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A random sample of 86 observations produced a mean x 25.4 and a standard deviation s 2.7 a. Find a 95% confidence interval fo

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From given data,

A random sample of 86 observations produced a mean \bar{x} = 25.4 and a standard deviation s = 2.7.

sample size = n= 86

mean = \bar{x} = 25.4

standard deviation = s = 2.7

Formula for for the population mean µ

\bar{x}  \pm Z\alpha/2\times (\sigma / \sqrt{n} )

(a) Find a 95% confidence interval for \mu

95% confidence interval

Confidence interval is 95%

95% = 95/100 = 0.95

\alpha = 1 - Confidence interval = 1-0.95 = 0.05

\alpha/2 = 0.05 / 2

= 0.025

Z\alpha/2 = Z0.025 = 1.96

for the population mean µ

\bar{x}  \pm Z\alpha/2\times (s / \sqrt{n} )

\bar{x}  \pm Z0.025\times (s / \sqrt{n} )

25.4 \pm 1.96\times (2.7 / \sqrt{86} )

25.4 \pm 1.96\times (2.7 / 9.2736 )

25.4 \pm 0.57065

(24.82935, 25.97065)

24.829 < \mu < 25.971

(b) Find a 90% confidence interval for \mu

90% confidence interval

Confidence interval is 90%

90% = 90/100 = 0.90

\alpha = 1 - Confidence interval = 1-0.90 = 0.10

\alpha/2 = 0.10 / 2

= 0.05

Z\alpha/2 = Z0.05 = 1.64

for the population mean µ

\bar{x}  \pm Z\alpha/2\times (s / \sqrt{n} )

\bar{x}  \pm Z0.05\times (s / \sqrt{n} )

25.4 \pm 1.64\times (2.7 / \sqrt{86} )

25.4 \pm 1.64\times (2.7 / 9.2736 )

25.4 \pm 0.477484

(24.922516, 25.877484)

24.923 < \mu < 25.877

(c) Find a 99% confidence interval for \mu ​​​​​​​

99% confidence interval

Confidence interval is 99%

99% = 99/100 = 0.99

\alpha = 1 - Confidence interval = 1-0.99 = 0.01

\alpha/2 = 0.01 / 2

= 0.005

Z\alpha/2 = Z0.005  = 2.57

for the population mean µ

\bar{x}  \pm Z\alpha/2\times (s / \sqrt{n} )

\bar{x}  \pm Z0.05\times (s / \sqrt{n} )

25.4 \pm 2.58\times (2.7 / \sqrt{86} )

25.4 \pm 2.58\times (2.7 / 9.2736 )

25.4 \pm 0.751164

(24.648836, 26.151164)

24.649 < \mu < 26.151

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