Week 6 Homework

What is the shape of the distribution of the sample mean as the sample size increases?

It becomes approximately normal as the sample size, n, increases, regardless of the shape of the underlying population.

What is the shape of the distribution of the sample proportion as the sample size increases?

It becomes approximately normal.

What is the value z v α/2 called?

Critical value. It represents the number of standard deviations the sample statistic can be from the parameter and still result in an interval that includes the parameter.

Spread: As the sample size n increases, what happens to the standard deviation of the distribution of the sample proportion?

It decreases.

What is the sampling distribution of the sample mean overbar x?

All possible values of the random variable computed from a sample size of n from a population with the mean µ and standard deviation σ.

A confidence interval for an unknown parameter consists of what?

An interval of numbers based on a point estimate.

Spread (σ sub-overbar x): As the sample size n increases, what happens to the standard deviation of the distribution of the sample mean?

As n increases, the standard deviation will decrease.

What type of data is needed to construct a confidence interval for a population proportion p?

Qualitative data with two outcomes (success or failure).

What type of data is needed to construct a confidence interval for a population mean µ?

Quantitative data

When constructing a 99% confidence interval, α = ______.

0.01

What are the requirements for constructing a confidence interval (t-interval) for a population mean µ?

1. The data must come from a random experiment (simple random sample). 2. n < 0.05N 3. It must be normally distributed (without outliers) or n ≥ 30.

A trade magazine routinely checks the​ drive-through service times of​ fast-food restaurants. A 95​% confidence interval that results from examining 579 customers in one​ fast-food chain's​ drive-through has a lower bound of 160.1 seconds and an upper bound of 163.1 seconds. What does this​ mean?

One can be 95​% confident that the mean​ drive-through service time of this​ fast-food chain is between 160.1 seconds and 163.1 seconds.

A​ ________ ________ is the value of a statistic that estimates the value of a parameter.

Point estimate

What does "90​% confidence" mean in a 90​% confidence​ interval?

If 100 different confidence intervals are​ constructed, each based on a different sample of size n from the same​ population, then we expect 90 of the intervals to include the parameter and 10 to not include the parameter.

Center: As the sample size n increases, what does the mean of the sampling distribution of the sample proportion equal?

It equals the population, p.

Center: As the sample size n increases, what does the mean of the distribution of the sample mean, overbar x, equal?

It is equal to the mean of the underlying population. (The center is not affected.)

How are upper and lower bounds calculated?

Lower bound: overbar x - t v α/2 * (s/√n), or point estimate - margin of error Upper bound: overbar x + t v α/2 * (s/√n), or point estimate + margin of error

What are the formulas for the mean and standard deviation of the sampling distribution of ^p.

Mean: µ sub-p-hat = p Standard deviation: σ sub-p-hat = √(p(1-p)/n)

T/F: The mean of the sampling distribution of ^p is p.

True. The mean of the sample distribution of the sampling proportion equals the population​ proportion, p.

In a trial of 163 patients who received​ 10-mg doses of a drug​ daily, 39 reported headache as a side effect. Obtain a point estimate for the population proportion of patients who received​ 10-mg doses of a drug daily and reported headache as a side effect.

^p = 39/163 = 0.24

When constructing​ 95% confidence intervals for the mean when the parent population is right skewed and the sample size is​ small, the proportion of intervals that include the population mean approaches​ _____ as the sample​ size, n, increases.

0.95

Compute the critical value z v α/2 that corresponds to a 85​% level of confidence.

1-0.85 = 0.15 0.15/2 = 0.075 0.075+0.85 = 0.925 invNorm (0.925, 0, 1) = 1.44

What are the three steps for determining the sample distribution of the sample mean?

1. Obtain a simple random sample of size n. 2. Compute the sample mean. 3. Assuming that we are sampling from a finite population, repeat steps 1 and 2 until all distinct simple random samples of size n have been obtained.

What are the six properties of a t-distribution?

1. The t-distribution is different for different degrees of freedom. 2. The t-distribution is centered at 0 and is symmetric about 0. 3. The area under the curve is 1. The area under the curve to the right of 0 equals the area under the curve to the left of 0, which equals 1/2. 4. As t increases or decreases without bound, the graph approaches, but never equals, 0. 5. The area in the tails of the t-distribution is a little greater than the area in the tails of the standard normal distribution because we are using s as an estimate of σ, thereby introducing further variability into the t-statistic. 6. As the sample size n increases, the density curve of t gets closer to the standard normal density curve. This result occurs because as the sample size increases, the values of s get closer to the value of σ by the Law of Large Numbers. See the figure to the right, which shows the t-distribution for samples of size n=5 and n=15, along with the standard normal density curve.

As the sample size n increases, what happens to the margin of error?

It decreases. If the sample size is quadrupled, the margin of error will be cut in half.

Suppose a polling agency reported that 45.0​% of registered voters were in favor of raising income taxes to pay down the national debt. The agency states that results are based on telephone interviews with a random sample of 1017 registered voters. Suppose the agency states the margin of error for 90​% confidence is 2.6​%. Determine and interpret the confidence interval for the proportion of registered voters who are in favor of raising income taxes to pay down the national debt.

Point estimate ± margin of error: 0.45 - 0.026 = 0.424 0.45 + 0.026 = 0.476 We are 90% confident that the proportion of registered voters in favor of raising income taxes to pay down the national debt is between 0.424 and 0.476.

The​ _____ _____, denoted ^p​, is given by the formula ^p=​_____, where x is the number of individuals with a specified characteristic in a sample of n individuals.

Sample, Proportion, x/n

For the​ following, indicate whether a confidence interval for a proportion or mean should be constructed to estimate the variable of interest. A researcher with a golf association obtained a random sample of 25 rounds of golf on a Saturday morning and recorded the time it took to complete the round. The goal of the research was to estimate the amount of time it typically takes to complete a round of golf on Saturday morning.

The confidence interval for a population mean should be constructed because the variable of interest is time to complete the round, which is a quantitative variable.

Researchers within an organization asked a random sample of 1016 adults aged 21 years or​ older, "Right​ now, do you think the state of moral values in the country as a whole is getting​ better, or getting​ worse?"

The confidence interval for a proportion should be constructed because the variable of interest is an individual's opinion, which is a qualitative variable.

Suppose a simple random sample of size n is obtained from a population whose distribution is skewed right. As the sample size n​ increases, what happens to the shape of the distribution of the sample​ mean?

The distribution becomes approximately normal. According to the Central Limit Theorem, if the mean values for increasing sample sizes are obtained, the distribution of sample means will be normally distributed, even if the individual samples do not have normal distributions. Typically, sample sizes of 30 or greater are recommended.

A group conducted a poll of 2049 likely voters just prior to an election. The results of the survey indicated that candidate A would receive 45​% of the popular vote and candidate B would receive 44​% of the popular vote. The margin of error was reported to be 2​%. The group reported that the race was too close to call. Use the concept of a confidence interval to explain what this means.

The margin of error suggests candidate A may receive between 43​% and 47​% of the popular vote and candidate B may receive between 42​% and 46​% of the popular vote. Because the poll estimates overlap when accounting for margin of​ error, the poll cannot predict the winner.

What is the point estimate for the population mean?

The point estimate for the population mean, µ, is the sample mean, overbar x.

Whether a confidence interval contains the population parameter depends solely on what?

The value of the sample statistic. Any sample statistic that is in the tails of the sampling distribution will result in a confidence interval that does not include the population parameter.

Explain why the​ t-distribution has less spread as the number of degrees of freedom increases.

The​ t-distribution has less spread as the degrees of freedom increase​ because, as n​ increases, s becomes closer to σ by the law of large numbers. As the sample size n​ increases, the density curve of t gets closer to the standard normal density curve. The variability introduced into the​ t-statistic becomes less.

In a survey of 1019 adults, a polling agency​ asked, "When you​ retire, do you think you will have enough money to live comfortably or not?" Of the 1019 surveyed, 534 stated that they were worried about having enough money to live comfortably in retirement. Construct a 99​% confidence interval for the proportion of adults who are worried about having enough money to live comfortably in retirement.

^p = 534/1019 = 0.524 α = 0.005 (1-0.99 = 0.01/2 = 0.005) Critical value for 99% = 2.575 Lower Bound: 0.524-(2.576*√(0.524(1-0.524)/1019)) = 0.524-0.04030152 = 0.484 Upper Bound: 0.524+(2.576*√(0.524(1-0.524)/1019)) = 0.524+0.04030152 = 0.564 There is 99​% confidence that the true proportion of worried adults is between 0.484 and 0.564.

Put the following in order for the most area in the tails of the distribution. ​(a) Standard Normal Distribution ​(b) Student's​ t-Distribution with 25 degrees of freedom. ​(c) Student's​ t-Distribution with 40 degrees of freedom.

b, c, a

What is the Central Limit Theorem?

n>30 for the sample mean (overbar x) to be normal if the parent population is not known to be normal.

Aside from the fact that the sample must be obtained by simple random sampling or through a randomized experiment, list the two conditions that must be met when constructing a confidence interval for a population proportion p.

np(1-p)≥10 n≤0.05N

Aside from the fact that the sample must be obtained by simple random sampling or through a randomized experiment and the sample size must be small relative to the size of the population, what other condition must be satisfied?

n≥30 (Central Limit Theorem invoked) n<30 (needs to be normally distributed with no outliers)

Determine μ overbar x and σ overbar x from the given parameters of the population and sample size. μ=85​ σ=22​ n=29

µ overbar x = 85 σ overbar x = σ/√n = 22/√29 = 4.085

When α = 0.003, we are constructing a ______ confidence interval.

99.7%

What is the sampling distribution of a statistic?

A probability distribution for all possible values of the statistic computed from a sample of size n.

When constructing​ 95% confidence intervals for the mean when the parent population is right skewed and the sample size is​ small, the proportion of intervals that include the population mean is​ _______ 0.95.

Below

How is margin of error (E) calculated?

E = (upper bound - lower bound)/2

T/F: A​ 95% confidence interval may be interpreted by saying there is a​ 95% probability that the interval includes the unknown parameter.

False. A​ 95% confidence interval does not mean that there is a​ 95% probability that the interval contains the parameter. The​ 95% in a​ 95% confidence interval represents the proportion of all samples that will result in intervals that include the population proportion.

Two​ researchers, Jaime and​ Mariya, are each constructing confidence intervals for the proportion of a population who is​ left-handed. They find the point estimate is 0.26. Each independently constructed a confidence interval based on the point​ estimate, but​ Jaime's interval has a lower bound of 0.204 and an upper bound of 0.301​, while​ Mariya's interval has a lower bound of 0.258 and an upper bound of 0.262. Which interval is​ wrong? Why?

Jaime​'s interval is wrong because it is not centered on the point estimate.

The​ _______ represents the expected proportion of intervals that will contain the parameter if a large number of different samples of size n is obtained. It is denoted​ _______.

Level of confidence, (1-α) * 100%

Suppose a random sample of n=320 teenagers 13 to 17 years of age was asked if they use social media. Of those​ surveyed, 243 stated that they do use social media. Find the sample proportion of teenagers 13 to 17 years of age who use social media.

^p = x/n = 243/320 = 0.759

Fill in the blanks to complete the sentences below. (a) As the number of samples​ increases, the proportion of​ 95% confidence intervals that include the population proportion approaches​ ______. (b) If a​ 95% confidence interval results in a sample proportion that does not include the population​ proportion, then the sample proportion is more than​ ______ standard errors from the population proportion.

(a) 0.95 (b) 1.96

Fill in the blanks to complete the following statements. (a) For the shape of the distribution of the sample proportion to be approximately​ normal, it is required that ​np(1−​p)≥​______. (b) Suppose the proportion of a population that has a certain characteristic is 0.3. The mean of the sampling distribution of ^p from this population is μ^p=​______.

(a) 10 (b) 0.3

Complete parts ​(a​) through ​(d) for the sampling distribution of the sample mean shown in the accompanying graph. (a) What is the value of μ v overbar x​? ​(b) What is the value of σ v overbar x​? ​(c) If the sample size is n=16​, what is likely true about the shape of the​ population? ​(d) If the sample size is n=16​, what is the standard deviation of the population from which the sample was​ drawn? The standard deviation of the population from which the sample was drawn is 4040.

(a) 400 (b) 10 (c) The shape of the population is approximately normal. (d) (√16) * 10 = 40

Determine the point estimate of the population​ proportion, the margin of error for the following confidence​ interval, and the number of individuals in the sample with the specified​ characteristic, x, for the sample size provided. Lower bound=0.108​ Upper bound=0.422​ n=1200 (a) The point estimate of the population proportion is _______. (b) The margin of error is _______. (c) The number of individuals in the sample with the specified characteristic is _______.

(a) p = (0.108+0.422)/2 = 0.265 (b) 0.422-0.265 = 0.157 (c) p = x/n 0.265 = x/1200 x = 0.265 * 1200 = 318

A survey of 2276 adults in a certain large country aged 18 and older conducted by a reputable polling organization found that 401 have donated blood in the past two years. (a) Obtain a point estimate for the population proportion of adults in the country aged 18 and older who have donated blood in the past two years. (b) Verify that the requirements for constructing a confidence interval about p are satisfied. (c) Construct and interpret a 90​% confidence interval for the population proportion of adults in the country who have donated blood in the past two years.

(a) 401/2276 = 0.176 (b) n = 2276, x = 401, ^p = 0.176 n^p(1-^p) = 2276*0.176*(1-0.176) = 330.349 The sample can be assumed to be a simple random​ sample, the value of n^p(1−^p) is 330.349​, which is greater than or equal to​ 10, and the sample size can be assumed to be less than or equal to​ 5% of the population size. (c) STAT-->TEST-->1-PropZInt (401, 2276, 0.90) W e are 90​%confident the proportion of adults in the country aged 18 and older who have donated blood in the past two years is between 0.163 and 0.189.

The horizontal axis in the sampling distribution of ^p represents all possible sample proportions from a simple random sample of size n. (a) What percent of sample proportions results in a 95​% confidence interval that includes the population​ proportion? (b) What percent of sample proportions results in a 95​% confidence interval that does not include the population​ proportion?

(a) 95% (b) 5%

A doctor wants to estimate the mean HDL cholesterol of all​ 20- to​ 29-year-old females. (a) How many subjects are needed to estimate the mean HDL cholesterol within 4 points with 99% confidence assuming s=16.1 based on earlier​ studies? (b) Suppose the doctor would be content with 90% confidence. How does the decrease in confidence affect the sample size​ required?

(a) 99% critical value = 2.575 n = ((2.575*16.1)/4)² = 108 (b) 90% critical value = 1.645 n = ((1.645*16.1/4)² = 44 Decreasing the confidence level decreases the sample size needed.

Suppose a simple random sample of size n=75 is obtained from a population whose size is N=25,000 and whose population proportion with a specified characteristic is p=0.4. (a) Describe the sampling distribution of ^p. (b) Determine the mean of the sampling distribution of ^p. (c) Determine the standard deviation of the sampling distribution of ^p. (d) What is the probability of obtaining x=33 or more individuals with the​ characteristic? That​ is, what is P(^p≥0.44​)? (e) What is the probability of obtaining x=21 or fewer individuals with the​ characteristic? That​ is, what is P(^p≤0.28​)?

(a) Approximately normal because n≤0.05N and np(1-p)≥10. (b) 0.4 (same as p) (c) σ^p = √(0.4(1-0.4)/(75)) = 0.056569 (d) normalcdf (0.44, 9999999, 0.4, 0.056569) = 0.2398 (e) normalcdf (-9999999, 0.28, 0.4, 0.56569) = 0.0169

Describe the sampling distribution of ^p. Assume the size of the population is 25,000. n=700​, p=0.6 (a) Choose the phrase that best describes the shape of the sampling distribution of ^p below. (b) Determine the mean of the sampling distribution of ^p. (c) Determine the standard deviation of the sampling distribution of ^p.

(a) Approximately normal because n≤0.05N and np(1-p)≥10. N = 25,000 25,000*0.05 = 1250. n = 700, so it is ≤ 0.05N np(1-p) = (700*0.6)*(1-0.6) = 160, so it is ≥ 10 (b) µ ^p = p = 0.6 (c) σ ^p = √p(1-p)/n = √0.6(1-0.6)/700 = 0.019

Suppose a simple random sample of size n=37 is obtained from a population with μ=67 and σ=17. ​(a) What must be true regarding the distribution of the population in order to use the normal model to compute probabilities regarding the sample​ mean? (b) Assuming the normal model can be​ used, describe the sampling distribution overbar x. ​(c) Assuming the normal model can be​ used, determine ​P(overbar x<71.1​). ​(d) Assuming the normal model can be​ used, determine ​P(overbar x≥69.1​).

(a) Since the sample size is large enough, the population distribution does not need to be normal. (b) Approximately normal​, with µ v overbar x=67 and σ v overbar x=17/(√37) (c) normalcdf (-9999999, 71.1, 67, (17/(√37)) = 0.9288 (d) normalcdf (69.1, 9999999, 67, (17/√(37)) = 0.2262

A simple random sample of size n is drawn from a population that is normally distributed. The sample​ mean, overbar x​, is found to be 115​, and the sample standard​ deviation, s, is found to be 10. ​(a) Construct a 90​% confidence interval about μ if the sample​ size, n, is 21. ​(b) Construct a 90​% confidence interval about μ if the sample​ size, n, is 16. How does decreasing the sample size affect the margin of​ error, E? ​(c) Construct an 80​% confidence interval about μ if the sample​ size, n, is 21. Compare the results to those obtained in part​ (a). How does decreasing the level of confidence affect the size of the margin of​ error, E? ​(d) Could we have computed the confidence intervals in parts​ (a)-(c) if the population had not been normally​ distributed?

(a) TInterval Input: Stats (115, 10, 21, 0.9) = 111.2, 118.8 (b) TInterval Input: Stats (115, 10, 16, 0.9) = 110.6, 119.4 As the sample size decreases​, the margin of error increases. (c) TInterval Input: Stats (115, 10, 21, 0.8) = 112.1, 117.9. As the level of confidence decreases, the size of the interval decreases. (d) ​No, the population needs to be normally distributed.

In a survey of 2055 adults in a certain country conducted during a period of economic​ uncertainty, 65​% thought that wages paid to workers in industry were too low. The margin of error was 2 percentage points with 95​% confidence. For parts​ (a) through​ (d) below, which represent a reasonable interpretation of the survey​ results? For those that are not​ reasonable, explain the flaw. ​(a) We are 95​% confident 65​% of adults in the country during the period of economic uncertainty felt wages paid to workers in industry were too low. (b) We are 93​% to 97​% confident 65​% of adults in the country during the period of economic uncertainty felt wages paid to workers in industry were too low. (c) We are 95​% confident that the interval from 0.63 to 0.67 contains the true proportion of adults in the country during the period of economic uncertainty who believed wages paid to workers in industry were too low. (d) In 95​% of samples of adults in the country during the period of economic​ uncertainty, the proportion who believed wages paid to workers in industry were too low is between 0.63 and 0.67.

(a) The interpretation is flawed. The interpretation provides no interval about the population proportion. (b) The interpretation is flawed. The interpretation indicates that the level of confidence is varying. (c) The interpretation is reasonable. (d) The interpretation is flawed. The interpretation suggests that this interval sets the standard for all the other​ intervals, which is not true.

According to a​ study, the proportion of people who are satisfied with the way things are going in their lives is 0.76. Suppose that a random sample of 100 people is obtained. (a) Suppose the random sample of 100 people is​ asked, "Are you satisfied with the way things are going in your​ life?" Is the response to this question qualitative or​ quantitative? Explain. (b) Explain why the sample​ proportion, ^p​, is a random variable. What is the source of the​ variability? (c) Describe the sampling distribution of ^p​, the proportion of people who are satisfied with the way things are going in their life. Be sure to verify the model requirements. (d) In the sample obtained in part​ (a), what is the probability that the proportion who are satisfied with the way things are going in their life exceeds 0.79​? (e) Using the distribution from part​ (c), would it be unusual for a survey of 100 people to reveal that 70 or fewer people in the sample are satisfied with their​ lives?

(a) The response is qualitative because the responses can be classified based on the characteristic of being satisfied or not. (b) The sample proportion ^p is a random variable because the value of ^p varies from sample to sample. The variability is due to the fact that different people feel differently regarding their satisfaction. (c) np(1-p) = (100*0.76)*(1-0.76) = 18.240 µ^p = ^p = 0.76 σ^p = √(p*(1-p)/n) = √(0.76*(1-0.76)/100) = 0.0427 Since the sample size is no more than​ 5% of the population size and np(1−p)=18.240≥​10, the distribution of ^p is approximately normal with μ^p=0.760 and σ^p=0.043. (d) normalcdf (0.79, 9999999, 0.76, 0.043) = 0.2427 (e) normalcdf (-9999999, 0.7, 0.76, 0.043) = 0.0815 The probability that 70 or fewer people in the sample are satisfied is 0.0815​, which is not unusual because this probability is not less than 5​%.

The shape of the distribution of the time required to get an oil change at a 15​-minute ​oil-change facility is unknown.​ However, records indicate that the mean time is 16.1 minutes​, and the standard deviation is 4.8 minutes. (a) To compute probabilities regarding the sample mean using the normal​ model, what size sample would be​ required? (b) What is the probability that a random sample of n=45 oil changes results in a sample mean time less than 15 minutes? (c) Suppose the manager agrees to pay each employee a​ $50 bonus if they meet a certain goal. On a typical​ Saturday, the​ oil-change facility will perform 45 oil changes between 10 A.M. and 12 P.M. Treating this as a random​ sample, at what mean​ oil-change time would there be a​ 10% chance of being at or​ below? This will be the goal established by the manager.

(a) The sample size needs to be greater than 30. (b) normalcdf (-9999999, 15, 16.1, (4.8/(√45) = 0.0621 (c) invNorm (0.10, 16.1, (4.8/(√45)) = 15.2

The acceptable level for insect filth in a certain food item is 3 insect fragments​ (larvae, eggs, body​ parts, and so​ on) per 10 grams. A simple random sample of 40 ​ten-gram portions of the food item is obtained and results in a sample mean of overbar x=3.2 insect fragments per​ ten-gram portion. (a) Why is the sampling distribution of overbar x approximately​ normal? (b) What is the mean and standard deviation of the sampling distribution of overbar x assuming μ=3 and σ=√3​? (c) What is the probability a simple random sample of 40 ten-gram portions of the food item results in a mean of at least 3.2 insect​ fragments? Is this result​ unusual? What might we​ conclude?

(a) The sampling distribution of overbar x is approximately normal because the sample size is large enough. (b) µ v overbar x = 3 σ v overbar x = (√3)/(√40) = 0.274 (c) ​P(overbar x≥3.2​) normalcdf (3.2, 9999999, 3, 0.274) = 0.2327 This result is not unusual because its probability is large. Since this result is not unusual, it is not reasonable to conclude that the population mean is higher than 3.

Determine the​ t-value in each of the cases. (a) Find the​ t-value such that the area in the right tail is 0.025 with 15 degrees of freedom. (b) Find the​ t-value such that the area in the right tail is 0.20 with 11 degrees of freedom. (c) Find the​ t-value such that the area left of the​ t-value is 0.02 with 14 degrees of freedom.​ (d) Find the critical​ t-value that corresponds to 90​% confidence. Assume 11 degrees of freedom.

(a) invT (1-0.025, 15) = 2.131 (b) invT (1-0.20, 11) = 0.876 (c) invT (0.02, 14) = -2.264 (d) 1-0.90 = 0.10/2 = 0.05 per tail invT (1-0.05, 11) = 1.796

In a trial of 150 patients who received​ 10-mg doses of a drug​ daily, 27 reported headache as a side effect. (a) Are the requirements for constructing a confidence​ satisfied? (b) Construct and interpret a 95​% confidence interval for the population proportion of patients who receive the drug and report a headache as a side effect.

(a) n^p = 150 * 0.27 = 40.5 n(1-^p) = 150(1-0.27) = 109.5 Both 40.5 and 109.5 ≥ 10, so yes, the requirements for constructing a confidence interval are satisfied. (B) STAT-->TEST-->1-PropZInt (27, 150, 0.95) = (0.11852, 0.24148) One can be 95​% confident that the proportion of patients who receive the drug and report a headache as a side effect is between 0.119 and 0.241.

Suppose the lengths of the pregnancies of a certain animal are approximately normally distributed with mean μ=206 days and standard deviation σ=10 days. (a) What is the probability that a randomly selected pregnancy lasts less than 203 days? Interpret this probability. (b) Suppose a random sample of 23 pregnancies is obtained. Describe the sampling distribution of the sample mean length of pregnancies. (c) What is the probability that a random sample of 23 pregnancies has a mean gestation period of 203 days or​ less? Interpret this probability. (d) What is the probability that a random sample of 50 pregnancies has a mean gestation period of 203 days or​ less? Interpret this probability. (e) What might you conclude if a random sample of 50 pregnancies resulted in a mean gestation period of 203 days or​ less? (f) What is the probability a random sample of size 19 will have a mean gestation period within 9 days of the​ mean?

(a) normalcdf (-9999999, 203, 206, 10) = 0.3821 If 100 pregnant individuals were selected independently from this​ population, we would expect 38 pregnancies to last less than 203 days. (b) n = 23, σ = σ/√n = 10/√23 = 2.0851 The sampling distribution of overbar x is normal with μx=206 and σx=2.0851 (c) normalcdf (-9999999, 203, 206, (10/(√23)) = 0.0751 If 100 independent random samples of size n=23 pregnancies were obtained from this​ population, we would expect 8 sample(s) to have a sample mean of 203 days or less. (d) normalcdf (-9999999, 203, 206, (10/(√50)) = 0.0169 If 100 independent random samples of size n=50 pregnancies were obtained from this​ population, we would expect 2 sample(s) to have a sample mean of 203 days or less. (e) This result would be unusual, so the sample likely came from a population whose mean gestation period is less than 206 days. (f) Within 9 days of the mean = 197 to 215 days normalcdf (197, 215, 206, (10/√19)) = 0.9999

A researcher wishes to estimate the proportion of adults who have​ high-speed Internet access. What size sample should be obtained if she wishes the estimate to be within 0.03 with 99​% confidence if: (a) she uses a previous estimate of 0.58​? (b) she does not use any prior​ estimates?

(a) α = 1-0.99 = 0.01 0.01/2 = 0.005 Critical value of 0.005 = 2.575 n = 0.58(1-0.58)*(2.575/0.03)^2 = 1795 (b) Use 0.5 for estimated value. n = 0.5(1-0.5)*(2.575/0.03)^2 = 1842

How is point estimate calculated?

(upper bound + lower bound)/2

When constructing a 95% confidence interval, α = ______.

0.05

Katrina wants to estimate the proportion of adults who read at least 10 books last year. To do​ so, she obtains a simple random sample of 100 adults and constructs a​ 95% confidence interval. Matthew also wants to estimate the proportion of adults who read at least 10 books last year. He obtains a simple random sample of 400 adults and constructs a​ 99% confidence interval. Assuming both Katrina and Matthew obtained the same point​ estimate, whose estimate will have the smaller margin of​ error? Justify your answer.

Matthew's estimate will have the smaller margin of error because the larger sample size more than compensates for the higher level of confidence.

List the formulas for the mean, standard deviation, and checking for independence of the sampling distribution of overbar x.

Mean: µ sub-overbar x = µ Standard deviation: σ sub-overbar x = σ/√n Independence: n≤0.05N

A simple random sample of size n=31 is obtained from a population that is skewed left with μ=70 and σ=99. Does the population need to be normally distributed for the sampling distribution of overbar x to be approximately normally​ distributed? Why? What is the sampling distribution of overbar x?

No. The central limit theorem states that regardless of the shape of the underlying​ population, the sampling distribution of overbar x becomes approximately normal as the sample​ size, n, increases.

For what type of variable does it make sense to construct a confidence interval about a population​ proportion?

Qualitative with 2 possible outcomes.

A t-distribution is _______.

Symmetric about 0.

The level of confidence represents the expected proportion of intervals that will contain what?

The parameter if a large number of different samples is obtained.

As the sample size n​ increases, what happens to the standard error of the​ mean?

The standard error of the mean decreases.

Under what condition is the shape of the sampling distribution of ^p approximately normal?

n*p(1-p) ≥ 10