stats

8.3.15-T 8.3.22-T Students estimated the length of one minute without reference to a watch or clock, and the times (seconds) are listed below. Assume that a simple random sample has been selected. Use a 0.10 significance level to test the claim that these times are from a population with a mean equal to 60 seconds. Does it appear that students are reasonably good at estimating one minute? Adata set includes data from student evaluations of courses. The summary statistics are n=89, x=4.17, $=0.53. Use a 0.01 significance level to test the claim that the population of student course evaluations has a mean equal to 4.25. Assume that a simple random sample has been selected. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim. Sample mean = 62.8, Sample St Dev = 19.21, n=15 Ho:p =4.25 . Step 1 Ha:p =/ 4.25 Two-Tailed Ho:pu =60 ; Step 1 Ha:p =/ 60 Two-Tailed Step 2 What Distribution to Use? Row 4 in the Table says Use Z What Distribution to Use? Step 2 Row 3 in the Table says Use T (assume Normality of parent distribution). df = 14 Step 3 Calculate Z score. Z=(4.17-4.25) /(0.53 / sqrt(89)) =-1.42 Step 3 Calculate T score. T=(62.8-60)/(19.21/sqrt(15)) = 0.565 Step 4 Calculate P-value. P(Z<-1.42) + P(Z > 1.42) =2*0.0778 = 0.157 Step 4 Calculate P-value. P(T(14) <-0.565) + P(Z > 0.565) = 2*0.291 = 0.582 Do Not Reject Ho. There is no evidence in the data to support the Lt aboaar toias e 8.3.23-T Stenb Do Not Reject Ho. There is no evidence in the data to support the claim that the average time w/o clock is 60 seconds Step 5 Assume that a slmple random sample h: ted from a normally distributed population and test the given claim. Identify the null nnd al!srv\auvo hypotheses, test statistic, P-value, and state th final conciusion that addresses the nal cla A safety administration conducted cmsh tests of child booster seats for cars. Listed below are resuns T level to test the claim that the results suagest that all of the child booster seats meet the requirement Sample mean = 757, Sample St Dev = 256.62, n=6 Ho: u = 1000 . Step 1 Ha:p < 1000 Lower Tailed What Distribution to Use? Step 2 Row 3 in the Table says Use T (Normality of parent distribution given). df =5 Step 3 Calculate T score. T=(757 -1000) / ( 256.2 / sqrt(6)) =-2.323 Step 4 Calculate P-value. P(T(5) <-2.323) =0.034 Cannot reject Ho at 1% level of significance, but can reject the null hypothesis at 5% level of significance 7.2.7-T Step 5 6.4.5-T Assume that females have pulse rates that are normally distributed with a mean of 1 = 76.0 beats per minute and a standard deviation of o = 12.5 beats per minute. Complete parts (a) through (c) below. Here are summary statistics for the weights of Pepsi in randomly selected cans: n =36, x=0.82414 Ib, $=0.00572 Ib. Use a confidence level of 99% to complete parts (a) through (d) below. a. |dentify the critical value t,, used for finding the margin of error. tyyp= 272 (Round to two decimal places as needed.) a. If 1 adult female is randomly selected, find the probability that her pulse rate is less than 82 beats per minute. b. Find the margin of error. The probability is 0.6844". (Round to four decimal places as needed.) E= 0.00260 Ib (Round to five decimal places as needed.) b. If 16 adult females are randomly selected, find the probability that they have pulse rates with a mean less than 82 beats per minute. c. Find the confidence interval estimate of . The probability is 0.9726' (Round to four decimal places as needed.) 0.82154' Ib<p < 0.82674 Ib (Round to five decimal places as needed.) d. Write a brief that i the interval. ¢. Why can the normal distribution be sed in part (b), even though the sample size does not exceed 307 * One has 99% confidence that the interval from the lower bound to the upper bound contains the true value of the population mean weight of Pepsi in a can. Because parent distribution is normal