Chapter 7 - Estimates & Sample Size

Which of the following is NOT an equivalent expression for the confidence interval given by 161.7 < mu < ​189.5?

161.7 +- 27.8

The number of individuals in the sample with the specified characteristic is

n * point estimate

A chocolate chip cookie manufacturing company recorded the number of chocolate chips in a sample of 30 cookies. The mean is 25.14 and the standard deviation is 3.12. Construct a 95​% confidence interval estimate of the standard deviation of the numbers of chocolate chips in all such cookies.

(find Standard deviation confidence interval calculator) http://www.statskingdom.com/40_confidence_interval.html

Express the confidence interval (.038, .086) in the form of p-hat - E < p < p-hat + E

.038 , p < .086

Find the critical value z_a/2 that corresponds to a = 0.02

1-0.02 = .98 find score on statcrunch

When σ is unknown, use _S__

1. As n increases, the t-distributions approach the standard normal z 2. Since we don't have n yet (we don't have df either) so we

Properties of the t-Distribution

1. Like z, the t distribution is bell-shaped & symmetric 2. The standard-t has mean still at 0 but std. dev. is > 1 3. The total area under the curve is still 1 (or 100%) 4. The t-distribution has larger standard deviation than the standard normal. t ->("squished from top"): thicker tails & lower peak 5. The t-distribution is a family of curves, where the shape is determined by the degrees of freedom: d.f. = n - 1 (one less than the sample size) 6. As n, and the degrees of freedom increase, the t-distribution gets closer and closer to the normal.

Sampling Distributions of p-hat

1. The distribution of sample proportions, p-hat , will be normally distributed 2. The mean of the sample proportion is centered at the population proportion 3. The standard deviation of the sample proportion is

Requirements Checks

1. The sample is a simple random sample. 2. The conditions for the binomial distribution are satisfied: we are counting a characteristic that does/doesn't occur from a fixed number (n) of independent trials: Check: n <= 5% of the population size 3. Sample size, n, is large enough for the sampling distribution to be normal (called the normal approximation to the binomial): Check: There are at least 5 successes and 5 failures. (x = # of successes) ≥ 5 (n - x = # of failures) ≥ 5

Procedure for Constructing a Confidence Interval for μ (With x bar & estimated s)

1. Verify that the requirements are satisfied: -- n ≥30, or parent population must be normally distributed -- no outliers 2. Using n - 1 degrees of freedom, refer to t-table or use technology to find the critical value t_a/2 that corresponds to the desired confidence level 3. Evaluate the margin of error 4. Find the values of x bar - E and x bar + E. Substitute those values in the general format for the confidence interval: 5. Write a sentence to interpret your result. Label the resulting confidence interval limits with the correct units.

Procedure for Constructing a Confidence Interval for p (Using p-hat and E)

1. Verify that the requirements are satisfied: --can treat each person as independent binomial trials: n<.05 N. -- #successes, np ≥ 5 and #failures, nq ≥5 2. Using standard normal, z-table or use technology to find the critical value corresponding to the desired confidence level. 3. Evaluate the margin of error 4. Find the values of p-hat + E and p-hat - E. Substitute those values in the general format for the confidence interval: 5. Write a sentence to interpret: Convert the resulting confidence interval limits to percent, rounded to tenths of a percent.

Assume that the sample is a simple random sample obtained from a normally distributed population of flight delays at an airport. Use the table below to find the minimum sample size needed to be 99​% confident that the sample standard deviation is within 30 % of the population standard deviation. A histogram of a sample of those arrival delays suggests that the distribution is​ skewed, not normal. How does the distribution affect the sample​ size?

38 The computed minimum sample size is not likely correct.

Confidence Interval

A confidence interval (or interval estimate) is a range (or an interval) of values used to estimate the true value of a population parameter Confidence interval estimates are of the form: point estimate ± "a margin of error"

Confidence Level

A confidence level is the probability that the confidence interval captures the true value of the population parameter... assuming that the estimation process is repeated a large number of times. The most common choices for confidence level are 90% Confidence Level 95% Confidence Level or 99% Confidence Level

Point Estimate

A point estimate is a single value (or point) used to approximate a population parameter. For example: What info from sample could we use to estimate the ->percent of the population that has a particular characteristic?

A survey was taken of the​ heights, in​ inches, of​ 20- to​ 29-year-old males. A simple random sample of n = 15 males 20 to 29 years old results in the data in the table. Construct a​ 95% t-interval about the population mean. Decide if a​ 95% t-interval can be constructed about the population mean. If​ not, state the reason why.

A​ t-interval can be constructed.

A police officer hides behind a billboard to catch speeders. The following data represent the number of minutes he waits before first observing a car that is exceeding the speed limit by more than 10 miles per hour on 10 randomly selected days. Construct a 95 % ​t-interval about the population mean. If it cannot be​ constructed, state the reason why. For​ convenience, a normal probability plot and boxplot are given.

A​ t-interval cannot be constructed because there is an outlier. There is no solution.

Interpreting a Confidence Interval

Example: 95% Confidence Interval: 0.858 < p < 0.924 means: "We are 95% confident that the interval from 85.8% to 92.4% actually does contain the true value of the population proportion p." The level of confidence is the proportion of intervals created by this method that would be expected to capture the target parameter. if a large number of repeated samples are obtained.

Notation

First, we need notation to distinguish between a population proportion p and the proportion found within a sample: p = population proportion (percent) p-hat = sample proportion

When estimating a population mean

If n is less than 30 and the data is "strange" (i.e. not normal or contains outliers) we can't use the methods of this chapter

Round-Off Rule for Determining Sample Size

If the computed sample size n is not a whole number, round the value of n up to the next larger whole number.

Sample Size Needed to Reduce E

If you have a preliminary estimate, p-hat, the minimum sample size needed to create a confidence interval for p within a specified margin for Error, E: A "preliminary" p-hat means we've already done at least a small sample (or we have a previous sample) in order to have an initial to start from.

​(c) What could be done to increase the accuracy of the interval without changing the level of​ confidence?

Increase the sample size.

Do one of the​ following, as appropriate.​ (a) Find the critical value z_a/2, ​(b) find the critical value t_a/2​, ​(c) state that neither the normal nor the t distribution applies. Confidence level 99​%; n=29​; sigma is known​; population appears to be very skewed.

Neither normal nor t distribution applies

Listed below are speeds​ (mi/h) measured from traffic on a busy highway. This simple random sample was obtained at​ 3:30 P.M. on a weekday. Use the sample data to construct an 80​% confidence interval estimate of the population standard deviation.

No. The confidence interval is an estimate of the standard deviation of the population of speeds at​ 3:30 on a​ weekday, not other times.

CI for u --- o unknown

Note: For sample size smaller than 30 - need to check to see if the population is approximately normal - and check that there are no outliers

Find the sample proportion of candy that are red.

Number of red candy = 7 Pieces of candy in sample = 32 7/32 = .219 (proportion of red candy) z_0.025 = 1.96 1.96 * sqrt((.219*.781) / 32) = .143294 .076 < p < .362 Yes, because the confidence interval includes 32%

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

One has 90​% confidence that the interval from the lower bound to the upper bound actually does contain the true value of the population proportion.

Point Estimate for p

Since sample proportion targets the value of the true population proportion. The sample proportion p-hat is the best point estimate of the population proportion p.

Sample Size

Suppose we want to collect sample data in order to estimate some population proportion with a desired accuracy. The question is how many sample items must be obtained?

a. What does it mean to say that the confidence interval methods of this section are robust against departures from​ normality?

The confidence interval methods of this section are robust against departures from​ normality, meaning they work well with distributions that​ aren't normal, provided that departures from normality are not too extreme.

Critical Values

The confidence level is used to find: critical values. The critical values are symmetric z-scores that create a probability area in the middle of the standard normal curve equal to the confidence level % a is complement of the confidence level

When to use which method?

The main question that determine the method - What are we trying to estimate, μ or p? -- estimating the population mean, μ? --- Data is about a continuous random variable (measured) and the question talks about mean or average -- estimating a population proportion, p? --- Data is a counted (discrete) random variable we can turn into a percent

Margin of Error for Proportions

The margin of error, E, can be found by multiplying the critical value by the standard deviation of the sample proportion:

Margin of Error

The margin of error, E, is based on three things: - The std. deviation of the sampling distribution (how much the sample estimator normally varies) - The sample size, n - The "level of confidence" we choose

A study of the ages of motorcyclists killed in crashes involves the random selection of 168 drivers with a mean of 37.53 years. Assuming that sigma=8.4 ​years, construct and interpret a 90 ​%confidence interval estimate of the mean age of all motorcyclists killed in crashes.

The mean age of the population will most likely not be less than 20 years old

Assume that the sample is a simple random sample obtained from a normally distributed population of IQ scores of statistics professors. Use the table below to find the minimum sample size needed to be 95​% confident that the sample standard deviation s is within 40​% of sigma. Is this sample size​ practical?

The minimum sample size needed is 12 Yes, because the sample size is small enough to be practical for most applications

The following data represent the asking price of a simple random sample of homes for sale. Construct a 99​% confidence interval with and without the outlier included. Comment on the effect the outlier has on the confidence interval.

The outlier caused the width of the confidence interval to increase.

Overall Goal of Confidence Intervals

The overall goal of estimating with confidence interval estimates is to have - A high confidence level in our prediction - While still providing as narrow a prediction interval as possible (a narrow prediction interval gives a more precise estimate)

Which of the following is NOT a property of the Student t​ distribution?

The standard deviation of the Student t distribution is s=1

Interpret the results found in the previous parts. Do the confidence intervals suggest a difference in the variation among waiting​ times? Does the​ single-line system or the​ multiple-line system seem to be a better​ arrangement?

The variation appears to be significantly lower with a single line system. The​ single-line system appears to be better.

A poll of 1567 adults in a certain country found that 55​% identified themselves as the followers of some religion. The margin of error was 4 percentage points with 95​% confidence.

There is 95​% confidence that the proportion of adults in a certain country who identify themselves as the followers of some religion is between 51​% and 59​%

Interpret a​ 90% confidence interval for the mean commute time of adults employed full or part time. Choose the correct answer below. ​(b) Is it possible that the mean commute time is less than 40​ minutes?

There is​ 90% confidence that the mean commute time of adults from this​ country, employed full or part​ time, lies between the lower and upper bounds of the interval. b. Yes not likely

Finding t-critical values

To find t-critical values, ta/2, we need to know the area in the tail (a/2) ...and the degrees of freedom (df = n-1) In t-table (t-distribution area to right) locate the row with the correct df, move across to the column with the correct area in the tail

Which of the following would be a correct interpretation of a​ 99% confidence interval such as 4.1 < u < 5.6?

We are​ 99% confident that the interval from 4.1 to 5.6 actually does contain the true value of mu.

Analyzing Polls

When analyzing polls consider: 1. The sample should be a simple random sample, not an inappropriate sample (such as a voluntary response sample). 2. The confidence level should be provided. (It is often 95%, but media reports often neglect to identify it.) 3. The sample size should be provided. (It is usually provided by the media, but not always.) 4. Except for relatively rare cases, the quality of the poll results depends on the sampling method and the size of the sample, but the size of the population is usually not a factor.

Round-Off Rule for Confidence Interval Estimates of p

While computing the confidence interval limits for p use 4 or more decimal places in your computations In the final result, round the confidence interval for p to three decimal places (tenths of a percent).

Randomly selected students participated in an experiment to test their ability to determine when one minute​ (or sixty​ seconds) has passed. Forty students yielded a sample mean of 57.8 seconds. Assuming that sigma=11.3 ​seconds, construct and interpret a 95​% confidence interval estimate of the population mean of all students.

Yes​, because the confidence interval includes sixty seconds

In a poll of 610 human resource​ professionals, 48.5​% said that body piercings and tattoos were big grooming red flags. Complete parts​ (a) through​ (d) below. ​a) Among the 610 human resource professionals who were​ surveyed, how many of them said that body piercings and tattoos were big grooming red​ flags? 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. c) Repeat part​ (b) using a confidence level of​ 80% and rounding to three decimal places. ​d) Compare the confidence intervals from parts​ (b) and​ (c) and identify the interval that is wider. Why is it​ wider?

a. .485*610 = 296 b. .433 < p < .537 c. .459 < p < .511 d. The 99​% confidence interval is wider than the 80​% confidence interval. As the confidence interval​ widens, the probability that the confidence interval actually does contain the population parameter increases.

Using the simple random sample of weights of women from a data​ set, we obtain these sample​ statistics: n = 55 and x overbar = 150.31 lb. Research from other sources suggests that the population of weights of women has a standard deviation given by sigma = 31.95 lb. a. Find the best point estimate of the mean weight of all women. b. Find a 99​% confidence interval estimate of the mean weight of all women.

a. 150.31 b. z_0.005 = 2.575 E = 2.575 * 31.95/sqrt(55) = 11.093 139.22 < u < 161.40

A data set includes 107 body temperatures of healthy adult humans for which x overbar= 98.7 degrees F and s = 0.72 degrees F. Complete parts​ (a) and​ (b) below. a. What is the best point estimate of the mean body temperature of all healthy​ humans? b. Using the sample​ statistics, construct a 99​% confidence interval estimate of the mean body temperature of all healthy humans. Do the confidence interval limits contain 98.6 degrees ​F? What does the sample suggest about the use of 98.6 degrees F as the mean body​ temperature?

a. 98.7 b. 98.517 < u < 98.883 This suggests that the mean body temperature could very possibly be 98.6 degrees F.

In the week before and the week after a​ holiday, there were 10,000 total​ deaths, and 4999 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?

a. p-hat = 4999/10000 = .4999 z_a/2 = z_0.05/2 = z_0.025 = 1.96 E = 1.96* sqrt((.4999*.5001)/10,000) = .0098 .4999 - .0098 = .490 .4999 + .0098 = .510 b. No​, because the proportion could easily equal 0.5. The interval is not less than 0.5 the week before the holiday.

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 equals 1111 and x equals 510 who said​ "yes." Use a 95 % confidence level. ​a) Find the best point estimate of the population proportion p. ​b) Identify the value of the margin of error E. c) Construct the confidence interval. ​d) Write a statement that correctly interprets the confidence interval.

a. p-hat = x/n = 510/1111 = .459 b. a = 0.05 (1-.95) z_a/2 = z_0.05/2 = z_0.025 = 1.96 E = 1.96*sqrt[.459(1-.459) / 1111] = .029 c. p-hat - E = .459 - 0.029 = .430 p-hat + E = .459 + 0.029 = .488 .430 < p < .488 ​d. One has 95​% confidence that the interval from the lower bound to the upper bound actually does contain the true value of the population proportion.

In a study designed to test the effectiveness of magnets for treating back​ pain, 32 patients were given a treatment with magnets and also a sham treatment without magnets. Pain was measured using a scale from 0​ (no pain) to 100​ (extreme pain). After given the magnet​ treatments, the 32 patients had pain scores with a mean of 6.0 and a standard deviation of 2.4. After being given the sham​ treatments, the 32 patients had pain scores with a mean of 6.3 and a standard deviation of 2.8. Complete parts​ (a) through​ (c) below. a. Construct the 95​% confidence interval estimate of the mean pain score for patients given the magnet treatment. b. Construct the 95​% confidence interval estimate of the mean pain score for patients given the sham treatment. c. Compare the results. Does the treatment with magnets appear to be​ effective?

a. t_a/2 = 2.040 E = 2.040 * 2.4/sqrt(32) = .865 5.1 < u < 6.9 b. E = 2.040 * 2.8/sqrt(32) = 1.010 5.3 < u < 7.3 c. Since the confidence intervals ​overlap, it appears that the magnet treatments are no more effective than the sham treatments.

An agricultural researcher is interested in estimating the mean length of the growing season in a region. Treating the last 10 years as a simple random​ sample, he obtains the following​ data, which represent the number of days of the growing season. ​(a) Because the sample size is​ small, we must verify that the data come from a population that is normally distributed and that the sample size does not contain any outliers. The normal probability plot and boxplot are shown below.

a. ​Yes, both conditions are met. b. (151.4, 176.6)

People were polled on how many books they read the previous year. Initial survey results indicate that s = 18.3 books. Complete parts ​(a) through ​(d) below

c. Doubling the required accuracy nearly quadruples the sample size. d. Increasing the level of confidence increases the sample size required. For a fixed margin of​ error, greater confidence can be achieved with a larger sample size.

Do one of the​ following, as appropriate.​ (a) Find the critical value z_a/2, ​(b) find the critical value t_a/2​, ​(c) state that neither the normal nor the t distribution applies. Confidence level 90​%; n=18​; sigma is known​; population appears to be very skewed.

df = 17 a/2 = .05 t_a/2 = 1.740 (use right tail)

Use the given information to find the number of degrees of​ freedom, the critical chi values, and the confidence interval estimate of sigma. It is reasonable to assume that a simple random sample has been selected from a population with a normal distribution. Nicotine in menthol cigarettes 98​% ​confidence; n=18​, s=0.27 mg.

df = 18-1 = 17 X_L ^ 2 = X_0.99 ^ 2 = 6.408 X_R ^ 2 = X_0.01 ^ 2 = 33.409 sqrt((n-1)s^2 / X_L^2) < o < sqrt((n-1)s^2 / X_R^2) 0.19 < o < 0.44

Listed below are the amounts of mercury​ (in parts per​ million, or​ ppm) found in tuna sushi sampled at different stores. The sample mean is 1.163 ppm and the sample standard deviation is 0.306 ppm. Use technology to construct a 90​% confidence interval estimate of the mean amount of mercury in the population.

find mean confidence interval

Use technology and the given confidence level and sample data to find the confidence interval for the population mean mu. Assume that the population does not exhibit a normal distribution.

find mean confidence interval ​No, because the sample size is large enough.

Listed below are measured amounts of lead​ (in micrograms per cubic​ meter, or mu g divided by m cubed​) in the air. The EPA has established an air quality standard for lead of 1.5 mu g divided by m cubed. The measurements shown below were recorded at a building on different days. Use the given values to construct a 95​% confidence interval estimate of the mean amount of lead in the air. Is there anything about this data set suggesting that the confidence interval might not be very​ good?

mean confidence interval ​Yes, the value of 5.40 appears to be an outlier.

A doctor wants to estimate the mean HDL cholesterol of all​ 20- to​ 29-year-old females. How many subjects are needed to estimate the mean HDL cholesterol within 5 points with 99 % confidence assuming s equals 9.4 based on earlier​ studies? Suppose the doctor would be content with 95 % confidence. How does the decrease in confidence affect the sample size​ required?

n = (2.575*9.4 / 5)^2 = 24 n = (1.960 * 9.4 / 5)^2 = 14 Decreasing the confidence level decreases the sample size needed.

A student wants to estimate the mean score of all college students for a particular exam. First use the range rule of thumb to make a rough estimate of the standard deviation of those scores. Possible scores range from 0 to 3000. Use technology and the estimated standard deviation to determine the sample size corresponding to a 99​% confidence level and a margin of error of 100 points. What​ isn't quite right with this​ exercise?

o = range/4 = 750 z_a/2 = 2.576 n = 374 A margin of error of 100 points seems too high to provide a good estimate of the mean score.

Express the confidence interval 0.111 < p < 0.333 in the form p-hat +- E

p-hat = (.333+.111) / 2 = .222 E = (.333-.111) / 2 = .111 p-hat +- E = .222 +- .111

During a period of 11 years 2650 of the people selected for grand jury duty were​ sampled, and 18​% 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, 68.3​% of them were​ immigrants, does it appear that the jury selection process was somehow biased against​ immigrants?

p-hat = .18 z_0.01/2 = z_0.005 = 2.575 E = 2.575* sqrt((.18*.82)/2650) = .01922 .18-.01922 = .161 .18+.01922 = .199 (0.161, 0.199) Yes​, the confidence interval does not include the true percentage of immigrants.

An online site presented this​ question, "Would the recent norovirus outbreak deter you from taking a​ cruise?" Among the 34,165 people who​ responded, 68​% 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?

p-hat = .68 z_a/2 = z_0.025 = 1.96 E = 1.96* sqrt((.68*.32)/34,165) = .0050 .675 < p < .685 ​No, the sample is a voluntary sample and might not be representative of the population.

A study of 419,875 cell phone users found that 145 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 0.0346​% for those not using cell phones. Complete parts​ (a) and​ (b).

p-hat = 145/419875 = 0.00034534 z_0.025 = 1.96 E = 1.98*sqrt((.00034534*.99965466)/419875) = 0.00005620 .00034534-0.00005620 = .00029 .00034534+0.00005620 = 0.00040 0.029% < p < 0.040% No, because 0.0346% is included in the confidence interval

A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 452 babies were​ born, and 339 of them were girls. Use the sample data to construct a 99​% confidence interval estimate of the percentage of girls born. Based on the​ result, does the method appear to be​ effective?

p-hat = 339/452 = .75 z_a/2 = z_.01/2 = z_.005 = 2.575 E = .05245 .698 < p < .802 Yes, the proportion of girls is significantly different from 0.5

In a survey of 1041 ​adults, a poll​ asked, "Are you worried or not worried about having enough money for​ retirement?" Of the 1041 ​surveyed, 551 stated that they were worried about having enough money for retirement. Construct a 99​% confidence interval for the proportion of adults who are worried about having enough money for retirement.

p-hat = 551/1041 = .5293 z = 2.58 LB = .5293 - 2.58 * sqrt(.5293(1-.5293))/2001) UP = .5293 + 2.58 * sqrt(.5293(1-.5293))/2001)

In a sample of seven​ cars, each car was tested for​ nitrogen-oxide emissions​ (in grams per​ mile) and the following results were​ obtained: 0.16, 0.13​, 0.07​, 0.15​, 0.14​, 0.18​, 0.11 .Assuming that this sample is representative of the cars in​ use, construct a 98​% confidence interval estimate of the mean amount of​ nitrogen-oxide emissions for all cars. If the EPA requires that​ nitrogen-oxide emissions be less than 0.165 g/mi​, can we safely conclude that this requirement is being​ met?

s = 0.036 t_a/2 = 3.143 E = 3.143 * 0.036/sqrt(7) = 0.043 sample mean = .134 x - E < u < x + e 0.019 < u < 0.177 ​No, it is possible that the requirement is being​ met, but it is also very possible that the mean is not less than 0.165 g/mi.

The​ _______ is the best point estimate of the population mean.

sample mean

Twelve different video games showing substance use were observed and the duration times of game play​ (in seconds) are listed below. The design of the study justifies the assumption that the sample can be treated as a simple random sample. Use the data to construct a 95​% confidence interval estimate of mu​, the mean duration of game play 4061 3886 3865 4023 4309 4821 4645 4021 5024 4812 4331 4325

sample mean = 4343.583 s = 396.344 t_a/2 = 2.201 E = 2.201*396.344/sqrt(12) = 251.827 4091.8 < u < 4595.4

Listed below are measured amounts of lead​ (in micrograms per cubic​ meter, or µg/m3​) in the air. The EPA has established an air quality standard for lead of 1.5 µg/m3. The measurements shown below were recorded at a building on different days. 5.40 1.10 0.39 0.77 0.84 0.90 Use the given values to construct a 95​% confidence interval estimate of the mean amount of lead in the air. Is there anything about this data set suggesting that the confidence interval might not be very​ good?

sample std. dev. = 1.892 t_a/2 = 2.571 (one-tail) E = 2.571 * 1.892/sqrt(6) = 1.986 sample mean = 1.567 1.567 - 1.986 < u < 1.567 + 1.986 -0.419 < u < 3.553 In order to calculate the confidence​ interval, we assumed that the population was normally distributed. This assumption is not correct because of which​ value? 5.4

Do one of the​ following, as appropriate.​ (a) Find the critical value z_a/2, ​(b) find the critical value t_a/2​, ​(c) state that neither the normal nor the t distribution applies. Confidence level 90​%; n=400​; sigma is unknown​; population appears to be very skewed.

t_a/2 = 1.649 (use two-tailed t-value)

A genetic experiment with peas resulted in one sample of offspring that consisted of 370 green peas and 143 yellow peas. a. Construct a 95 ​% confidence interval to estimate of the percentage of yellow peas. b. It was expected that​ 25% of the offspring peas would be yellow. Given that the percentage of offspring yellow peas is not​ 25%, do the results contradict​ expectations?

total peas = 513 p-hat = 143/513 = .2788 z_0.025 = 1.96 E = .0388 .240 < p < .318 No, the confidence interval includes​ 0.25, so the true percentage could easily equal​ 25%

A study asked​ respondents, "If ever​ married, how old were you when you first​ married?" The results are summarized in the MINITAB excerpt that follows. Complete parts​ (a) and​ (b) below. MEAN = 21.450 95.0% CI (21.400, 21.500)

upper bound = point estimate + margin of error 21.450= 21.500 + E E = 0.05 One can be 90​% confident that the mean age of people when first married is between 21.400 and 21.500 years.

Salaries of 47 college graduates who took a statistics course in college have a​ mean, x overbar​, of $61,000. Assuming a standard​ deviation, sigma​, of ​$13,424​, construct a 90​% confidence interval for estimating the population mean mu

z_0.05 = 1.645 E = 1.645 * 13424/sqrt(47) = 3221.061 5779 < u < 64221

Do one of the​ following, as appropriate.​ (a) Find the critical value z_a/2, ​(b) find the critical value t_a/2​, ​(c) state that neither the normal nor the t distribution applies. Confidence level 90​%; n=27​; sigma=31.2​; population appears to be very skewed.

z_a/2 = 1.645

In order to estimate the mean amount of time computer users spend on the internet each​ month, how many computer users must be surveyed in order to be 95​% confident that your sample mean is within 15 minutes of the population​ mean? Assume that the standard deviation of the population of monthly time spent on the internet is 250 min. What is a major obstacle to getting a good estimate of the population​ mean? Use technology to find the estimated minimum required sample size.

z_a/2 = 1.96 o = 250 E = 15 n = 1067.111 minimum required is 1068 It is difficult to precisely measure the amount of time spent on the​ internet, invalidating some data values.

An IQ test is designed so that the mean is 100 and the standard deviation is 12 for the population of normal adults. Find the sample size necessary to estimate the mean IQ score of statistics students such that it can be said with 99​% confidence that the sample mean is within 5 IQ points of the true mean. Assume that sigma = 12 and determine the required sample size. Then determine if this is a reasonable sample size for a real world calculation.

z_a/2 = 2.58 n = 39 n = (z_a/2 * sigma / E)^2 Yes. This number of IQ test scores is a fairly small number.

Given that the required sample size is relatively​ small, could he simply survey the adults at the nearest​ college?

​No, a sample of students at the nearest college is a convenience​ sample, not a simple random​ sample, so it is very possible that the results would not be representative of the population of adults.

A physician wants to develop criteria for determining whether a​ patient's pulse rate is​ atypical, and she wants to determine whether there are significant differences between males and females. Use the sample pulse rates below.

​No, because the two confidence intervals​ overlap, we cannot conclude that the two population means are different.

b. Does the given dotplot appear to satisfy these​ requirements?

​Yes, because the dotplot resembles a normal distribution and the sample size is greater than 30.

​(a) Because the sample size is​ small, we must verify that the data come from a population that is normally distributed and that the sample size does not contain any outliers. Are the conditions for constructing a confidence interval about the mean​ satisfied?

​Yes, the population is normally distributed and the sample does not contain any outliers.