If X vector = 50, s = 15, n = 16, and assuming that the population is normally d

Question

If X vector = 50, s = 15, n = 16, and assuming that the population is normally distributed, construct a 99% confidence interval estimate of the population mean mu. 8.13 Construct a 95% confidence interval estimate for the population mean, based on each of the following sets of data, assuming that the population is normally distributed: Set 1: 1, 1, 1, 1, 8, 8, 8, 8 Set 2: 1, 2, 3, 4, 5, 6, 7, 8 Explain why these data sets have different confidence the intervals even though they have the same mean and range.

Explanation / Answer

Q 8.12

100(1 – )% confidence interval for µ when 2 is unknown is: {Xbar ± (s/n)(t/2)}, where

Xbar = sample mean,

= population standard deviation,

s = sample standard deviation,

n = sample size and

t/2 = upper (/2) % point of t-Distribution with (n - 1) degrees of freedom..

Given, n = 16, = 0.01, Xbar = 50, s = 15, and t/2= 2.947, [using Excel Function],

99% Confidence Interval for µ is: {50 ± (15/16)(2.947)} = (50 ± 11.051)

Lower Bound = 38.95, Upper Bound = 61.05

Q 8.13

From the given data, for Set 1, n = 8, Xbar = 4.5 and s = 3.742, = 0.05, t/2 = 2.365,

CI = 4.5 ± 3.129

Lower Bound = 1.37, Upper Bound = 7.63 ANSWER 1

For Set 2, n = 8, Xbar = 4.5 and s = 2.249, = 0.05, t/2 = 2.365,

CI = 4.5 ± 1.881

Lower Bound = 2.62, Upper Bound = 6.38 ANSWER 2

Though the means are the same, standard deviations are different and hence the difference in CI. In general, as the sample standard deviation gets smaller, the CI gets narrower.

DONE

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