Num 9 8 7 Pg Up 6 5 4 3 Enter 2 Po On 0 20. A simple random sample of size n is

Question

Num 9 8 7 Pg Up 6 5 4 3 Enter 2 Po On 0 20. A simple random sample of size n is drawn. The sample mean, T, is found to be 35.1, and the sample standard deviation, s,is found to be 8.7 (a) Construct a 90% confidence interval for if the sample size, n, is 40. (b) Construct a 90% confidence interval for if the sample size, n, is 100. How does increasing the sample size affect the margin of error, E? (c) Construct a 98% confidence interval for if the sample size, n, is 40. Compare the results to those obtained in part (a). How does increasing the level of confidence affect the margin of error, E?

Explanation / Answer

x = 35.1

s = 8.7

(a) Here n = 40

90% confidence interval for = x +- t39,0.10 (s/ sqrt(n) )

90% confidence interval for = 35.1 +- 1.6849 * (8.7/ sqrt (40)

= 35.1+- 2.32

= (32.78, 37.42)

(B)

Here n = 100

90% confidence interval for = x +- t99,0.10 (s/ sqrt(n) )

90% confidence interval for = 35.1 +- 1.6604 * (8.7/ sqrt (100)

= 35.1+- 1.44

= (33.66, 36.54)

here sample size affect the margin of error and it reduces the margin of error when sample size increses.

(c)

Here n = 40

confidence interval = 98%

98% confidence interval for = x +- t39,0.02 (s/ sqrt(n) )

98% confidence interval for = 35.1 +- 2.4258 * (8.7/ sqrt (40)

= 35.1+- 3.34

= (31.76, 38.44)

so increasing confidence level will increase the margin of error.

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