STAT FINAL PRACTICE
In clinical studies of an allergy drug, 81 of the 900 subjects experienced drowsiness. A competitor claims that at least 10% of the users of this drug experience drowsiness. a. Is there enough evidence at the 5% significance level to infer that the competitor is correct? b. Compute the p-value of the test in the part (a).
)A) Conclusion Since the test statistic does not fall into the rejection region we reject Ha at 5% significance level. That is, the data do not support the competitor's claim that 10% of users of this drug experience drowsiness B)p-value= P(z > −1) = .84 > .05. Since the p-value is larger than the significance level we reject Ha.
A sample of 49 measurements of tensile strength (roof hanger) are calculated to have a mean of 2.45 and a standard deviation of 0.25. Determine the 95% confidence interval for the measurements of all hangers.
2.45 ± 1.96 .25 √ 49 = 2.45 ± .07 or [2.38; 2.52].
A statistician wants to estimate the mean weekly family expenditure on clothes. He believes that the standard deviation is about $125. Determine with 99% confidence the number of families that must be sampled to estimate the mean weekly family expenditure on clothes to within ±$15.
461.
In a survey two days before election day, the Democratic candidate for the Senate was supported by 918 voters. His opponent was supported by 582 voters. Construct 99% confidence interval for the proportion of all voters who support the Democratic candidate.
918 1500 ± 2.576s 918 1500 582 1500 1500 = .612 ± .032 or, in form of interval, [.58; .644]
During the past month Debra stopped at the Soaring Eagle Casino in Mount Pleasant, Michigan, 6 times with no intention to play (since she has no money). Having nothing better to do, she counted how many jackpots the players received
RR = [tα,n−1; +∞) = [t.05,5; +∞) = [2.02; +∞) Conclusion Since the test statistic does not fall into the rejection region we reject Ha at 5% significance level. That is, the data do not support the Debra's believe that the average number of jackpots is over 15.
A marketing consultant was in the process of studying the perception of married couples concerning their weekly food expenditures. He believed that the husbands perception would be higher than the wifes. To judge his belief, he takes a random sample of ten married couples and asks each spouse to estimate the family food expenditure (in dollars) during the previous week. The data are shown below. Can the consultant conclude at the 5% significance level that the husbands estimate is higher than the wifes estimate. check exam for data
Since the test statistic falls into the rejection region we reject H0 at 5% significance level in favor of the alternative. That is, the data support the marketing consultant's claim that the husbands' estimate is higher than the wife estimate.
The air pumps at service stations come equipped with a gauge to regulate the air pressure of tires. A mechanic believes that the gauges are in error by at least 3 pounds per square inch. To test his belief he takes a random sample of 50 air pump gauges and determines the absolute difference between the true pressure (as measured by an accurate measuring device) and the pressure shown on the air pump gauge. The mean and the standard deviation of the sample are X¯ = 3.4 and s = 1.2. Can the mechanic infer that he is correct at the 5% significance level?
Since the test statistic is in the rejection region we accept Ha at 5% significance level. That is, the data support the mechanic's believe that the gauge are in error by at least 3 pounds per square inch. RR = [zα; +∞) = [z.05; +∞) = [1.645; +∞)
. Suppose that a random sample of size 64 is to be selected from a population with mean 40 and standard deviation 5. a. What is the mean and standard deviation of sample mean X¯? b. What is the probability that X¯ will be greater than 40.7?
The mean of X¯ is 40, and the standard deviation is 5/ √ 64 = .625. b. Since the sample size is above 30, approximately X¯ ∼ N(40, .6252 ). Therefore, P(X >¯ 40.7) = P Z > 40.7 − 40 .625 = P(Z > 1.12) = 1 − P(Z < 1.12) = 1 − .8686 = .1314
A random sample is to be selected from a population that has a proportion of successes p = 0.65. a. Determine the mean and standard deviation of the sample proportion ˆp for the sample size n = 100. b. When n = 100, what is P(0.6 < p <ˆ 0.7)?
a)The mean of ˆp is .65, and the standard deviation is p .65 × .35/100 ≈ .0477. b. Since the sample size is large (np(1 − p) > 5), approximately ˆp ∼ N(.65, .04772 ). Therefore, P(.6 < p < . ˆ 7) = P .6 − .65 .0477 < Z < .7 − .65 .0477 = P(−1.05 < Z < 1.05) = P(Z < 1.05) − P(Z < −1.05) = .8531 − .1469 = .7062
Do government employees take longer coffee breaks than private sector workers? That is a question that interested a management consultant. To examine the issue, he took a random sample of ten government employees and another random sample of ten private sector workers and measured the amount of time (in minutes) they spent in coffee breaks during the day. The results are listed below. Do these data provide sufficient evidence at the 5% significance level to support the consultants claim? check exam for data
n this case we have independent samples, small sample sizes, one-tailed alternative, and ν = 16, so the rejection region is RR = (−∞; −tα,ν] = (−∞; −t.05,16] = (−∞; −1.746]. 4. Conclusion Since the test statistic falls into the rejection region we reject H0 at 5% significance level in favor of the alternative. That is, the data support the consultant's claim that government employees take longer breaks than private sector workers.
