How do you measure Confidence Level?

Answer 1

That is with a confidence interval of approximately 95% the "true mean" is within the interval of [336.10, 353.90] and that the sample mean (which is an estimate of the "true mean") is $350.00.

SMALL-SAMPLE CONFIDENCE INTERVAL FOR A POPLATION MEAN, t-DISTRIBUTION

95% Confidence Interval = x-bar +/- (t-critical value) * s/SQRT(n)

x-bar = SAMPLE MEAN [350]

s = STANDARD DEVIATION [100]

n = NUMBER OF SAMPLES [200]

n - 1 = 199 df (DEGREES OF FREEDOM)

t-critical value = (approx) 1.972 from "look-up Table for "two-sided interval" df = 200 [CLOSED df IN TABLE]

95% Confidence Interval: 350+/- 1.972 *100 / SQRT(200) = [336.10, 353.90]

That is with a confidence interval of approximately 95% the "true mean" is within the interval of [336.10, 353.90] and that the sample mean (which is an estimate of the "true mean") is $350.00.

c. ANSWER: A random selection of 1537 customers will provide 95% confidence for estimating the mean extended warranty price paid.

Why???

CHOOSING THE SAMPLE SIZE

n = [(z-critical value * s)/B]^2

z-critical value = 1.96 (associated with 95% confidence level)

s = STANDARD DEVIATION [100.00]

B = BOUND ON THE ERROR OF ESTIMATION [5.00]

n = [(1.96 * 100.00)/5.00]^2 = 1537 (ROUNDED - UP)

CONCLUSION: A random selection of 1537 customers will provide 95% confidence for estimating the mean extended warranty price paid.