DS - Chapter 15
15. 1. 41-T A sample of 150 calls to a customer help line during the week found callers there were kept waiting on average for 21 minutes with s=6. Complete a through d.
(a) Find the margin for error for this survey if we use a 95% confidence interval for the length of time all customers during this period are kept waiting. Answers: (a) Margin of Error = 0.971 (Round to three decimal places as needed.) REASON= use excel to calculate (b) Interpret for management the margin of error. Choose the correct answer below. Answers: A. Management can be 95% confident that average waiting time for all callers is within a number of minutes equal to the margin of error of the average wait of 21 minutes. (c) If we only need to be 90% confident, does the confidence interval become wider or narrower? Answers: The need to only be 90% confident makes the margin of error smaller, because the multiplier (or percentile) that comes from wanting 90% percent confidence and using a t-distribution (or normal model) to describe the sampling distribution is smaller than the multiplier for 95% interval. Thus, the confidence interval becomes narrower. (d) Find the 90% confidence interval. Answers: The confidence interval is (20.189,21.811). REASON= keep n=150 meaning don't n-1 to get df.
15. 1. 6 Match the item with its correct description. σ / (squareroot(n))
Answer: Actual standard error of Y bar REASON= Since σ is the population standard deviation, σ/ √n is the actual standard error of Y bar.
15. 1. 2 Match the item with its correct description. ^p±se(^p)
Answer: An interval with 68% coverage REASON= By the Empirical Rule, about 68%of the data are within one standard deviation of the mean for a normal distribution, so ^p±se(^p) is an interval with about 68% coverage.
15. 1. 5 Match the statement with its correct description. s/(squareroot(n))
Answer: Estimated standard error of Y bar REASON= The estimated standard error of Y is of the form below. s / √n
15. 1. 19 Match the statement with its correct description. √^p (1−^p) / n
Answer: Estimated standard error of ^p (p hat) REASON= The estimated standard error of ^p is of the form below.
15. 1. 20 Mark the statement true or false. If you believe that a statement is false, briefly explain why you think it is false. The 95% t-interval for μ only applies if the sample data are nearly normally distributed.
Answer: False. With a large enough sample, the central limit theorem implies a normal sampling distribution regardless of the distribution of the data. REASON= The sample data does not need to be nearly normally distributed. For large sample sizes, the central limit theorem implies a normal sampling distribution regardless of the distribution of the data.
15. 1. 23 What are the chances that X>μ?
Answer: The probability is 0.5, because the distribution of X is symmetric with mean μ. REASON= The random variable X that represents the mean of a randomly chosen sample is approximately normally distributed with mean μ and variance σ2/n, where σ2 is the population variance and n is the sample size. Since the normal distribution is symmetric about the mean, half of the values are less than the mean and half of the values are greater. Therefore, the probability that X>μis the same as the probability that X<μ, so both probabilities are 0.5.
15. 1. 12 Mark the following statement as True or False. If you believe that the statement is false, briefly explain why you think it is false. Ninety-five percent z-intervals have the form of a statistic plus or minus 3 standard errors of the statistic.
Answer: The statement is false. Ninety-five percent z-intervals have the form of a statistic plus or minus 2 standard error(s) of the statistic.
15. 1. 11 Mark the statement true or false. If you believe that a statement is false, briefly explain why you think it is false. All other things the same, a 90% confidence interval is shorter than a 99% confidence interval.
Answer: The statement is true. REASON= The general forms of a z-interval and a t-interval are shown below, where p is the sample proportion, zα/2 is the percentile from the normal distribution, se(p) is the standard error of p, x is the sample mean, tα/2,n−1 is the percentile from the t-distribution with n−1 degrees of freedom, and se(x) is the standard error of x. p±zα/2se(p) x±tα/2,n−1se(x) As the confidence level increases, the percentile of the distribution that must be covered increases. This increases the values of zα/2 and tα/2,n−1, which increases the width of the confidence interval. Therefore, an interval for a lower confidence level is shorter, so the statement is true.
15. 1. 17 Mark the statement true or false. If you believe that a statement is false, briefly explain why you think it is false. If zero lies inside the 95% confidence interval for μ, then zero is also inside the 99% confidence interval for μ.
Answer: True REASON= The confidence interval for 95% is smaller than the confidence interval for 99%. The 99% confidence interval has a larger range of values around the mean than does the 95% confidence interval. Therefore, if zero is in the smaller range, it has to be in the larger range.
15. 1. 39-T Hoping to lure more shoppers downtown, a city builds a new public parking garage in the central business district. The city plans to pay for the structure through parking fees. For a two-month period (43 weekdays), daily fees collected averaged $1,255, with a standard deviation of $160. Complete parts a through d.
(a) What assumptions must you make in order to use these statistics for inference? Select all that apply. Answers: C. The data from the sample are not extremely skewed. The sample size is larger than 10 times the squared skewness and 10 times the absolute value of the kurtosis. D. The sample is a simple random sample from the relevant population. (b) Find a 95% confidence interval for the mean daily income this parking garage will generate, rounded appropriately for your message. Answers: The 95% confidence interval for the mean daily income is $1205 to $1305. (Round to the nearest dollar as needed.) REASON= calculate using calculator (c) The consultant who advised the city on this project predicted that parking revenues would average $1,307 per day. On the basis of your confidence interval, do you think the consultant was correct? Why or why not? Answers: Since the 95% confidence interval does not contain the predicted average, the consultant's prediction is probably not accurate. REASON= It doesn't land between the interval (d) Give a 95% confidence interval for the total revenue earned during five weekdays. Answers: The 95% confidence interval for the total revenue earned during five weekdays is $6,030 to $6,520. (Round to the nearest dollar as needed.) REASON=multiply the confidence interval by 5
15. 1. 45 Find the 95% z-interval or t-interval for the indicated parameter.
(a) μ x-bar =164, s=34, n=27 Answer= The 95% confidence interval for μ is (150.55 to 177.45) REASON= Use calculator. (b) p ^p=0.4, n=73 Answer: The 95% confidence interval for p is . 288 to . 512. (Round to three decimal places as needed.) REASON= ^p − zα/2 √^p(1−^p) /n to p + zα/2 √^p(1−^p) /n
15. 1. 46-T Show the 95% z-interval or t-interval for the indicated parameter.
(a) μ x=−45, s=62, n=40 Answers: −64.83 to −25.17 (Round to one decimal place as needed.) REASON= use t-interval (tests) in calculator (b) μ x=280, s=19, n=24 Answers: 271.98 to 288.02 (Round to one decimal place as needed.) REASON= use t-interval (tests) in calculator (c) p p=0.32, n=50 0.1907 to 0.4493 (Round to four decimal places as needed.) REASON= use formula ^p − zα/2 √^p(1−^p) /n to p + zα/2 √^p(1−^p) /n (d) p p=0.84, n=45 Answers: 0.7329 to 0.9471 (Round to four decimal places as needed.) REASON= use formula ^p − zα/2 √^p(1−^p) /n to p + zα/2 √^p(1−^p) /n
