You are testing the null hypothesis that p = 0 versus the alternative ,u > 0 using a = .05. Assume a = 23. Suppose} = 5.5 and n = 12. Calculate the test statistic and its P-value. Repeat assuming the same value of} but with n = 22. Do the same for sample sizes of 32, 42, and 52. (Round the test statistic to two decimal places. Round the P-value to four decimal places.) a = 12: z :' Pvalue \\: n = 22: z :| P-value ':| n = 32: 2 ': P-value ':l n = 42: z [:1 Maine I: n = 52: z :| P-value I: In the Health ABC Study, 531 subjects owned a pet and 1975 subjects did not. Among the pet owners, there were 300 women; 954 of the non-pet owners were women. Find the proportion of pet owners who were women. Do the same for the non-pet owners. (Be sure to let Population 1 correspond to the group with the higher proportion so that the difference will be positive. Round your answers to three decimal places.) [31: 0.047 x L52 = 0,052 X Give a 95% condence interval for the difference in the two proportions. (Do not use rounded values. Round your nal answers to three decimal places.) (_x -x) A survey of Internet users reported that 21% downloaded music onto their computers. The ling of lawsuits by the recording industry may be a reason why this percent has decreased from the estimate of 27% (mm a survey taken two years before. Suppose we are not exactly sure about the sizes of the samples. Perform the calculations for the significance tests and 95% confidence intervals under each of the following assumptions. (Use previous - recent. Round your test statistics to two decimal places and your condence intervals to four decimal places.) (i) Both sample sizes are 1000. 2 = X ss%c.t. (.|:|) (ii) Both sample sizes are 1600. z = X 95% (:1. ( , | ) (Iii) The sample size for the survey reporting 27% is 1000 and the sample size for the survey reporting 21% Is 1600. 2 = x 95% or. ( , | ) Many food products contain small quantities of substances that would give an undesireable taste or smell if they are present in large amounts. An exaple is the 'off-odors' caused by sulfur compounds in wine. Oenologists (wine experts) have determined the odor threshold. the lowest concentration of a compound that the human nose can detect. For example. the odor threshold for dimethyl sulde (DMS) is given in the oenology literature as 25 micrograms per liter of wine (ugh). Untrained noses may be less sensitive. however. Here are the DMS odor thresholds for 10 beginning students of oenology. 35 33 42 37 20 30 31 25 23 32 Assume (this is not realistic) that the standard deviation of the odor threshold for untrained noses is known to be a = 7 ugfl. A normal quantile plot confirms that there are no systematic departures from normality. (a) Make a stemplot to verify that the distribution is roughly symmetric with no outliers. (A Normal quanlile plot conrms that there are no systematic departures from Normality. Enter numbers from smallest to largest separated by spaces. Enter NONE for stems with no values.) 2 03 J 2 5 .3 3 (I123 f 3 5? .f 4 2 ,' (b) Give a 95% confidence interval for the mean DMS odor threshold among all beginning oenology students. (Round your answers to three decimal places.) (:1:) (c) Are you convinced that the mean odor threshold for beginning students is higher than the published threshold. 25 ugfl? Carry out a signicance test to justify your answer. (Use a = 0.05. Round your value for z to two decimal places and round your P-value to four decimal places.) P-value = : You want to rent an unfurnished one-bedroom apartment in Dallas next year. The mean monthly rent for a random sample of 14 apartments advertised in the local newspaper is $960. Assume the monthly rents in Dallas follow a Normal distribution with a standard deviation of $270. Find a 95% condence interval for the mean monthly rent for unfurnished one-bedroom apartments available for rent in this community. (Round your answers to two decimal places.) (i-x .$-x)