The Bureau of Labor Statistics is planning the next yearly survey to determine t

Question

The Bureau of Labor Statistics is planning the next yearly survey to determine the average cost of a summer vacation for a US family. The following standards have been set: a confidence level of 95% and an error of less than $100. Past research has indicated that the standard deviation should be $656.

What is the required sample size?

What would happen to your results if you use a sample smaller than the required sample size?

What would be the required sample size if the BLS standards specify a 99% confidence level instead?

Explanation / Answer

Solution-

A. Step 1 -Divide the confidence interval by two, and look that area up in the z-table:
.95 / 2 = 0.475
The closest z-score for 0.475 is 1.96.

Step 2- Multiply Z score by Std Dev = 656*1.96 = 1285.76

Step 3: Divide Step 2 by the margin of error. Our margin of error (from the question), is $100

1285.76/100 = 12.8576

Step 4: Square Step 3 to get the sample size-
12.8576 * 12.8576 = 165.32 = 166

Required sample size = 166

B. Let us see how sample size affects different parameters-

Thus if we use a smaller sample than required, our margin of error gets large and our confidence level will decrease.

C. If confidence level is 99%, then Z would be 2.58 instead of 1.96. Rest of the calculation will remain same as shown above for part A.

Required Sample Size = [(656*2.58)/100] ^2 = 286.45 = 287.

As we can see our sample size increased with an increase in the confidence level.

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