Question

12 Consider a sample of willingness to donate for a social cause, as below:

x

290

175

100

240

175

13 Find the standard deviation of the above sample:

a. 72.23 b. 49.57 c. 95.7 d. 49.57

14 The margin of error for an interval estimate with 90% level of confidence is,

a. 79.28 b. 85.32 c. 68.87 d. 80.73

15 To obtain a margin of error of $10, what is the minimum number of data to be included in the sample? Use standard deviation of $75 as the planning value. Confidence level still is 90%.

a. 152 b. 208 c. 176 d. 127

Answer 1

13) The standard deviation of the above sample = 72.23

14)

Corresponding to 90% confidence interval and df = n - 1 = 4

the t value = 2.132

Thus margin of error = 2.132*72.23/√5 = 68.87

15)

Here the sample size needs to be high for a margin of error at 90% confidence level.

Thus, from Central Limit Theorem we can assume the sample mean to follow Normal distribution

For 90% confidence interval, the z score = 1.645

-> 10 = 1.645*75/√n

-> n = 152.21 ≈ 152