Question 6 Use the sample data and confidence level given below to complete part

Question

Question 6

Use the sample data and confidence level given below to complete parts (a) through (d). A research institute poll asked respondents if they felt vulnerable to identity theft. In the poll, n=981 and x=571 who said "yes." Use a 90% confidence level.

a) Find the best point estimate of the population proportion p.

(Round to three decimal places as needed.)

b) Identify the value of the margin of error E.

(Round to four decimal places as needed.)

c) Construct the confidence interval.

0.6240.624less than p less than

(Round to three decimal places as needed.)

d) Write a statement that correctly interprets the confidence interval. Choose the correct answer below.

A. One has 95 % confidence that the sample proportion is equal to the population proportion.

B. 95 % of sample proportions will fall between the lower bound and the upper bound.

C. One has 95 % that the interval from the lower bound to the upper bound actually does contain the true value of the population proportion.Your answer is correct.

D.There is a 95 %chance that the true value of the population proportion will fall between the lower bound and the upper bound.

Question 8.

In the week before and the week after a holiday, there were 10,000 total deaths, and 4959 of them occurred in the week before the holiday.

a. Construct a 95 % confidence interval estimate of the proportion of deaths in the week before the holiday to the total deaths in the week before and the week after the holiday.

b. Based on the result, does there appear to be any indication that people can temporarily postpone their death to survive the holiday?

Question 9.

An online site presented this question, "Would the recent norovirus outbreak deter you from taking a cruise?" Among the 34,425 people who responded, 65 % answered "yes." Use the sample data to construct a 95 % confidence interval estimate for the proportion of the population of all people who would respond "yes" to that question. Does the confidence interval provide a good estimate of the population proportion?

question 10.

In a poll of 532 human resource professionals, 44.4 % said that body piercings and tattoos were big grooming red flags. Complete parts (a) through (d) below.

a) Among the 532 human resource professionals who were surveyed, how many of them said that body piercings and tattoos were big grooming red flags?

(Round to the nearest integer as needed.)

b) Construct a 99% confidence interval estimate of the proportion of all human resource professionals believing that body piercings and tattoos are big grooming red flags.

(Round to three decimal places as needed.)

c) Repeat part (b) using a confidence level of 80%

d) Compare the confidence intervals from parts (b) and (c) and identify the interval that is wider. Why is it wider?

Select the correct choice below and fill in the answer boxes to complete your choice.

A. The ____% confidence interval is wider than the _____% confidence interval. As the confidence interval widens, the probability that the confidence interval actually does contain the population parameter increases.

B. The ____% confidence interval is wider than the _____% confidence interval. As the confidence interval narrows, the probability that the confidence interval actually does contain the sample parameter increases.

C. The ____% confidence interval is wider than the ____% confidence interval. As the confidence interval widens, the probability that the confidence interval actually does contain the sample parameter increases.

D. The ____ %confidence interval is wider than the ____% confidence interval. As the confidence interval narrows, the probability that the confidence interval actually does contain the population parameter increases.

Question 11.

Use the data in the table to the right to answer the following questions.

Find the sample proportion of candy that are red.

Weights (g) of a Sample Bag of Candy

Red   Blue   Brown   Green   Yellow
0.786   0.746   0.813   0.942   0.703
0.743   0.713   0.994   0.841   0.856
0.887   0.822   0.754   0.793   0.993
0.735   0.995   0.757   0.851   0.984
0.711   0.935   0.915   0.939   0.835
0.754   0.719   0.781 0.995
0.939   0.903 0.856
0.863   0.973 0.801
0.952

The proportion of red candy=

Question 12.

During a period of 11 years 2443 of the people selected for grand jury duty were sampled, and 21 % of them were immigrants. Use the sample data to construct a 99% confidence interval estimate of the proportion of grand jury members who were immigrants. Given that among the people eligible for jury duty, 55.2 % of them were immigrants, does it appear that the jury selection process was somehow biased against immigrants?

Quesion 13

A study of 420,094 cell phone users found that 134 of them developed cancer of the brain or nervous system. Prior to this study of cell phone use, the rate of such cancer was found to be 0.0324% for those not using cell phones. Complete parts (a) and (b).

a. Use the sample data to construct a 95 % confidence interval estimate of the percentage of cell phone users who develop cancer of the brain or nervous system.

__%

Question 15

Many states are carefully considering steps that would help them collect sales taxes on items purchased through the Internet. How many randomly selected sales transactions must be surveyed to determine the percentage that transpired over the Internet? Assume that we want to be 90 % confident that the sample percentage is within five percentage points of the true population percentage for all sales transactions.

n=__(Round up to the nearest integer.)

Question 15

Find the sample size, n, needed to estimate the percentage of adults who have consulted fortune tellers. Use a 0.06 margin of error, use a confidence level of 95 %,and use results from a prior poll suggesting that 20 % of adults have consulted fortune tellers.

n=__(Round up to the nearest integer.)

Explanation / Answer

6) a) Mean(x) = 571

Sample Size(n) = 981

Sample proportion = x/n =571/981 = 0.582

b) Margin of Error = Z a/2 Sqrt(p*(1-p)/n))

x = Meann = Sample Size

a = 1 - (Confidence Level/100)

Za/2 = Z-table value

CI = Confidence Interval

Sample proportion =0.582

Margin of Error = Z a/2 * ( Sqrt ( (0.582*0.418) /981) ) = 1.645* Sqrt(0.00025) = 0.0260

c) Confidence Interval For Proportion

CI = p ± Z a/2 Sqrt(p*(1-p)/n)))

x = Mean

n = Sample Size

a = 1 - (Confidence Level/100)

Za/2 = Z-table value

CI = Confidence Interval

Confidence Interval = [ 0.582 ± Z a/2 ( Sqrt ( 0.582*0.0.418) /981)]= [ 0.582 - 1.645* Sqrt(0.00025) , 0.582 + 1.645* Sqrt(0.00025) ]= [0.5559, 0.6080]

d) D option

There is a 95 %chance that the true value of the population proportion will fall between the lower bound and the upper bound.

Related Questions

Navigate

• Browse (All)
• Browse Q
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.