Based on a sample of 80 men, 45% owned cats Based on a sample of 20 women, 50% owned cats test statistic = p-value = [three decimal accuracy] [three decimal accuracy]
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- Using the dotplot and the difference in mean distances from the samples, is there convincing evidence that the new golf ball travels farther than the original golf ball? A company that manufactures golf balls produces a new type of ball that is supposed to travel significantly farther than the company's previous golf ball. To determine this, 40 new-style golf balls and 40 original-style golf balls are randomly selected from the company's production line on a specific day. A golf pro randomly selects a ball, not knowing which type is chosen, and hits it. The difference in mean distances traveled (new - original) for the samples was 2.6 feet. Assuming there is no difference in distance traveled between the two types of golf balls, 200 simulated differences in sample means are O Yes, because a difference in mean distances of 2.6 feet or more occurred only 34 out of 200 times, meaning the difference is statistically significant. There is convincing evidence the riew golf ball travels…It is believed that from among senior high school students, 15% would like to pursue engineering course. Suppose a random survey reveals that 85 out of 600 students choose engineering course, would you say that the percentage of those who would pursue engineering course is less than 15%? Use 0.05 level of significance.In an ANOVA, there are 5 groups and the between-groups sum of squares is 98. What is the mean square between groups? Round to three decimal places. Answer: Please give solution in correct and step by step detailed format thanku Don't give solution in image format
- 12) Many graduate schools require applicants to take the GRE for admission. The mean GRE score worldwide is 149.7 with a standard deviation of 8.2. The recommended overall GRE score for admission at UC Berkeley is 153. Q) What proportion of students taking the GRE would be accepted by UC Berkeley just based on GRE scores?Determine the value of ‘***’? b. Explain what is meant by Sampling Distribution ofthe Mean in the context of MBA graduates’ startingsalaries? c. What type of distribution will the SamplingDistribution of the Mean for MBA graduates’ startingsalaries follow? Give reasons for your answer. d. What are the available point estimator andestimate of the population mean starting salary? e. Which of these estimators in ‘Part (c)’ above isunbiased? Give reasons for your answer? f. Identify an available point estimate for thepopulation variance? g. Construct and interpret an 82% confidence interval of the population mean starting salary? Transcribed Image Text:A top graduate business school is interested in the starting salaries of their MBA graduates. A survey consisting of all the graduates from previous year was done and the results displayed in the table below: Mean Median Std. Dev. Std. Error MBA's 175 11,500 8,700 *** 1441a) Randomly pick 5 numbers between 1 - 100 and pick 5 numbers, remember, no repeats of numbers then find the average of the test scores that correlate to it. This is your sample average. List your 5 numbers and write down your average. 1b)The actual mean of the scores is 63.52, how off were you? The standard deviation of the data is 25.606 (think about if these are your sample statistics or population parameters). 1c) Find your Margin of Error using 95% confidence. I gave you the population standard deviation so think about if it's a z-test or t-test.
- Survey was conducted about real estate prices, Data collected is 106,427, 231,714, 372,706, 423,646, 575,866, 621,338, 723,083, 883,328, 994,802, 1.090,841, 1,147,051, 1,227,981, 1,387,149. What is the third quartile price?The mean response time on a task for 10 younger children (6-10 y.o.a.) is 45 and the mean response time for 14 older children (11-16 y.o.a.) is 30. What is the weighted mean for the two groups (round to nearest tenth)? -> SHOW YOUR WORK ON YOUR SCRAP PAPER!!! 36.25 37.5 38.75 answer not listed- see work on scrap paperIn 2015, the average duration of long-distance telephone calls from a certain town was 3.9 minutes.A telephone company wants to perform a test to determine whether this average duration of longdistance calls has changed. Fifty calls, originating from the town, was randomly selected and thefollowing summary minutes.∑ ? = 205 ∑(? − ?̅)∑(? − ?̅)2 = 56.43 a) Calculate the sample mean, ?̅. b) Calculate the sample standard deviation c) Using the p value approach at the 1% level of significance, test whether the mean duration oflong-distance calls from the town had increased. d) Construct the 99% confidence interval for the population mean duration of the long-distance calls from the town. Answer sub-parts (d)
- Do wearable devices that monitor diet and physical activity help people lose weight? Researchers had 237 subjects, already involved in a program of diet and exercise, use wearable technology for 24 months. They measured their weight (in kilograms) before using the technology and 24 months after using the technology. (a) Explain why the proper procedure to compare the mean weight before using the wearable technology and 24 months after using the wearable technology is a matched pairs t test. Choose the correct explanation. A person's weight before and after wearing the device would not be independent; thus, the data from this before and after study should be analyzed using a matched pairs t test. O A person's weight before and after wearing the device would be independent; thus, the data from this before and after study should be analyzed using a matched pairs t test. O A person's weight before and after wearing the device would be independent; thus, the data from this before and after…1) Randomly pick 5 numbers from the attached link (use a Random number generator - on Google/calculator) and say 1 - 100 and pick 5 numbers, remember, no repeats (kind of like the first discussion) then find the average of the test scores that correlate to it. This is your sample average. List your 5 numbers and write down your average. 2) The actual mean of the scores is 63.52, how off were you? The standard deviation of the data is 25.606 (think about if these are your sample statistics or population parameters). 3) Find your Margin of Error using 95% confidence. I gave you the population standard deviation so think about if it's a z-test or t-test.what statsistical test, if any, would be typically associated with a bar gragh