AP Statistics Final Review: ANSWER KEY - #50-60 50. z = invnorm(.08) = 1.405 51. If you know the true standard deviation of the population ( ), use a z-critical value. If you have to use the sample standard deviation (s) to estimate , then you must use t. 52. Using an interval estimate gives the statistician a much better chance of getting a correct estimate. 53. The standard error of a statistic is the estimate of the standard deviation of a statistic. For example s SE(mean) is and describes how much x varies from , on average. n 54. Take many samples from the population and calculate the value of the statistic for each sample. Graph the values of the statistic and see if they are centered above the value of the parameter (the true value). 55a. 1. We are trying to estimate p = the true proportion of voters in this city who approve of the mayor’s performance. Our best guess is pˆ = .3375, but because of sampling variability this is probably incorrect. So, we will calculate a 95% z-interval for p. 2. Conditions: a. random sample of city voters? Given. b. sample < 10% of population? Assume population of city voters >4000 c. large sample size? Yes, npˆ = 135 > 10, nqˆ = 265 > 10.
.3375 1 .3375 = .3375 ± .0463 = (.2912, .3838) 400 4. We are 95% confident that the interval from .2912 to .3838 captures the true proportion of city voters who approve of the mayor’s performance. 55b. If we were to take many samples and compute many confidence intervals, approximately 95% of them would capture the true proportion. 55c. No, since all of the plausible values for p are below .5. 3. 95% CI = .3375 1.96
55d. .03 1.96 56.
.3375 1 .3375 n > 954.4. So we need 955 – 400 = 555 more people. n
1. We are trying to estimate = the true mean number of unpopped kernels for Pop Secret Microwave Popcorn. Our best guess is x = 23, but because of sampling variability it is unlikely to be correct. So, we will calculate a 90% t interval for . 2. Conditions: o Random sample of popcorn bags? Given. o Sample < 10% of population? Assuming > 70 bags of popcorn o Population approximately normal?
Since the normal probability plot is roughly linear, it is reasonable to assume that the population is approximately normal. 3.46 3. CI = 23 1.943 23 2.54 20.46, 25.54 7
4. I am 90% confident that the interval from 20.46 to 25.54 captures the true mean number of unpopped kernels per bag for Pop Secret Microwave Popcorn. 57a. 1. We are trying to estimate the true change in the proportion of registered voters who approve of the job George W. Bush is doing as president from February 2007 to February 2008 ( p2008 p2007 ). Our best guess is p 2008 p 2007 = .30 - .34 = -.04, but because of sampling variability this is probably incorrect. Instead, I will calculate a 90% 2 sample z-interval for p2008 p2007 . 2. Conditions: i. Independent random samples of registered voters in Feb 2007 and March 2008? Given. ii. Sample sizes < 10% of population sizes? Yes, there are more than 9000 registered voters in 2007 and 2008. iii. Large sample sizes? 1. 2007: n p = 306 > 10, n(1- p ) = 594 > 10 2. 2008: n p = 270 > 10, n(1- p ) = 630 > 10
.34(.66) .30(.70) .04 .036 (.076, .004) 900 900 4. I am 90% confident that the interval from -.086 to -.014 captures the true change in the proportion of registered voters who approve of the job George W. Bush is doing as president from February 2007 to February 2008. .34(.66) .30(.70) 57b. SE = = .022 ME = .036 900 900 57c. Yes, since all of the plausible values are negative. 58a. 1. At first glance, it appears that the true mean test score for students taking the test in silence ( s ) is higher than the true mean score for students taking the test with noise ( n ) since xs xn 1.7 0 . However, it is possible that in reality the true means are the same and the difference we observed was due to randomization variability. To decide I will conduct a 2 sample t-test for s n ( = .05). 2. Ho: s n = 0 Ha: s n > 0 3. Conditions: a) The two treatments were randomly assigned to individuals? Given b) Both treatment distributions are approximately normal? 3. 90% CI = .04 1.645
Since there are no outliers and no strong skewness in either sample, this is reasonable. 1.7 0 ) P(t 1.228) 4. p-value = P( xs xn 1.7) P(t 3.107 2 3.0842 10 10 Table: df = 9: p = .1253 Calculator: df = 18: p = .1176 5. Since p-value > alpha, we fail to reject the null hypothesis and cannot conclude that the presence of background noise has a negative effect on reading comprehension. 58b. The power is the probability of deciding that background noise has a negative effect on reading comprehension when background noise really does have a negative effect on reading comprehension.
58c. Pair students by intelligence, pretest score, GPA, etc. and randomly split each pair to the different treatments OR give each student both treatments in a random order. 59a. We should use a matched pairs t-test, since each subject had both treatments (5.5 hours and 8.5 hours). 59b. The treatments should be randomly assigned—in this case, the order of the 5.5 and 8.5 days. This was not stated, so we must assume. Also, since it is not a large sample size (n = 11), we must assume the population of differences is approximately normal. We have no data to check, so we must assume this is ok. 59c. H 0 : d = 0 H a : d 0 (Since it is not clear which direction the researchers suspected, we will use a two-sided test by default). 59d. If the amount of sleep has no effect on the amount of calories consumed, there is a .001 probability of getting a sample mean difference as or more extreme as 220 calories simply by chance. 59e. Since the p-value < alpha (.001 < .05), they should reject H 0 and conclude that there is a relationship between amount of sleep and amount of calories consumed. 59f. Type I. They decided that there is a relationship when in reality there may not be one. 59g. A 95% confidence interval would not include 0, since we rejected 0 as a plausible value for the true mean difference at alpha = .05. 59h. To make a 2-sample t-test appropriate, we would randomly assign half of the volunteers to sleep 8.5 hours and the other half to sleep 5.5 hours and compare the mean calories consumed per day. To randomly assign them, put all the names in a hat, mix them, and pick out half. 60a. H 0 : old new 0 H a : old new 0 60b. sold snew 6.056 3.03498 3.02102 . This gives evidence that the new machine is more consistent, since the difference of the sample standard deviations is greater than 0 ( sold snew ). 60c. p-value 13/100 = .13 (13% of the simulated ratios are 3.02102). 60d. Since the p-value > alpha (.13 > .05) we fail to reject H 0 and cannot conclude that the blocks produced by the new machine have more consistent weights (are less variable). 60e. Since we failed to reject H 0 , it is possible we made a Type II error: not choosing the new machine when it is really better.
