Question

A simple random sample of 90 is drawn from a normally distributed population, and the mean is found to be 138, with a standard deviation of 34. What is the 90% confidence interval for the population mean? Use the table below to help you answer the question. Confidence Level 90% 95% 99% z*-score 1. 645 1. 96 2. 58 Remember, the margin of error, ME, can be determined using the formula M E = StartFraction z times s Over StartRoot n EndRoot EndFraction. 128. 75 to 147. 25 130. 98 to 145. 02 132. 10 to 143. 90 137. 38 to 138. 62.

Answer 1

The 90% confidence interval for the population mean of the considered population from the given sample data is given by: Option C:  [130.10, 143.90]

How to find the confidence interval for population mean from large samples (sample size > 30)?

Suppose that we have:

• Sample size n > 30
• Sample mean = $\overline{x}$
• Sample standard deviation = s
• Population standard deviation = $\sigma$
• Level of significance = $\alpha$

Then the confidence interval is obtained as

• Case 1: Population standard deviation is known

$\overline{x} \pm Z_{\alpha /2}\dfrac{\sigma}{\sqrt{n}}$

• Case 2: Population standard deviation is unknown.

$\overline{x} \pm Z_{\alpha /2}\dfrac{s}{\sqrt{n}}$

For this case, we're given that:

• Sample size n = 90 > 30
• Sample mean = $\overline{x}$ = 138
• Sample standard deviation = s = 34
• Level of significance = $\alpha$ = 100% - confidence = 100% - 90% = 10% = 0.1 (converted percent to decimal).

At this level of significance, the critical value of Z is: $Z_{0.1/2}$ = ±1.645

Thus, we get:

$CI = \overline{x} \pm Z_{\alpha /2}\dfrac{s}{\sqrt{n}}\\CI = 138 \pm 1.645\times \dfrac{34}{\sqrt{90}}\\\\CI \approx 138 \pm 5.896\\CI \approx [138 - 5.896, 138 + 5.896]\\CI \approx [132.104, 143.896] \approx [130.10, 143.90]$

Thus, the 90% confidence interval for the population mean of the considered population from the given sample data is given by: Option C:  [130.10, 143.90]

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