Find a 90% confidence interval for a population mean for these values. (Round yo

ID: 3062357 • Letter: F

Question

Find a 90% confidence interval for a population mean for these values. (Round your answers to three decimal places.)

(a) n = 105, x = 0.86, s2 = 0.082

(b) n = 60, x = 24.6, s2 = 3.29

90% of all values will fall within the interval.

There is a 10% chance that an individual sample proportion will fall within the interval.

In repeated sampling, 90% of all intervals constructed in this manner will enclose the population mean.

There is a 90% chance that an individual sample proportion will fall within the interval.

In repeated sampling, 10% of all intervals constructed in this manner will enclose the population proportion.

a) n = 105, x = 0.86, s^2 = 0.082

DF = 105 - 1 = 104

With 104 degrees of freedom and 90% confidence interval the critical value is t0.05, 104 = 1.6595

The 90% confidence interval is

x +/- t0.05, 104 * sqrt(s^2/n)

= 0.86 +/- 1.6595 * sqrt(0.082/105)

= 0.86 +/- 0.046

= 0.814, 0.906

b) n = 60, x = 24.6, s^2 = 3.29

df = 60 - 1 = 59

With 104 degrees of freedom and 90% confidence interval the critical value is t0.05, 59 = 1.6711

The 90% confidence interval is

x +/- t0.05, 59 * sqrt(s^2/n)

= 0.86 +/- 1.6711 * sqrt(0.082/105)

= 0.86 +/- 0.047

= 0.813, 0.907

In repeated sampling , 90% of all intervals constructed in this manner will enclose the population mean.

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