? (Show work) Referring to Table A-4 using degrees of freedom = 10 – 1 = 9, the critical value is 3.325 which corresponds to the “area to the right” of 0.95. Since test statistic 11.198 is greater than 3.325, we fail to reject null hypothesis. There is not sufficient evidence to support that the songs are from a population with a standard deviation less than 1 minute (or 60 seconds).

5. (6 pts) Assume the population is normally distributed. Given a sample size of 25,

with sample mean 736.2 and sample standard deviation 82.3, we perform the following hypothesis test.

0 : 750H µ =

1 : 750H µ <

What is the conclusion of the test at the 0.10α = level? Explain your answer. (Show work)

We want to test a claim about population mean with σ unknown, so we will use the “Test Statistic for Testing a Claim About a Mean (with σ unknown)” formula on page 412 of the textbook.

3.325 Critical Value

11.198 Test Statistic

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736.2 750 13.8 0.838396 -0.838

82.3 16.46 25

xx t

s n

µ− − − = = = = − ≈

Referring to Table A-3 using degrees of freedom = 25 – 1 = 24, the critical value is -1.318 which corresponds to the “area to the left” of 0.10. Since test statistic -0.838 is greater than -1.318, we fail to reject null hypothesis. There is not sufficient evidence to support that the population mean is less than 750.

-1.318 Critical Value

-0.838 Test Statistic