4. Use the values in the table below, which represent the heights, in inches, of

Question

4. Use the values in the table below, which represent the heights, in inches, of 36 randomly-selected students enrolled in BMGT 230, this semester, to answer/complete Parts a through c: 68 72 62 70 70 72 70 697 6471 68 62 65 65 63 67 62 64 67 63 70 65 63 69 77 66 7471 6977 77 65 70 71 60 a. Estimate, with 90 percent confidence, the proportion of all students enrolled in BMGT 230, this semester, who are at least 70 inches tall. b. How would the width of the interval in Part a change if you wanted to be 99 percent confident, holding all other values/ideas constant? c. How would the width of the interval in Part a change if the sample size equaled 49, holding all other values/ideas constant?

Explanation / Answer

There are X = 15 students out of n = 36 students whose height is at least 70 inches.

So number of success = X = 15

Sample size = n = 36

a) In this part we want to obtain 90% confidence interval for population proportion ( P )

Let' write given information.

So confidence = c = 90%

Let's used minitab to construct confidence interval for population proportion ( P )

Step 1) Click on Stat>>>Basic statistics>>>1-proportion...

Step 2) select summarized data

number of events = x = 15

Number of trials = n = 36

Step 3) click on option

The given confidence level is = 90.0

so put "Confidence level " = 90.0

Alternative = not equal

then click on "Use test and interval based on normal approximation"

Then click on OK and again click on OK

So we get the following output

Test and CI for One Proportion

Sample X N Sample p 90% CI
1 15 36 0.416667 (0.281513, 0.551821)

From the above output the 90% confidence interval is (0.281513, 0.551821)

b) Using same procedure we can get 99% confidence interval by replacing level of confidence.

The minitab output for 99% confidence interval is

Test and CI for One Proportion

Sample X N Sample p 99% CI
1 15 36 0.416667 (0.205017, 0.628317)

From the above output the 90% confidence interval is   (0.205017, 0.628317)

So the width of 99% confidence interval is wider than 90% confidence interval

c) Proportion P = 15/36 = 0.41667

So we can expect that there are X = n*P = 49 * 0.41667 = 20 students out of n = 49 students whose height is at least 70 inches.

Using minitab

We want to obtain 90% confidence interval for population proportion ( P )

n = sample size = 49

Number of success = x = 20

Confidence = c = 90%

Step 1) Click on Stat>>>Basic statistics>>>1-proportion...

Step 2) select summarized data

number of events = x = 20

Number of trials = n = 49

Step 3) click on option

so put "Confidence level " = 90.0

Alternative = not equal

then click on "Use test and interval based on normal approximation"

Then click on OK and again click on OK

So we get the following output

Test and CI for One Proportion

Sample X N Sample p 90% CI
1 20 49 0.408163 (0.292673, 0.523654)

From the above output the 90% confidence interval is   (0.292673, 0.523654)

So the width of 90% confidence interval when n = 36 is wider than 90% confidence interval when n = 49.

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