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You are opening your own wedding planning business and are conducting some marke

ID: 3334013 • Letter: Y

Question

You are opening your own wedding planning business and are conducting some market research. According to The Knot Real Weddings Study, the average cost of a wedding in the U.S. is $35,000.

You are skeptical about whether these numbers apply to the Piedmont Triad area—your target market—and so you decide to conduct your own survey. After randomly sampling 40 recently married couples, you find an average cost of $32,000 with a standard deviation of $10,000.

When you conduct your survey, you also ask whether the couple had a destination wedding. Seven out of the 40 couples reported having a destination wedding.

A. What are the point estimates for mean wedding cost and proportion of destination weddings for the Piedmont Triad?

B. Find the 90% confidence interval for the mean wedding cost in the Piedmont Triad.. Find the standard error of the mean.

B2. Find the critical value. (You’ll need to decide whether to use a z-score or a t-score.)

B3.Put it together to find the confidence interval.

C. . What is the margin of error for the 90% confidence interval?

D. Interpret the 90% confidence interval for the mean wedding cost in the Piedmont Triad. Does the interval include the average for the U.S.? Would you conclude that the cost of weddings in the Piedmont Triad is different from the rest of the U.S.?

E. Find the 95% confidence interval for the mean wedding cost for the Piedmont Triad. How does it compare to the 90% confidence interval (narrower or wider)? Does your conclusion about the Piedmont Triad being different change?

F. Suppose you wanted to keep the confidence level at 90%, but you wanted to decrease the width of the interval. How could you do this?

G. Find the 80% confidence interval for the proportion of couples that have a destination wedding.

Find the standard error of the proportion.

G2. Find the critical value.

G3. Put it together to find the confidence interval.

H. You find out that respondents to The Knot’s survey were recruited from members of TheKnot.com. Do you trust that the survey results come from a representative sample? Why or why not?

Explanation / Answer

A.

The point estimates for mean wedding cost is sample average cost = $32,000

The proportion of destination weddings for the Piedmont Triad = 7 / 40 = 0.175

B.

Standard error of mean = SD / sqrt(n) = 10,000 / sqrt(40) = 1581.139

As, the sample size is greater than 30, we use z score.

Z-score for 90% confidence interval is 1.64

90% confidence interval of mean wedding cost is

($32,000 - 1.64 * 1581.139, $32,000 + 1.64 * 1581.139)

($29406.93, $34593.07)

C.

Margin of error for the 90% confidence interval = 1.64 * 1581.139 = 2593.068

D.

When we take multiple random samples of recently married couples, and find an average cost, 90% of the times, average cost will lie in the 90% confidence interval ($29406.93, $34593.07)

No, the interval does not include the average for the U.S ($35,000)

As, 90% confidence interval does not include the average for the U.S , we can conclude that the cost of weddings in the Piedmont Triad is different from the rest of the U.S.

E.

Z-score for 95% confidence interval is 1.96

95% confidence interval of mean wedding cost is

($32,000 - 1.96 * 1581.139, $32,000 + 1.96 * 1581.139)

($28900.97, $35099.03)

Width of 90% confodence interval = 34593.07 - 29406.93 = 5186.14

Width of 95% confodence interval = 35099.03 - 28900.97 = 6198.06

We see that 95% confidence interval is wider than the 90% confidence interval.

As, 95% confidence interval include the average for the U.S , we can conclude that the cost of weddings in the Piedmont Triad is similar to the rest of the U.S. So, the conclusion about the Piedmont Triad being different change.

F.

To decrease the width of the interval, we need to decrease the matgin of error and hence we need to increase the standard error of the mean. To decrease the  standard error of the mean, we need to increase the sample size.

Hence, to decrease the width of the interval, we need to increase the sample size.

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