Question

The distribution of heights of 18-year-old men in the United States is approximately normal, with mean...

The distribution of heights of 18-year-old men in the United States is approximately normal, with mean 68 inches and standard deviation 3 inches (U.S. Census Bureau). In Minitab, we can simulate the drawing of random samples of size 20 from this population (⇒ Calc ⇒ Random Data ⇒ Normal, with 20 rows from a distribution with mean 68 and standard deviation 3). Then we can have Minitab compute a 95% confidence interval and draw a boxplot of the data (⇒ Stat ⇒ Basic Statistics ⇒ 1 -- Sample t, with boxplot selected in the graphs). The boxplots and confidence intervals for four different samples are shown in the accompanying figures. The four confidence intervals are as follows. VARIABLE N MEAN STDEV SEMEAN 95.0 % CI Sample 1 20 68.050 2.901 0.649 (66.692 , 69.407) Sample 2 20 67.958 3.137 0.702 (66.490 , 69.426) Sample 3 20 67.976 2.639 0.590 (66.741 , 69.211) Sample 4 20 66.908 2.440 0.546 (65.766 , 68.050) (a) Examine the figure [parts (a) to (d)]. How do the boxplots for the four samples differ? Why should you expect the boxplots to differ? They differ in interquartile length and whisker length. The boxplots differ because they come from different samples. They differ in interquartile length, median location, and whisker length. The boxplots differ because they come from the same sample. They differ in interquartile length and median location. The boxplots differ because they come from different samples. They differ in interquartile length, median location, and whisker length. The boxplots differ because they come from different samples. (b) Examine the 95% confidence intervals for the four samples shown in the printout. Do the intervals differ in length? Do the intervals all contain the expected population mean of 68 inches? ---Select--- If we draw more samples, do you expect all of the resulting 95% confidence intervals to contain μ = 68? Why or why not? We expect 95% of the samples to generate intervals that contain the mean of the population. We expect none of the samples to generate intervals that contain the mean of the population. We expect all of the samples to generate intervals that contain the mean of the population. We expect 5% of the samples to generate intervals that contain the mean of the population

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