A quality inspector is labeling the products at production line with labels either "non- conforming" or "conforming." In order to estimate the proportion of non-conforming products, a random sample of 100 products are selected from the production line and 10 are found to be non-conforming products. To monitor the quality of the production line, another set of 100 samples is taken and observed that only 6 non-conforming products are found. (a) Give a 95% confidence interval on P1 - P2, where P, is the population proportion non- conforming before improvement, and P2 is the proportion non-conforming after improvement. (b) Is there information in the confidence interval found in (a) that would suggest that P, > P2? Explain.
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- A cellphone provider has the business objective of wanting to determine the proportion of subscribers who would upgrade to a new cellphone with improved features if it were made available at a substantially reduced cost. Data are collected from a random sample of 500 subscribers. The results indicate that 135 of the subscribers would upgrade to a new cellphone at a reduced cost. At the 0.05 level of significance, is there evidence that more than 20% of the customers would upgrade to a new cellphone at a reduced cost? How would the manager in charge of promotional programs concerning residential customers use the results in (a)?According to a certain government agency for a large country, the proportion of fatal traffic accidents in the country in which the driver had a positive blood alcohol concentration (BAC) is 0.33.Suppose a random sample of 109 traffic fatalities in a certain region results in 47that involved a positive BAC. Does the sample evidence suggest that the region has a higher proportion of traffic fatalities involving a positive BAC than the country at the α=0.05 level of significance?In 2016, the US Department of Agriculture established a “maximum acceptable” salmonella prevalence of 15.4 percent for chicken parts sampled at the end of a slaughterhouse’s kill line. 54 of the 154 largest chicken-slaughter facilities in the US failed to meet this standard in 2018. Assume that these results are representative for all large chicken-slaughter facilities. Suppose that in a random sample of 60 chicken parts from the large ACME chicken-slaughter facility (which produces 1,000,000 chicken parts per year), 45 chicken parts test negative for salmonella. c) Estimate the number of chicken parts that are needed for a follow-up study with 90% power to test thenull hypothesis that the prevalence of salmonella among chicken parts at the ACME facility is equal to the maximum acceptable value.
- In 2016, the US Department of Agriculture established a “maximum acceptable” salmonella prevalence of 15.4 percent for chicken parts sampled at the end of a slaughterhouse’s kill line. 54 of the 154 largest chicken-slaughter facilities in the US failed to meet this standard in 2018. Assume that these results are representative for all large chicken-slaughter facilities. Suppose that in a random sample of 60 chicken parts from the large ACME chicken-slaughter facility (which produces 1,000,000 chicken parts per year), 45 chicken parts test negative for salmonella. b) Estimate the number of facilities that are needed in a study to obtain 80% power to test the null hypothesis that the prevalence of failing to meet the chicken part salmonella standard among large chicken-slaughter facilities is 20%.In 2016, the US Department of Agriculture established a “maximum acceptable” salmonella prevalence of 15.4 percent for chicken parts sampled at the end of a slaughterhouse’s kill line. 54 of the 154 largest chicken-slaughter facilities in the US failed to meet this standard in 2018. Assume that these results are representative for all large chicken-slaughter facilities. Suppose that in a random sample of 60 chicken parts from the large ACME chicken-slaughter facility (which produces 1,000,000 chicken parts per year), 45 chicken parts test negative for salmonella. a) Test the null hypothesis that the prevalence of salmonella among ACME chicken parts is 15.4%, usingthe p-value method.A cellphone provider wants to determine the proportion of subscribers who would upgrade to a new cellphone with improved features if it were made available at a substantially reduced cost. Data are collected from a random sample of 500 subscribers. The results indicate that 135 of the subscribers would upgrade to a new cellphone at a reduced cost. At the 0.05 level of significance, is there evidence that more than 20% of the customers would upgrade to a new cellphone at a reduced cost?
- A low-level CDC bureaucrat wants to please his boss by gathering evidence that the current government mandated shut down of society is not causing people's mental health to deteriorate, so that it can safely be continued for several years if the expert says is necessary. He pulls a random sample of 1600 citizens, Gathering data on such items as income loss, weight gain, access to toilet paper, our spent binge watching Netflix, and number of injuries caused by household fight, and compiles all this into a significantly-weighted "misery index". The mean misery index from the sample is 99.2; it seems reasonable to use a population standard deviation o=19.1. A. Does this information provide significant evidence (at the 5% level) that the Nationwide mean misery index is less than 100? Set up appropriate null and alternative hypothesis, calculate the appropriate test statistic, find the p-value, and state your conclusion.For several years, evidence had been mounting that folic acid reduces major birth defects. In a study, doctors enrolled women prior to conception and divided them randomly into two groups. One group, consisting of 2783 women, took daily multivitamins containing 0.8 mg of folic acid; the other group, consisting of 2103 women, received only trace elements. Major birth defects occurred in 31 cases when the women took folic acid and in 47 cases when the women did not. a. At the 5% significance level, do the data provide sufficient evidence to conclude that women who take folic acid are at lesser risk of having children with major birth defects? b. Is this study a designed experiment or an observational study? Explain your answer. c. In view of your answers to parts (a) and (b), could you reasonably conclude that taking folic acid causes a reduction in major birth defects? Explain your answer. Question content area bottom Part 1 a. Use the two-proportions z-test…In analyzing the consumption of cottage cheese by members of various occupational groups, the United Dairy Industry Association found that 326 of 837 professionals seldom or never ate cottage cheese, versus 220 of 489 white-collar workers and 522 of 1243 blue-collar workers (Sheet 53). Assuming independent samples, use the 0.03 level in testing the null hypothesis that the population proportions could be the same for the three occupational groups. Sheet 53 Group 1 Group 2 Group 3 Total seldom or never 326 220 522 1068 often 511 269 721 1501 Total 837 489 1243 2569 Select one: a) chi-square stat = 4.81, crit. value = 7.01, fail to reject H0, population proportions are not different b) p-value = 0.09, reject H0, population proportions are not different c) chi-square stat = 4.81, crit. value = 9.2, fail to reject H0, population proportions are not different d) p-value = 0.029, reject H0, population proportions different
- Twenty middle-aged men with glucose readings between 90 milligrams per deciliter and 120 milligrams per deciliter of blood were selected randomly from the population of similar male patients at a large local hospital. Ten of the 20 men were assigned randomly to group A and received a placebo. The other 10 men were assigned to group B and received a new glucose drug. After two months, posttreatment glucose readings were taken for all 20 men and were compared with pretreatment readings. The reduction in glucose level (Pretreatment reading − Posttreatment reading) for each man in the study is shown here. Group A (placebo) reduction (in milligrams per deciliter): 12, 8, 17, 7, 20, 2, 5, 9, 12, 6 Group B (glucose drug) reduction (in milligrams per deciliter): 29, 31, 13, 19, 21, 5, 24, 12, 8, 21 Part A: Do the data provide convincing evidence, at the α = 0.02 level, that the glucose drug is effective in producing a reduction in mean glucose level? Part B: Create and interpret a 98%…A sports betting company is testing the effectiveness of its online ads and has developed a classification model that predicts whether a potential customer will respond to its offer using information in the potential customer's browser history (cookies, etc.). Out of a test dataset of 1,000 customers, the 100 customers most likely to respond (according to the model) are identified and it is discovered that 14 of these actually did respond to the online ads. In the entire test dataset of 1,000 customers, only 39 customers responded to the ad. What is the lift of the classification model for the set of 100 customers it identified as most likely to respond? If required, round your answer to two decimal places. I got 3.59 as the answer and its incorrect.According to a certain government agency for a large country, the proportion of fatal traffic accidents in the country in which the driver had a positive blood alcohol concentration (BAC) is 0.36. Suppose a random sample of 110 traffic fatalities in a certain region results in 49 that involved a positive BAC. Does the sample evidence suggest that the region has a higher proportion of traffic fatalities involving a positive BAC than the country at the α=0.05 level of significance?