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You measure 50 textbooks' weights, and find they have a mean weight of 52 ounces. Assume the population standard deviation is 14.6 ounces. Based on this, construct a 90% confidence interval for the true population mean textbook weight.

Respuesta :

The 90% confidence interval = 48.603 < μ < 55.397

Explanation:

Given:

Mean = 52

Standard deviation = 14.6

Sample size, n = 50

Confidence Interval, c = 90%

                                  c = 0.9

Significance level, α = 1 - c

                                  = 1 - 0.9

                                  = 0.1

Critical value, z(α/2) = z(0.05) = 1.645

Critical value = ± 1.645

Margin of error, E = [tex]z(\alpha /2) X \frac{SD}{\sqrt{n} }[/tex]

                             [tex]=1.645 X\frac{14.6}{\sqrt{50} } \\\\= 3.3965[/tex]

Limits of 90% confidence interval are given by:

Lower limit = μ - E

                  = 52 - 3.397

                  = 48.603

Upper limit = μ + E

                  = 52 + 3.397

                  = 55.397

Thus, 90% confidence interval = 48.603 < μ < 55.397

                 

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