Answer :
To answer this question, we will calculate the confidence intervals using the z-score associated with each confidence level. We are given that the sample size (n) is 36, the sample mean (\(\bar{x}\)) is 749, and the population standard deviation (\(\sigma\)) is 32. Since the population standard deviation is known and the sample size is large enough, we can use the z-distribution for our calculations.
Here are the steps to find the confidence intervals:
1. Find the z-score corresponding to each confidence level.
2. Calculate the margin of error for each confidence level.
3. Calculate the confidence interval for the population mean using the formula:
\[ \text{Confidence Interval} = \bar{x} \pm \text{Margin of Error} \]
**a. 90% Confidence Interval**
1. For the 90% confidence level, we need to find the z-score that cuts off the upper 5% of the normal distribution (since the middle 90% leaves 5% on each tail). It is commonly known that the z-score for 90% confidence is approximately 1.645.
2. Next, we calculate the margin of error:
\[ \text{Margin of Error} = z \cdot \frac{\sigma}{\sqrt{n}} \]
\[ \text{Margin of Error} = 1.645 \cdot \frac{32}{\sqrt{36}} \]
\[ \text{Margin of Error} = 1.645 \cdot \frac{32}{6} \]
\[ \text{Margin of Error} \approx 1.645 \cdot 5.333 \]
\[ \text{Margin of Error} \approx 8.777 \]
3. Now, we calculate the confidence interval:
\[ \text{CI}_{90\%} = 749 \pm 8.777 \]
\[ \text{CI}_{90\%} \approx (740.223, 757.777) \]
**b. 95% Confidence Interval**
1. For the 95% confidence level, the z-score that cuts off the upper 2.5% is commonly known to be approximately 1.96.
2. Calculate the margin of error using the z-score of 1.96:
\[ \text{Margin of Error} = 1.96 \cdot \frac{32}{\sqrt{36}} \]
\[ \text{Margin of Error} = 1.96 \cdot \frac{32}{6} \]
\[ \text{Margin of Error} \approx 1.96 \cdot 5.333 \]
\[ \text{Margin of Error} \approx 10.453 \]
3. The confidence interval is:
\[ \text{CI}_{95\%} = 749 \pm 10.453 \]
\[ \text{CI}_{95\%} \approx (738.547, 759.453) \]
**c. 99% Confidence Interval**
1. For the 99% confidence level, the z-score that cuts off the upper 0.5% is commonly known to be approximately 2.576.
2. Calculate the margin of error using the z-score of 2.576:
\[ \text{Margin of Error} = 2.576 \cdot \frac{32}{\sqrt{36}} \]
\[ \text{Margin of Error} = 2.576 \cdot \frac{32}{6} \]
\[ \text{Margin of Error} \approx 2.576 \cdot 5.333 \]
\[ \text{Margin of Error} \approx 13.744 \]
3. The confidence interval is:
\[ \text{CI}_{99\%} = 749 \pm 13.744 \]
\[ \text{CI}_{99\%} \approx (735.256, 762.744) \]
To summarize, the confidence intervals are:
a. The 90% confidence interval is approximately (740.223, 757.777)
b. The 95% confidence interval is approximately (738.547, 759.453)
c. The 99% confidence interval is approximately (735.256, 762.744)
