Questions: Construct a 95% confidence interval to estimate the population mean when x̄=138 and s=33 for the sample sizes below. a) n=30 b) n=70 c) n=90

Solution Steps

To construct a 95% confidence interval for the population mean, we use the formula: \[ \bar{x} \pm t_{\alpha/2} \left(\frac{s}{\sqrt{n}}\right) \] where:

• \(\bar{x}\) is the sample mean
• \(s\) is the sample standard deviation
• \(n\) is the sample size
• \(t_{\alpha/2}\) is the critical t-value for a 95% confidence level and \(n-1\) degrees of freedom

We will use Python to calculate the confidence intervals for the given sample sizes.

We are given:

• Sample mean \(\bar{x} = 138\)
• Sample standard deviation \(s = 33\)
• Sample sizes \(n = 30\), \(n = 70\), and \(n = 90\)

The formula for the 95% confidence interval for the population mean is: \[ \bar{x} \pm t_{\alpha/2} \left(\frac{s}{\sqrt{n}}\right) \]

For \(n = 30\):

• Degrees of freedom \(df = 30 - 1 = 29\)
• Critical t-value \(t_{\alpha/2} \approx 2.0452\) (from t-distribution table)

The margin of error is: \[ ME = 2.0452 \left(\frac{33}{\sqrt{30}}\right) \approx 12.32 \]

Thus, the confidence interval is: \[ 138 \pm 12.32 \] \[ (138 - 12.32, 138 + 12.32) \] \[ (125.68, 150.32) \]

For \(n = 70\):

• Degrees of freedom \(df = 70 - 1 = 69\)
• Critical t-value \(t_{\alpha/2} \approx 2.0003\) (from t-distribution table)

The margin of error is: \[ ME = 2.0003 \left(\frac{33}{\sqrt{70}}\right) \approx 7.87 \]

Thus, the confidence interval is: \[ 138 \pm 7.87 \] \[ (138 - 7.87, 138 + 7.87) \] \[ (130.13, 145.87) \]

For \(n = 90\):

• Degrees of freedom \(df = 90 - 1 = 89\)
• Critical t-value \(t_{\alpha/2} \approx 1.9873\) (from t-distribution table)

The margin of error is: \[ ME = 1.9873 \left(\frac{33}{\sqrt{90}}\right) \approx 6.91 \]

Thus, the confidence interval is: \[ 138 \pm 6.91 \] \[ (138 - 6.91, 138 + 6.91) \] \[ (131.09, 144.91) \]

Final Answer