(a) State the type of data collection involved in the given case study. Justify your answer. (b) Construct 90% confidence interval for the different mean between previous and new engine version. (c) Suppose the sampling for both engine versions were reduced to 15. Test the engineer claim at 0.05 significance level.
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- Intellectual development (Perry) scores were determined for 21 students in a first-year, project-based design course. (Recall that a Perry score of 1 indicates the lowest level of intellectual development, and a Perry score of 5 indicates the highest level.) The average Perry score for the 21 students was 3.27 and the standard deviation was .40. Apply the confidence interval method to estimate the true mean Perry score of all undergraduate engineering students with 90% confidence. Interpret the results.We have developed a new drug used to reduce fever and want to test its efficacy. Ten ER patients who are running a fever agree to take the new drug instead of the traditional treatment. Their temperatures are taken before the drug is administered and 30 minutes after taking the drug. The results are recorded in the Excel file below. Use the data to construct and interpret a 99% confidence interval for the mean change in temperatures of patients after taking this drug using Excel. Be sure to include any explanation necessary along with a sentence that explains what the interval means. First Temp Second Temp 100.1 98.9 101.3 99.1 102.1 99.2 102.7 99 101.9 98.7 100.8 98.6 103.1 99.4 102.5 99.2 103.5 100.1 101.7 99.2Suppose IQ scores were obtained for 20 randomly selected sets of siblings. The 20 pairs of measurements yield x = 101.13, y = 103.5, r=0.867, P-value = 0.000, and y = 6.25 +0.96x, where x represents the IQ score of the younger child. Find the best predicted value of y given that the younger child has an IQ of 101? Use a significance level of 0.05. Click the icon to view the critical values of the Pearson correlation coefficient r. The best predicted value of ŷ is. (Round to two decimal places as needed.)
- Suppose that we are interested in comparing the academic success of college students who belong to fraternal organizations and the academic success of those who do not belong to fraternal organizations. Random samples of size 40 are taken from each population where the variances and standard deviations are known. Sample Size Mean GPA Variance (2) 2.03 2.21 Fraternity member (f): 40 Nonmembers (n): 40 .46 .35 Construct a 96% confidence interval estimate for the difference between the mean GPAS (uf- un). Interpret this interval estimate. What can you conclude about the mean GPAs of the two groups?The percentage of Texans not covered by health care insurance in 2015 was 17%. The Texas Health and Human Services Commission (HHSC) has been charged with conducting a sample survey to obtain more current information. What sample size would you recommend if the HHSC’s goal is to estimate the current proportion of Texans without health care insurance with a margin of error of .03? Use a 95% confidence level. Repeat part (a) using a 99% confidence level.Two brands of batteries are tested, and their voltages are compared. The summary statistics follow. For sample 1, the sample size is 27; the population standard deviation is 0.3 volts and the mean is 9.2 volts. For sample 2, the sample size is 30; the population standard deviation is 0.1 volts and the mean is 8.8 volts. Is the mean of sample 1 smaller than the mean of sample 2? Using a significance level of 0.05. a) Identify the claim and state the hypotheses. b) What is the test statistics? Find the critical values (s). c) Make the decision to reject or not reject the null hypothesis and explain why. d) Graph your decision and include all the values. That is, the mean, the standard deviation, the critical value(s), the rejection region. e) Set up the formula with the correct numbers for the 95% confidence interval of the mean number of jobs.
- You were told that the mean daily cost of a hotel room in downtown St. Louis is $120. You will be needing to rent a hotel room there in the near future and you want to find the best rate. You randomly select a sample of 18 hotel rooms and find that the average cost is $121.33 with a standard deviation of $18.046. At the α = 0.05 significance level, determine if there is enough evidence to reject the claim. What is the decision for this hypothesis test? You were told that the mean daily cost of a hotel room in downtown St. Louis is $120. You will be needing to rent a hotel room there in the near future and you want to find the best rate. You randomly select a sample of 18 hotel rooms and find that the average cost is $121.33 with a standard deviation of $18.046. At the α = 0.05 significance level, determine if there is enough evidence to reject the claim. What is the decision for this hypothesis test? Reject the null because the test statistic is in the critical region and the…Suppose you work for the Department of Environment and National Resources (DENR). You want to estimate, with 95% confidence, the mean length of all tilapia in a fish hatchery pond. You take a random sample of 10 tilapia and determine the average length is 10.5 inches and the sample standard deviation is 2.9 inches. Construct a confidence interval for this sample and interpret the result.do B AND C
- QUESTION 1 a) Two machines are used to fill glass bottles with a net volume of 13.0ml. The filling processes are assumed to be normally distributed, with common but unknown standard deviations. The quality department suspects that both machines fill to the same net volume, whether or not this volume is 13.0ml. An experiment is performed by taking a random sample from the output of each machine. 13.01 12.98 13.02 13.05 12.99 Machine 1 13.03 13.01 12.96 12.97 13.04 13.04 13.02 13.01 13.05 13.02 Machine 2 13.00 12.96 12.99 13.02 13.03Please help with 1 a,b,c,d,e, and f.Suppose that you are interested in the number of people in each household in the U.S. Your sample includes 10 households, with a mean 2.5 and a standard deviation 1.2. Construct a 95% confidence interval (CI) of the mean number of household members. (A) 1.2-3.8 (B) 1.4-3.6 (C)1.6-3.4 (D) 1.7-3.3 (E) 1.8-3.2