CH 8 HW

For this problem, carry at least four digits after the decimal in your calculations. Answers may vary slightly due to rounding. A random sample of 5562 physicians in Colorado showed that 3386 provided at least some charity care (i.e., treated poor people at no cost). (a) Let p represent the proportion of all Colorado physicians who provide some charity care. Find a point estimate for p. (Round your answer to four decimal places.) ______ (b) Find a 99% confidence interval for p. (Round your answers to three decimal places.) lower limit = _____ upper limit = _____ Give a brief explanation of the meaning of your answer in the context of this problem. (c) Is the normal approximation to the binomial justified in this problem? Explain.

(a) 0.6088 p̂ = r/n p̂ = 3386/5562 p̂ = 0.60877 (b) lower limit = 0.5918 = p̂ - E = 0.6088 - 0.017 upper limit = 0.6258 = p̂ + E = 0.6088 + 0.017 E = zc√p̂q̂/n E = 2.58√0.6088(0.3912)/5562 E = 0.0168 --> 0.017 We are 99% confident that the true proportion of Colorado physicians providing at least some charity care falls within this interval. (c) Yes; np > 5 and nq > 5.

For this problem, carry at least four digits after the decimal in your calculations. Answers may vary slightly due to rounding. In a random sample of 60 professional actors, it was found that 42 were extroverts. (a) Let p represent the proportion of all actors who are extroverts. Find a point estimate for p. (Round your answer to four decimal places.) _____ (b) Find a 95% confidence interval for p. (Round your answers to two decimal places.) lower limit = _____ upper limit = _____ Give a brief interpretation of the meaning of the confidence interval you have found. (c) Do you think the conditions np > 5 and nq > 5 are satisfied in this problem? Explain why this would be an important consideration.

(a) 0.7000 p̂ = r / n p̂ = 42/60 (b) lower limit = 0.58 = p̂ - E = 0.70 - 0.12 upper limit = 0.82 = p̂ + E = 0.70 + 0.12 E = zc√p̂q̂/n E = 1.96√0.7(0.3)/60 E = 0.115 --> 0.12 We are 95% confident that the true proportion of actors who are extroverts falls within this interval. (c) Yes, the conditions are satisfied. This is important because it allows us to say that p̂ is approximately normal.

For this problem, carry at least four digits after the decimal in your calculations. Answers may vary slightly due to rounding. How hard is it to reach a businessperson by phone? Let p be the proportion of calls to businesspeople for which the caller reaches the person being called on the first try. (a) If you have no preliminary estimate for p, how many business phone calls should you include in a random sample to be 80% sure that the point estimate p̂ will be within a distance of 0.05 from p? (Round your answer up to the nearest whole number.) _____ phone calls (b) A report states that businesspeople can be reached by a single phone call approximately 19% of the time. Using this (national) estimate for p, answer part (a). (Round your answer up to the nearest whole number.) _____ phone calls

(a) 164 phone calles n = 0.25 (zc/E)^2 n = 0.25 (1.28/0.05)^2 n = 163.84 (b) 101 phone calls n = p(1-p) (zc/E)^2 n = 0.19(1-0.19) (1.28/0.05)^2 n = 100.8

What is the minimal sample size needed for a 95% confidence interval to have a maximal margin of error of 0.1 in the following scenarios? (Round your answers up the nearest whole number.) (a) a preliminary estimate for p is 0.19 (b) there is no preliminary estimate for p

(a) 60 n = p(1-p) (zc/E)^2 n = 0.19(1-0.19) (1.96/0.1)^2 n = 59.12 (b) 97 n = 0.25(zc/E)^2 n = 0.25(1.96/0.1)^2 n = 96.04

For this problem, carry at least four digits after the decimal in your calculations. Answers may vary slightly due to rounding. What percentage of your campus student body is female? Let p be the proportion of women students on your campus. (a) If no preliminary study is made to estimate p, how large a sample is needed to be 99% sure that a point estimate p̂ will be within a distance of 0.05 from p? (Round your answer up to the nearest whole number.) _____ students (b) A report indicates that approximately 59% of college students are females. Answer part (a) using this estimate for p. (Round your answer up to the nearest whole number.) _____ students

(a) 666 phone calls n = 0.25 (zc/E)^2 n = 0.25 (2.58/0.05)^2 n = 665.64 (b) 645 phone calls n = p(1-p) (zc/E)^2 n = 0.59(1-0.59) (2.58/0.05)^2 n = 644.07

What price do farmers get for their watermelon crops? In the third week of July, a random sample of 43 farming regions gave a sample mean of x = $6.88 per 100 pounds of watermelon. Assume that 𝜎 is known to be $1.96 per 100 pounds. (a) Find a 90% confidence interval for the population mean price (per 100 pounds) that farmers in this region get for their watermelon crop. What is the margin of error? (Round your answers to two decimal places.) (b) Find the sample size necessary for a 90% confidence level with maximal error of estimate E = 0.43 for the mean price per 100 pounds of watermelon. (Round up to the nearest whole number.) (c) A farm brings 15 tons of watermelon to market. Find a 90% confidence interval for the population mean cash value of this crop. What is the margin of error? Hint: 1 ton is 2000 pounds. (Round your answers to two decimal places.)

(a) lower limit = $6.39 = x̄ - E = 6.88 - 0.49 upper limit = $7.37 = x̄ + E = 6.88 + 0.49 margin of error = $0.49 E = zc (𝜎 /√n) E = 1.645 (1.96 /√43) = 0.49 (b) 57 farming regions n = (zc (𝜎) / E)^2 n = (1.645(1.96) / 0.43)^2 n = 56.22 (c) lower limit = 1917 15 tons = 30000 lbs = 6.39/100 lbs = x/30000 lbs x = (300)6.39 upper limit = 2211 15 tons = 30000 lbs = 7.37/100 lbs = x/30000 lbs x = (300)7.37 margin of error = 147 15 tons = 30000 lbs = 0.49/100 lbs = x/30000 lbs x = (300)0.49

Thirty-four small communities in Connecticut (population near 10,000 each) gave an average of x = 139.2 reported cases of larceny per year. Assume that 𝜎 is known to be 44.3 cases per year. (a) Find a 90% confidence interval for the population mean annual number of reported larceny cases in such communities. What is the margin of error? (Round your answers to one decimal place.) (b) Find a 95% confidence interval for the population mean annual number of reported larceny cases in such communities. What is the margin of error? (Round your answers to one decimal place.) (c) Find a 99% confidence interval for the population mean annual number of reported larceny cases in such communities. What is the margin of error? (Round your answers to one decimal place.) (d) Compare the margins of error for parts (a) through (c). As the confidence levels increase, do the margins of error increase? (e) Compare the lengths of the confidence intervals for parts (a) through (c). As the confidence levels increase, do the confidence intervals increase in length?

(a) lower limit = 126.7 = x̄ - E = 139.2 - 12.5 upper limit = 151.7 = x̄ + E = 139.2 + 12.5 margin of error = 12.5 E = zc (𝜎 /√n) E = 1.645 (44.3 /√34) = 12.49 (b) lower limit = 124.3 = x̄ - E = 139.2 - 14.9 upper limit = 154.1 = x̄ + E = 139.2 + 14.9 margin of error = 14.9 E = zc (𝜎 /√n) E = 1.96 (44.3 /√34) = 14.89 (c) lower limit = 119.6 = x̄ - E = 139.2 - 19.6 upper limit = 158.8 = x̄ + E = 139.2 + 19.6 margin of error = 19.6 E = zc (𝜎 /√n) E = 2.58 (44.3 /√34) = 19.60 (d) As the confidence level increases, the margin of error increases. (e) As the confidence level increases, the confidence interval increases in length.

Overproduction of uric acid in the body can be an indication of cell breakdown. This may be an advance indication of illness such as gout, leukemia, or lymphoma.† Over a period of months, an adult male patient has taken seven blood tests for uric acid. The mean concentration was x̄ = 5.39 mg/dl. The distribution of uric acid in healthy adult males can be assumed to be normal, with 𝜎 = 1.93 mg/dl. (a) Find a 95% confidence interval for the population mean concentration of uric acid in this patient's blood. What is the margin of error? (Round your answers to two decimal places.) (b) What conditions are necessary for your calculations? (Select all that apply.) (c) Interpret your results in the context of this problem. (d) Find the sample size necessary for a 95% confidence level with maximal margin of error E = 1.10 for the mean concentration of uric acid in this patient's blood. (Round your answer up to the nearest whole number.)

(a) lower limit = 3.96 = x̄ - E = 5.39 - 1.43 upper limit = 6.82 = x̄ + E = 5.39 + 1.43 margin of error = 1.43 E = zc (𝜎 /√n) E = 1.96 (1.93 /√7) = 1.43 (b) 𝜎 is known normal distribution of uric acid (c) We are 95% confident that the true uric acid level for this patient falls within this interval. (d) 12 blood tests n = (zc (𝜎) / E)^2 n = (1.96(1.93) / 1.10)^2 n = 11.61

What percentage of hospitals provide at least some charity care? Based on a random sample of hospital reports from eastern states, the following information is obtained (units in percentage of hospitals providing at least some charity care): 56.7, 55.5, 53.5, 66.1, 59, 64.7, 70.1, 64.7, 53.5, 78.2 Assume that the population of x values has an approximately normal distribution. (a) Use a calculator with mean and sample standard deviation keys to find the sample mean percentage x and the sample standard deviation s. (Round your answers to four decimal places.) x̄ = ____ % s = ____ % (b) Find a 90% confidence interval for the population average 𝜇 of the percentage of hospitals providing at least some charity care. (Round your answers to one decimal place.) lower limit = ____ % upper limit = ____ %

(a) x̄ = 62.2 % x̄ = Σx/n x̄ = 622/10 s = 8.0476 % s^2 = √Σxi^2 - (Σxi)^2/n / n - 1 s^2 = √39271.28 - 386884/10 / 9 s = 8.04764 (b) lower limit = 57.5 % = x̄ - E = 62.2 - 4.7 upper limit = 66.9 % = x̄ + E = 62.2 + 4.7 E = tcs /√n = 1.833(8.0476)/√10 = 4.66 --> 4.7

How much do wild mountain lions weigh? Adult wild mountain lions (18 months or older) captured and released for the first time in the San Andres Mountains had the following weights (pounds): 69, 101, 125, 128, 60, 64 Assume that the population of x values has an approximately normal distribution. (a) Use a calculator with mean and sample standard deviation keys to find the sample mean weight x and sample standard deviation s. (Round your answers to one decimal place.) x̄ = ____ lb s = ____ lb (b) Find a 75% confidence interval for the population average weight 𝜇 of all adult mountain lions in the specified region. (Round your answers to one decimal place.) lower limit = ____ lb upper limit = ____ lb

(a) x̄ = 91.2 lb x̄ = Σx/n x̄ = 547/6 x̄ = 91.16 s = 31 lb s^2 = √Σxi^2 - (Σxi)^2/n / n - 1 s^2 = √54667 - 399209/6 / 5 s = 30.98 (b) lower limit = 74.7 lb = x̄ - E = 91.2 - 16.5 upper limit = 107.7 lb = x̄ + E = 91.2 + 16.5 E = tcs /√n = 1.301(31)/√6 = 16.46 --> 16.5

As the degrees of freedom increase, what distribution does the Student's t distribution become more like?

0

Use the Student's t distribution to find tc for a 0.90 confidence level when the sample is 21. (Round your answer to three decimal places.)

1.725 n = 21 d.f. = n - 1 = 21 - 1 = 20 c = 0.90 tc = t0.90

Consider a 90% confidence interval for 𝜇. Assume 𝜎 is not known. For which sample size, n = 10 or n = 20, is the critical value tc larger?

10

Use the Student's t distribution to find tc for a 0.99 confidence level when the sample is 21.

2.845 n = 21 d.f. = n - 1 = 21 - 1 = 20 c = 0.99 tc = t0.99

You want to conduct a survey to determine the proportion of people who favor a proposed tax policy. How does increasing the sample size affect the size of the margin of error?

As the sample size increases, the margin of error decreases.

Sam computed a 95% confidence interval for 𝜇 from a specific random sample. His confidence interval was 10.1 < 𝜇 < 12.2. He claims that the probability that 𝜇 is in this interval is 0.95. What is wrong with his claim?

Either 𝜇 is in the interval or it is not. Therefore, the probability that 𝜇 is in this interval is 0 or 1.

True or false? A larger sample size produces a longer confidence interval for 𝜇.

False. As the sample size increases, the maximal error decreases, resulting in a shorter confidence interval.

True or false? Every random sample of the same size from a given population will produce exactly the same confidence interval for 𝜇.

False. Different random samples may produce different x̄ values, resulting in different confidence intervals.

Lorraine computed a confidence interval for 𝜇 based on a sample of size 41. Since she did not know 𝜎, she used s in her calculations. Lorraine used the normal distribution for the confidence interval instead of a Student's t distribution. Will her interval be longer or shorter than one obtained by using an appropriate Student's t distribution? Explain.

Shorter. For d.f. = 40, zc is less than tc.

Use the Student's t distribution to find tc for a 0.95 confidence level when the sample is 28. Step 1: Recall that a Student's tc distribution table lists critical values tc for a c confidence level. The table is arranged by column headings for c and by row headings for degrees of freedom d.f. Decide which column to use by identifying the confidence level value. c = _____ Step 2: We are given that the sample size is 28. Therefore, n = ___. Decide which row of the table to use by calculating the degrees of freedom using the formula d.f. = n − 1. d.f. = n − 1 = ___ - 1 = ___ Step 3: From the distribution table, find the column heading for c = 0.950 and read down the column until you reach the entry in the row headed by d.f. = 27. The intersection of the column and row in the table is the value of t0.95 for a sample of size 28. tc = _____

Step 1: c = 0.95 Step 2: d.f. = n - 1 = 28 - 1 = 27 Step 3: tc = 2.052

True or false? If the sample mean x̄ of a random sample from an x distribution is relatively small, when the confidence level c is reduced, the confidence interval for 𝜇 becomes shorter.

True. As the level of confidence decreases, the maximal error of estimate decreases.

True or false? If the original x distribution has a relatively small standard deviation, the confidence interval for 𝜇 will be relatively short.

True. As 𝜎 decreases, E decreases, resulting in a shorter confidence interval.

True or false? The value zc is a value from the standard normal distribution such that P(-zc < z < zc) = c.

True. By definition, critical values zc are such that 100c% of the area under the standard normal curve falls between -zc and zc.

True or false? The point estimate for the population mean 𝜇 of an x distribution is x̄, computed from a random sample of the x distribution.

True. The mean of the x̄ distribution equals the mean of the x distribution and the standard error of the x̄ distribution decreases as n increases.

Sam computed a 90% confidence interval for 𝜇 from a specific random sample of size n. He claims that at the 90% confidence level, his confidence interval contains 𝜇. Is this claim correct? Explain.

Yes. The proportion of all confidence intervals based on random samples of size n that contain 𝜇 is 0.90.

In order to use a normal distribution to compute confidence intervals for p, what conditions on np and nq need to be satisfied?

np > 5; nq > 5

For a binomial experiment with r successes out of n trials, what value do we use as a point estimate for the probability of success p on a single trial?

p̂ = r/n

As the degrees of freedom increase, what distribution does the Student's t distribution become more like?

standard normal