a research firm conducted a survey to determine

SOLUTION: A research firm conducted a survey to determine the mean amount Americans spend on coffee during a week. They found the distribution of weekly spending followed the normal dist

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Question: Suppose a research firm conducted a survey to determine the mean amount steady smokers spend on cigarettes during a week. A sample of 100 steady smokers revealed that the sample mean is $20. The population standard deviation is $7. What is the probability that a sample of 100 steady smokers spend between $19 and $21? 0.4236 1.0000 0.0764 0.8472

According to the given information provided the,

Standard deviation for the population is $ 7 .

Sample siz...

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Suppose a research firm conducted a survey to determine the mean amount steady smokers spend on cigarettes during a week. A sample of 100 steady smokers revealed that the sample mean is $20. The population standard deviation is $5. What is the probability that a sample of 100 steady smokers spend between $19 and $21

Expert-verified answer.

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95.44% probability that a sample of 100 steady smokers spend between $19 and $21

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

What is the probability that a sample of 100 steady smokers spend between $19 and $21

This is the pvalue of Z when X = 21 subtracted by the pvalue of Z when X = 19. So

By the Central Limit Theorem

0.9772 - 0.0228 = 0.9544

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Final answer:

To find the probability that a sample of 100 steady smokers spend between $19 and $21, calculate the Z-score and use a standard normal distribution table or calculator. The probability is approximately 0.3413.

Explanation:

To find the probability that a sample of 100 steady smokers spend between $19 and $21, we can use the Z-score formula. The Z-score is calculated as the difference between the sample mean and the desired value (in this case, $20), divided by the population standard deviation, multiplied by the square root of the sample size.

Z = (x - μ) / (σ / √n)

Plugging in the values we have:

Z = (21 - 20) / (5 / √100) = 1

We can then use a standard normal distribution table or a calculator to find the probability associated with a Z-score of 1. The probability of obtaining a Z-score of 1 or less is approximately 0.8413. Since we want the probability between $19 and $21, we subtract the probability of getting a Z-score of less than 1 from the probability of getting a Z-score of less than or equal to 0. This gives us:

Probability = 0.8413 - 0.5000 = 0.3413

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Chapter Questions

A sample of 49 observations is taken from a normal population with a standard deviation of $10 .$ The sample mean is $55 .$ Determine the $99 \%$ confidence interval for the population mean.

A sample of 81 observations is taken from a normal population with a standard deviation of $5 .$ The sample mean is $40 .$ Determine the $95 \%$ confidence interval for the population mean.

A sample of 250 observations is selected from a normal population with a population standard deviation of $25 .$ The sample mean is 20 . a. Determine the standard error of the mean. b. Explain why we can use formula $(9-1)$ to determine the $95 \%$ confidence interval. c. Determine the $95 \%$ confidence interval for the population mean.

Suppose you know $\sigma$ and you want an $85 \%$ confidence level. What value would you use as $z$ in formula $(9-1) ?$

A research firm conducted a survey of 49 randomly selected Americans to determine the mean amount spent on coffee during a week. The sample mean was 20 per week. The population distribution is normal with a standard deviation of 5 a. What is the point estimate of the population mean? Explain what it indicates. b. Using the 95% level of confidence, determine the confidence interval for mu . Explain what it indicates.

Refer to the previous exercise. Instead of $49,$ suppose that 64 Americans were surveyed about their weekly expenditures on coffee. Assume the sample mean remained the same a. What is the $95 \%$ confidence interval estimate of $\mu ?$ b. Explain why this confidence interval is narrower than the one determined in the previous exercise.

Bob Nale is the owner of Nale's Quick Fill. Bob would like to estimate the mean number of gallons of gasoline sold to his customers. Assume the number of gallons sold follows the normal distribution with a population standard deviation of 2.30 gallons. From his records, he selects a random sample of 60 sales and finds the mean number of gallons sold is $8.60 .$ a. What is the point estimate of the population mean? b. Develop a $99 \%$ confidence interval for the population mean. c. Interpret the meaning of part (b).

Dr. Patton is a professor of English. Recently she counted the number of misspelled words in a group of student essays. She noted the distribution of misspelled words per essay followed the normal distribution with a population standard deviation of 2.44 words per essay. For her 10 a.m. section of 40 students, the mean number of misspelled words was $6.05 .$ Construct a $95 \%$ confidence interval for the mean number of misspelled words in the population of student essays.

Use Appendix $\mathrm{B} .5$ to locate the value of $t$ under the following conditions. a. The sample size is 12 and the level of confidence is $95 \%$. b. The sample size is 20 and the level of confidence is $90 \%$. c. The sample size is 8 and the level of confidence is $99 \%$.

Use Appendix B.5 to locate the value of $t$ under the following conditions. a. The sample size is 15 and the level of confidence is $95 \%$. b. The sample size is 24 and the level of confidence is $98 \%$. c. The sample size is 12 and the level of confidence is $90 \%$.

The owner of Britten's Egg Farm wants to estimate the mean number of eggs produced per chicken. A sample of 20 chickens shows they produced an average of 20 eggs per month with a standard deviation of 2 eggs per month. a. What is the value of the population mean? What is the best estimate of this value? b. Explain why we need to use the $t$ distribution. What assumption do you need to make? c. For a $95 \%$ confidence interval, what is the value of $t ?$ d. Develop the $95 \%$ confidence interval for the population mean. e. Would it be reasonable to conclude that the population mean is 21 eggs? What about 25 eggs?

The U.S. Dairy Industry wants to estimate the mean yearly milk consumption. A sample of 16 people reveals the mean yearly consumption to be 45 gallons with a standard deviation of 20 gallons. Assume the population distribution is normal. a. What is the value of the population mean? What is the best estimate of this value? b. Explain why we need to use the $t$ distribution. What assumption do you need to make? c. For a 90\% confidence interval, what is the value of $t ?$ d. Develop the $90 \%$ confidence interval for the population mean. e. Would it be reasonable to conclude that the population mean is 48 gallons?

Merrill Lynch Securities and Health Care Retirement Inc. are two larqe employers in downtown Toledo, Ohio. They are considering jointly offering child care for their employees. As a part of the feasibility study, they wish to estimate the mean weekly child care cost of their employees. A sample of 10 employees who use child care reveals the following amounts spent last week. $$\left.\begin{array}{llllllll}\$ 107 & \$ 92 & \$ 97 & \$ 95 & \$ 105 & \$ 101 & \$ 91 & \$ 99 & \$ 95 & \$ 104\end{array}\right)$$ Develop a $90 \%$ confidence interval for the population mean. Interpret the result.

The Buffalo. New York, Chamber of Commerce wants to estimate the mean time workers who are employed in the downtown area spend getting to work. A sample of 15 workers reveals the following number of minutes spent traveling. $$\begin{array}{llllllll}14 & 24 & 24 & 19 & 24 & 7 & 31 & 20 \\ 26 & 23 & 23 & 28 & 16 & 15 & 21 &\end{array}$$ Develop a $98 \%$ confidence interval for the population mean. Interpret the result.

The owner of the West End Kwick Fill Gas Station wishes to determine the propor tion of customers who pay at the pump using a credit card or debit card. He surveys 100 customers and finds that 80 paid at the pump. a. Estimate the value of the population proportion. b. Develop a $95 \%$ confidence interval for the population proportion. c. Interpret your findings.

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A population's standard deviation is $10 .$ We want to estimate the population mean within $2,$ with a $95 \%$ level of confidence. How large a sample is required?

We want to estimate the population mean within $5,$ with a $99 \%$ level of confidence. The population standard deviation is estimated to be $15 .$ How large a sample is required?

The estimate of the population proportion should be within plus or minus. $05,$ with a $95 \%$ level of confidence. The best estimate of the population proportion is .15. How large a sample is required?

The estimate of the population proportion should be within plus or minus $.10,$ with a 99\% level of confidence. The best estimate of the population proportion is . 45 . How large a sample is required?

A large on-demand video streaming company is designing a large-scale survey to determine the mean amount of time corporate executives watch on-demand television. A small pilot survey of 10 executives indicated that the mean time per week is 12 hours, with a standard deviation of 3 hours. The estimate of the mean viewing time should be within one-quarter hour. The $95 \%$ level of confidence is to be used. How many executives should be surveyed?

A processor of carrots cuts the green top off each carrot, washes the carrots, and inserts six to a package. Twenty packages are inserted in a box for shipment. Each box of carrots should weigh 20.4 pounds. The processor knows that the standard deviation of box weight is 0.5 pound. The processor wants to know if the current packing process meets the 20.4 weight standard. How many boxes must the processor sample to be $95 \%$ confident that the estimate of the population mean is within 0.2 pound?

Suppose the U.S. president wants to estimate the proportion of the population that supports his current policy toward revisions in the health care system. The president wants the estimate to be within . 04 of the true proportion. Assume a $95 \%$ level of confidence. The president's political advisors found a similar survey from two years ago that reported that $60 \%$ of people supported health care revisions. a. How large of a sample is required? b. How large of a sample would be necessary if no estimate were available for the proportion supporting current policy?

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Near the time of an election, a cable news service performs an opinion poll of 1,000 probable voters. It shows that the Republican contender has an advantage of $52 \%$ to $48 \% .$ a. Develop a $95 \%$ confidence interval for the proportion favoring the Republican candidate. b. Estimate the probability that the Democratic candidate is actually leading. c. Repeat the above analysis based on a sample of 3,000 probable voters.

A sample of 352 subscribers to Wired magazine shows the mean time spent using the Internet is 13.4 hours per week, with a sample standard deviation of 6.8 hours. Find the $95 \%$ confidence interval for the mean time Wired subscribers spend on the Internet.

The Tennessee Tourism Institute (TTI) plans to sample information center visitors entering the state to learn the fraction of visitors who plan to camp in the state. Current estimates are that $35 \%$ of visitors are campers. How many visitors would you sample to estimate the population proportion of campers with a $95 \%$ confidence level and an allowable error of $2 \% ?$

Refer to the North Valley Real Estate data, which report information on homes sold in the area during the last year. Select a random sample of 20 homes. a. Based on your random sample of 20 homes, develop a $95 \%$ confidence interval for the mean selling price of the homes. b. Based on your random sample of 20 homes, develop a $95 \%$ confidence interval for the mean days on the market. c. Based on your random sample of 20 homes, develop a $95 \%$ confidence interval for the proportion of homes with a pool. d. Suppose that North Valley Real Estate employs several agents. Each agent will be randomly assigned 20 homes to sell. The agents are highly motivated to sell homes based on the commissions they earn. They are also concerned about the 20 homes they are assigned to sell. Using the confidence intervals you created, write a general memo informing the agents about the characteristics of the homes they may be assigned to sell. e. What would you do if your confidence intervals did not include the mean of all 105 homes? How could this happen?

Refer to the Baseball 2016 data, which report information on the 30 Major League Baseball teams for the 2016 season. Assume the 2016 data represent a sample. a. Develop a $95 \%$ confidence interval for the mean number of home runs per team. b. Develop a $95 \%$ confidence interval for the mean batting average by each team. c. Develop a $95 \%$ confidence interval for the mean earned run average (ERA) for each team.

Refer to the Lincolnville School District bus data. a. Develop a $95 \%$ confidence interval for the mean bus maintenance cost. b. Develop a $95 \%$ confidence interval for the mean bus odometer miles. c. Write a business memo to the state transportation official to report your results.