STATS TEST 3 - This Exam contains: -Guarantee passing score -2...

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STATS TEST 3

Qs and Ans Expert- Explanation

This Exam contains:

-Guarantee passing score -27 Questions and Answers -format set of multiple-choice -Expert- Explanation Question 1: State the confidence interval for the POPULATION PROPORTION of voters who support the proposed levy in a local school district if we have a sample proportion of 0.53 and a margin of error of 0.04

Answer:

CI= p-hat + or - error So, CI= .53 + or - .04

= (.49, .57)

Question 2: Compute the Z-SCORE necessary to construct a confidence interval with the

following levels of confidence:

• 0.90
• 0.85
• 0.78

Answer:

• invNorm(1-(1-c/2)) = +- 1.645

• invNorm(1-(1-c/2)) = +- 1.440

• invNorm(1-c/2) = +- 1.227

*use the longer formula for bigger numbers?

Question 3: Compute the T-SCORE necessary to construct a confidence interval with the

following levels of confidence and sample sizes:

• 0.95, 20
• 0.80, 10

Answer:

**

USE PDF TABLE

**

• C=.95 (95%), n=20

df=20-1= 19 use t chart and find intersection t= 2.093 b.C=.80 (80%), n=10 df=10-9= 9 use t chart to find intersection t= 1.383 Question 4: A random sample of 200 restaurants in the Covington, KY area indicated that 40% of them did not allow smoking. Answer the following questions as they relate to estimation of the

POPULATION PROPORTION:

• State the point estimate for the population proportion of Covington, KY restaurants that allow

smoking, using correct notation.

• Construct a 95% confidence interval for the population proportion of restaurants in

Covington, KY that do not allow smoking

• Use your interval to compute the margin of error.
• Write a sentence interpreting your 95% confidence interval.

Answer:

• q-hat = .60 ( POE is either qh or ph)

b.stats>test>A

(POPULATION PROPORTION))

x=80, n=200, c=.95

=(.332, .4679)

• MOE= interval-interval/2

=.4679-.332/2 =.068

• We are 95% confident that this interval ly contains the true population proportion of restaurants

in KY that do not allow smoking.Question 5: According to the 117th edition of Statistical Abstracts, 27% of adults in the United States visited an art museum at least once last year. A random sample of 200 adults in a large city showed that 50 of them had been to an art museum in the past year. Use the P-VALUE method to test to see if the proportion of people in this large city who visit an art museum is at least 27% using ?=0.05.

• Can we use the NORMAL APPROXIMATION? Explain.
• State the null and alternative hypotheses that you would use to test the claim made in

the problem.

• Compute the appropriate test statistic and its corresponding p-value.
• Using your p-value, make decision about rejecting or not rejecting your null hypothesis

and state your decision about the original claim.

Answer:

• yes, np and nq are >5

b. Ho= p>or=.27, Ha: p<.27

• don't need zc. use

stat

(NORMAL APPROXIMATION, NO SD)

p-value= .262 ztest= -.6371

• p-value (.262) > alpha (.05)= fail to reject Ho. Not enough evidence that the proportion of people in

this large city who visit an art museum is at least 27% using ?=0.05

Question 6: When do we conduct a T-test instead of a Z-test?

Answer:

When we have the sample standard deviation (S) instead of the population SD.

Question 7: Claim: The mean number of pepperoni on a large pizza is different than 75.

• What parameter are you being asked to test?
• Is this claim giving us the null hypothesis or the alternative hypothesis? Explain your

answer.

• State the null and alternative hypothesis
• State the Type I and Type II errors

Answer:

• mean (mu)

• alternative because "different than"

• Ho: mu = 75, Ha: mu =/ 75

• Type I: Reject that the mean number of pepperoni on a large pizza is different than 75, when it's

ly true.Type II: Fail to reject that the mean number of pepperoni on a large pizza is equal to 75, when it is ly different than 75.Question 8: A sample of 20 retired NFL wide receivers was asked about the number of concussions they had experienced throughout their careers. The sample showed a mean of 3.7 with a standard deviation of 0.6.

• State a point estimate for POPULATION MEAN NUMBER concussions experienced.
• Construct a 75% confidence interval for the population mean number concussions

experienced.

Answer:

• ??=3.7 (given)

b.stats

(POPULATION MEAN NUMBER)

n=20, Sx=.6, ??=3.7, C=.75

=(3.541, 3.859)