# Standard Deviation Randam Sampling

Suppose that weekly change in the price of a stock is normally distributed with a mean of \$17.43 and a standard deviation of \$1.88. Use MINITAB to generate 50 different random samples from this population, each of size n = 9.
• Click on CALC → RANDOM DATA → NORMAL.
• In the box for “Number of rows of data to generate”, enter the sample size (in this case, 9).
• In the box for “Store in columns”, enter “C1-C50”. (without the quotes)
• In the boxes for mean and standard deviation, enter the values 17.43 and 1.88, respectively. Click OK.
Each of the first 50 columns should now contain a random sample of 9 observations from the population under consideration. Based on each sample, we will then use MINITAB to compute an 90% confidence interval for the population mean.
• Click on STAT → BASIC STATISTICS → 1-SAMPLE Z.
• Click the “Options” button. Enter a value of 90 for the “Confidence level”. Click OK.
• Ensure that the drop box at the top of the dialogue box says “One or more samples, each in a column”.
• In the next box, specify the columns by entering “C1-C50”. (without the quotes)
• In the box for “Known standard deviation” enter a value of 1.88. Click OK. You must include the resulting confidence intervals in your assignment submission. Do NOT include the spreadsheet

Based on the 50 confidence intervals you obtained, answer the following questions.
(a) Prior to generating the random samples, how many of these 50 intervals would you expect to contain the true population mean value of \$17.43? Explain.
(b) How many of these 50 intervals actually contain the true population mean value of \$17.43? Should this value equal the value you gave in part (a)? Why or why not?
(c) Are all 50 confidence intervals of exactly the same length? Would this still be the case if the population standard deviation was unknown? Explain.

Solution

1. Prior to generating these random samples we expect 45 of these intervals to contain the true population mean value 17.43 .

Because we are calculating 90% confidence intervals for the mean which implies by repeated sampling from the population,90 out of 100 intervals are expected to contain the true value of the popnmean.In this case there are 50 samples,so 45 of them are expected to contain the popn mean.

1. The CIs thus obtained are as follows.

Variable  N    Mean  StDev  SE Mean       90% CI

C1        9  17.825  1.444    0.627  (16.794, 18.855)

C2        9  18.057  1.275    0.627  (17.026, 19.088)

C3        9  18.194  2.202    0.627  (17.163, 19.225)

C4        9  17.030  3.039    0.627  (15.999, 18.061)

C5        9  18.193  1.094    0.627  (17.162, 19.224)

C6        9  17.744  1.952    0.627  (16.713, 18.775)

C7        9  16.736  2.112    0.627  (15.705, 17.767)

C8        9  18.394  1.615    0.627  (17.363, 19.425)

C9        9  17.048  1.418    0.627  (16.017, 18.079)

C10       9  17.935  2.070    0.627  (16.905, 18.966)

C11       9  17.164  2.035    0.627  (16.133, 18.195)

C12       9  18.215  1.897    0.627  (17.184, 19.246)

C13       9  18.764  2.334    0.627  (17.733, 19.795)

C14       9  17.927  1.720    0.627  (16.896, 18.958)

C15       9  17.904  1.945    0.627  (16.874, 18.935)

C16       9  18.080  1.742    0.627  (17.049, 19.111)

C17       9  17.750  1.465    0.627  (16.720, 18.781)

C18       9  18.079  1.734    0.627  (17.048, 19.110)

C19       9  17.553  1.468    0.627  (16.522, 18.584)

C20       9  18.022  3.059    0.627  (16.991, 19.052)

C21       9  18.490  2.524    0.627  (17.460, 19.521)

C22       9  17.683  1.364    0.627  (16.653, 18.714)

C23       9  17.709  2.341    0.627  (16.678, 18.739)

C24       9  17.179  1.352    0.627  (16.148, 18.210)

C25       9  17.337  2.111    0.627  (16.306, 18.368)

C26       9  17.058  2.110    0.627  (16.027, 18.089)

C27       9  17.752  1.350    0.627  (16.721, 18.783)

C28       9  19.082  1.829    0.627  (18.051, 20.113)

C29       9  17.376  2.113    0.627  (16.346, 18.407)

C30       9  18.023  1.433    0.627  (16.992, 19.054)

C31       9  18.371  2.800    0.627  (17.340, 19.401)

C32       9  18.389  1.094    0.627  (17.358, 19.420)

C33       9  18.765  2.528    0.627  (17.734, 19.796)

C34       9  17.693  3.184    0.627  (16.662, 18.724)

C35       9  17.407  1.952    0.627  (16.377, 18.438)

C36       9  17.621  1.156    0.627  (16.590, 18.651)

C37       9  17.288  1.826    0.627  (16.257, 18.319)

C38       9  17.097  2.217    0.627  (16.066, 18.128)

C39       9  18.347  1.838    0.627  (17.316, 19.377)

C40       9  18.209  2.676    0.627  (17.179, 19.240)

C41       9  17.911  2.288    0.627  (16.880, 18.942)

C42       9  18.183  1.544    0.627  (17.152, 19.214)

C43       9  16.490  1.837    0.627  (15.459, 17.521)

C44       9  16.770  1.756    0.627  (15.739, 17.801)

C45       9  16.876  1.351    0.627  (15.845, 17.907)

C46       9  19.433  1.886    0.627  (18.402, 20.464)

C47       9  19.370  2.217    0.627  (18.339, 20.400)

C48       9  17.287  1.840    0.627  (16.256, 18.318)

C49       9  18.025  2.721    0.627  (16.994, 19.056)

C50       9  17.262  2.292    0.627  (16.231, 18.293)

We see that out of these,46 of them contain the true value of the mean.This value is not equal to 45.

90% CI essentially means that by repeated sampling from the  concerned distribution,one is expected to get 90 out of 100 of them to actually contain the true value,it does not guarantee that we will get exactly 90 out of them containing the true value.Since we  only have 50 such samples,wewont get exactly 45 of them to contain the turevalue.but the no of intervals actually containing will be around that expected value.

1. The 50 intervals calculated are of the same length. This is because the length of a CI of the mean depends on the standard deviations of each sample, which is a known quantity here as population sd is known and samples are all of the same size.

This wouldn’t be the case if popnsd was unknown since then we would have to estimate the sd with the sample sd obtained from a sample that would be different for different samples.