A sample of 20 pages was taken from a Yellow Pages directory. On each page, the

Question

A sample of 20 pages was taken from a Yellow Pages directory. On each page, the mean area devoted to display ads was measured (a display ad is a large block of multicolored illustrations, maps, and text). The data (in square millimeters)are shown below: 0 260 356 403 536 268 369 428 536 268 396 469 536 162 338 403 536 536 130 (a) Construct a 95percent confidence interval for the true mean. (b)What sample size would be needed to obtain an error (If t 20 square millimeters with95 percent confidence? (Data are based on a project by MBA student Daniel R. Dalach.) Display Ads

Explanation / Answer

Solution:

Part a

Here, we have to find 90% confidence interval for the population mean.

Confidence interval = Xbar -/+ t*S/sqrt(n)

From the given data, we have

Sample mean = Xbar = 346.5

Sample standard deviation = S = 170.3783715

Sample size = n = 20

Degrees of freedom = n - 1 = 20 – 1 = 19

Confidence level = 90%

Critical t value = 1.7291

Lower limit = Xbar - t*S/sqrt(n)

Lower limit = 346.5 - 1.7291*170.3783715/sqrt(20) = 280.6252

Upper limit = Xbar + t*S/sqrt(n)

Upper limit = 346.5 + 1.7291*170.3783715/sqrt(20) =412.3748

Confidence interval = (280.62, 412.38)

Part b

Here, we have to find sample size.

Sample size formula is given as below:

n = (Z*/E)^2

Here, we assume estimate for = 170.3783715

We are given

Confidence level = 90%

Critical Z value = 1.6449

Sampling error = E = 20

n = (1.6449*170.38/20)^2 = 196.3615

n = 197

Required sample size = 197

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---