please answer all the questions and show work For tha I For Problems Let\'s admi

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please answer all the questions and show work

For tha I For Problems Let's admit it. You speed on Hwy 34, I speed on Hwy 34, we all speed on Hwy 34. We have people to see, things to do, and that road is boring A cop was interested in estimating the proportion of people that speed along Hwy 34 and out of those that are speeding, she would like average speed people are travelling. Out of the randomly selected 174 people that passed her in an 8-hour time period, only 22 of them were speeding. Out of the 22 that were speeding, their average speed was 72 mph with a sample standard deviation of 5 mph. 20-25, refer to the statement below: to estimate the 20. What distribution should you use to compute a confidence interval for the average rate that people speeding travel? a. The t distribution with 21 degrees of freedom. b. The t distribution with 22 degrees of freedom. c. The t distribution with 174 degrees of freedom. d. The t distribution with 173 degrees of freedom. e. The z distribution. 21. What is a correct 98% confidence interval for the average rate that people speeding travel? a. 72 +/- 2.518 5/sqrt(21) 72 +/-2.747 b. 72+/-2.518 5/sqrt(22)-72+/-2.684 c. 72 +/-2.326 5/sqrt(174) 72 +/-0.882 d. 72 +/-2.326*5/sqrt(173) 72+-0.884 distribution should you use to compute a confidence interval for the proportion of people that speed along Hwy 34? a. The t distribution with 21 degrees of freedom. b. The t distribution with 22 degrees of freedom. c. The t distribution with 174 degrees of freedom. d. The t distribution with 173 degrees of freedom. e. The z distribution 23 Give an approximate 90% confidence interval for the proportion of people that speed along Hwy 34 a. 0.126 +-1.960 sqrt(0.126*0.874/22) b. 0.126+1.645*sqrt(0.126*0.874/22) c. 0.126 +-1.960*sqrt(0.126*0.874/174) d. 0.126+1.645 sqrt(0.126*0.874/174) 24. Suppose the cop wants to repeat this experiment several months later. How many drivers must she sample in order to produce a 90% confidence interval for the proportion who re speeding along Hwy 34, with a margin of error of two percent? a. 312 b. 745 c. 220 d. 2130 e. None of these 25. Suppose the cop wants to repeat this experiment several months later. How many people speeding must she sample in order to produce a 95% confidence interval for the average rate that people speeding travel, with a margin of error of 0.5 mph? b. 1501 d. 385 e. None of these

Explanation / Answer

Question 20

Answer:

a. The t distribution with 21 degrees of freedom

Explanation:

Required sample size of people who were speeding is given as

n = 22, so degrees of freedom = df = n – 1 = 21.

For this sample, we are given a sample mean and sample standard deviation. We are not given a population standard deviation. So, we use t distribution.

Question 21

We are given

Confidence level = 98%

Sample size = n = 22

Sample mean = Xbar = 72

Sample standard deviation = S = 5

df = 21

Critical t value = 2.518

(by using t-table)

Confidence interval = Xbar +/- t*S/sqrt(n)

Confidence interval = 72 +/- 2.518*5/sqrt(22)

Confidence interval = 72 +/- 2.684

Correct Answer: b.

Question 22

Answer: e. The Z distribution

Explanation: For confidence interval for the proportion, we always use normal z distribution; because sampling distribution of sample proportion follows an approximate normal distribution.

Question 23

We are given

X = 22

N = 174

P = X/N = 22/174 = 0.126437

P = 0.126

Confidence level = 90%

Critical Z value = 1.645

(by using z-table)

Confidence interval = P +/- Z*sqrt(P*(1 – P)/N)

Confidence interval = 0.126 +/- 1.645*sqrt(0.126*(1 – 0.126)/174)

Confidence interval = 0.126 +/- 1.645*sqrt(0.126*0.874/174)

Correct Answer: d.

Question 24

Here, we have to find sample size.

We are given

Estimate of true proportion = p = 0.126

Sampling error = margin of error = E = 0.02

Confidence level = 90%

So, Z = 1.6449

(by using z-table)

Sample size = n = p*(1 – p)*(Z/E)^2

Sample size = 0.126*(1 - 0.126)*(1.6449/0.02)^2 = 744.9052

Required sample size = n = 745

Correct Answer: b.

Question 25

We are given

? = 5

E = 0.5

Confidence level = 95%

Critical Z value = 1.96

(by using Z table)

Sample size = n = (Z*?/E)^2 = (1.96*5/0.5)^2 = 384.16

Required sample size = n = 385

Correct Answer: d

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