The questions are in the file. It is a broken down problem using Minitab
1. Survey questions where respondents are asked to rate items on a scale (often from 1 to 10) are called Likert scale questions. Because respondents often have either extreme or indifferent opinions yielding many responses of 1, 5, and 10, responses to these questions tend to be not normal. For this problem, you are going to simulate responses to a Likert scale question by sampling from a uniform distribution where each response is assumed to be equally likely. Using Minitab or Minitab Express, generate 36 random samples of size 50 from an integer distribution with minimum value 1 and maximum value 10. (This is simulating 50 random people answering one Likert scale question per day for 36 days.) • In Minitab, choose Calc > Random Data > Integer. In the pop-up window, type 50 for “Number of rows of data to generate” and type “C1-C36” in “Store in column(s)”. The minimum value is 1 and the maximum value is 10. • In Minitab Express, choose Data > Generate Random Data from the menu at the top of the program. Type 36 for “Number of columns to generate” and 50 for “Number of rows in each column”. From the dropdown menu, select “Integer” and type 1 for the minimum value and 10 for the maximum value. a. (2Pts) Create and insert a histogram of the sample in the first column ‘C1’. b. (1Pt) Describe the histogram in part (a) according to its symmetry and modality. Calculate the mean of each of your 36 samples using the descriptive statistics functionality. Type ‘Means’ as the title of column C37. Copy and paste your 36 sample means into this column. • In Minitab, choose Stat > Basic Statistics > Display Descriptive Statistics. Select columns C1 through C36, move them all into the “Variables” box, and click OK. • In Minitab Express, you can only perform calculations on 12 columns at a time. Click Summary > Descriptive Statistics. From the dropdown menu, select “Data are in more than one column”. Select columns C1 through C12, move them into the “Variables” box, and click OK to find the descriptive statistics. Repeat this for columns C13 through C24 and C25 through C36. c. (2Pts) Create and insert a histogram of the 36 sample means. d. (1Pt) Describe the histogram in part (c). e. (2Pts) Calculate the mean and standard deviation of the 36 sample means in column C37 using the descriptive statistics functionality. The distribution you sampled these observations from is a uniform distribution on the integers ? through ? where ? and ? represent the minimum and maximum values an observation can take on. As the number of observations sampled becomes large, the population mean is given by ? = ?+? 2 and the population standard deviation is ? = √ (?−?+1)2−1 12 . f. (2Pts) Calculate and report the population mean and the population standard deviation of the uniform distribution above. (This is the theoretical standard deviation of how spread out the individual observations are within a single sample.) g. (1Pt) Determine the sampling distribution of the sample mean. h. (2Pts) Suppose samples of size 100 were taken instead. Explain what would happen to the mean and the standard error of the sampling distribution. i. (2Pts) Calculate the probability that the sample mean of 75 responses exceeds 6. 2. In bowling, there are two categories of oil patterns that are placed on the lane that make the game easier or harder. House shots tend to create easier conditions where scores are higher while sport shots create harder conditions where scores are lower. Over the course of one bowling season, an amateur bowler carried an average of 203 in a league with a house shot with a population standard deviation of 25. He decides to join a league that uses a sport shot on the lanes instead. After 7 weeks of bowling on a sport shot, he has 21 observations, which are contained in the Minitab dataset. a. (1Pt) Report a point estimate for the bowler’s average in the sport shot league. b. (2Pts) Use Minitab to calculate a 95% confidence interval for the bowler’s average on a sport shot pattern. Assume he has the same population standard deviation in the sport shot league as in the house shot league. c. (1Pt) Interpret the interval in part (b) in the context of the problem. d. (1Pt) Is there evidence suggesting that the bowler is performing significantly better or worse on the sport shot compared to the house shot using the interval in part (b)? Briefly explain why or why not. For parts (e), (f), and (g), determine if the resulting interval would be wider, narrower, or the same width as the interval calculated in part (b) without performing the calculation. e. (1Pt) A 99% confidence interval using the same mean, standard deviation, and sample size f. (1Pt) A 95% confidence interval with a population standard deviation of 20 using the same mean and sample size g. (1Pt) A 95% confidence interval with a sample size of 36 using the same mean and standard deviation Suppose the bowler’s true bowling average on a sport shot is 184. Use Minitab to generate 20 samples of size 21 from a normal distribution with mean 184 and population standard deviation 25. Use Minitab to generate a 90% confidence interval for each. • In Minitab, choose Calc > Random Data > Normal. In the pop-up window, type 21 for “Number of rows of data to generate” and type “C1-C20” in “Store in column(s)”. The mean is 185 and the standard deviation is 25. • In Minitab Express, choose Data > Generate Random Data from the menu at the top of the program. Type 20 for “Number of columns to generate” and 21 for “Number of rows in each column”. From the dropdown menu, select “Normal” and type 184 for the mean and 25 for the standard deviation. h. (1Pt) Count the number of intervals that contain the true population mean of 184 and report this number. Also, report any confidence intervals that do not contain 184. i. (1Pt) If 20 samples of size 21 were taken from the population of games on a sport shot pattern for this bowler and a 90% confidence interval was calculated for each, how many of these intervals would we expect to contain 184? j. (1Pt) Are your simulated results similar to the theoretical result from? Briefly explain why or why not.