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Week 6 Homework full solution guide 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. 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. What are the three steps for determining the sample distribution of the sample mean?
- Obtain a simple random sample of size n.
- Compute the sample mean.
- 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 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 Central Limit Theorem? n>30 for the sample mean (overbar x) to be normal if the parent population is not known to be normal. 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.) 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. 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 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 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. We have an expert-written solution to this problem! As the sample size n increases, what happens to the standard error of the mean? The standard error of the mean decreases. We have an expert-written solution to this problem! 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?
(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. 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. The sampling distribution of overbar x is normal with μx=206 and σx=2. (c) normalcdf (-9999999, 203, 206, (10/(√23)) = 0. If 100 independent random samples of size n= 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. If 100 independent random samples of size n= 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.
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= 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. (c) invNorm (0.10, 16.1, (4.8/(√45)) = 15. 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. (c) P(overbar x≥3.2)
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) Describe the sampling distribution of ^p. Assume the size of the population is 25,000. n=700, p=0. (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, 25,0000.05 = 1250. n = 700, so it is ≤ 0.05N np(1-p) = (7000.6)*(1-0.6) = 160, so it is ≥ 10 (b) μ ^p = p = 0. (c) σ ^p = √p(1-p)/n = √0.6(1-0.6)/700 = 0. 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. (d) normalcdf (0.44, 9999999, 0.4, 0.056569) = 0. (e) normalcdf (-9999999, 0.28, 0.4, 0.56569) = 0. 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) = (1000.76)(1-0.76) = 18.
(b) What percent of sample proportions results in a 95% confidence interval that does not include the population proportion? (a) 95% (b) 5% 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. (b) 1. 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% We have an expert-written solution to this problem! A confidence interval for an unknown parameter consists of what? An interval of numbers based on a point estimate. 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. 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. 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. 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. We have an expert-written solution to this problem! When constructing a 95% confidence interval, α = ______.
When constructing a 99% confidence interval, α = ______.
When α = 0.003, we are constructing a ______ confidence interval. 99.7% 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. Compute the critical value z v α/2 that corresponds to a 85% level of confidence. 1-0.85 = 0. 0.15/2 = 0. 0.075+0.85 = 0.
(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. 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. 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. 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. 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. n(1-^p) = 150(1-0.27) = 109. 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.
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. 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. (b) n = 2276, x = 401, ^p = 0. n^p(1-^p) = 22760.176(1-0.176) = 330. 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. 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. Upper bound=0. n= (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. (b) 0.422-0.265 = 0. (c) p = x/n 0.265 = x/
- The t-distribution is different for different degrees of freedom.
- The t-distribution is centered at 0 and is symmetric about 0.
- 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.
- As t increases or decreases without bound, the graph approaches, but never equals, 0.
- 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.
- 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. 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 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. We have an expert-written solution to this problem!
Determine the t-value in each of the cases. (a) Find the t-value such that the area in the right tail is 0. with 15 degrees of freedom. (b) Find the t-value such that the area in the right tail is 0. with 11 degrees of freedom. (c) Find the t-value such that the area left of the t-value is 0. 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. (b) invT (1-0.20, 11) = 0. (c) invT (0.02, 14) = -2. (d) 1-0.90 = 0.10/2 = 0.05 per tail invT (1-0.05, 11) = 1. What are the requirements for constructing a confidence interval (t-interval) for a population mean μ?
- The data must come from a random experiment (simple random sample).
- n < 0.05N
- It must be normally distributed (without outliers) or n ≥ 30. 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 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.
How is margin of error (E) calculated?
As the level of confidence decreases, the size of the interval decreases. (d) No, the population needs to be normally distributed. 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. n = ((2.57516.1)/4)² = 108 (b) 90% critical value = 1. n = ((1.64516.1/4)² = 44 Decreasing the confidence level decreases the sample size needed. What type of data is needed to construct a confidence interval for a population proportion p? Qualitative data with two outcomes (success or failure). 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)≥ n≤0.05N What type of data is needed to construct a confidence interval for a population mean μ? Quantitative data 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)
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. 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. α = 0.005 (1-0.99 = 0.01/2 = 0.005) Critical value for 99% = 2. Lower Bound: 0.524-(2.576√(0.524(1-0.524)/1019)) = 0.524-0.04030152 = 0. Upper Bound: 0.524+(2.576√(0.524(1-0.524)/1019)) = 0.524+0.04030152 = 0. There is 99% confidence that the true proportion of worried adults is between 0.484 and 0.564. For what type of variable does it make sense to construct a confidence interval about a population proportion? Qualitative with 2 possible outcomes.
