Sure! Let's solve this step by step.
### Part (a): Compute the 95% confidence interval for the population mean with a sample of 60 items.
1. Given Information:
- Sample size ([tex]\( n \)[/tex]) = 60
- Sample mean ([tex]\( \bar{x} \)[/tex]) = 80
- Population standard deviation ([tex]\( \sigma \)[/tex]) = 15
- Confidence level = 0.95
2. Determine the z-score for a 95% confidence level:
A 95% confidence level corresponds to a z-score of approximately 1.96 (from the standard normal distribution table).
3. Calculate the standard error (SE):
[tex]\[
SE = \frac{\sigma}{\sqrt{n}} = \frac{15}{\sqrt{60}}
\][/tex]
4. Calculate the margin of error (ME):
[tex]\[
ME = z \times SE = 1.96 \times \frac{15}{\sqrt{60}}
\][/tex]
5. Determine the confidence interval:
[tex]\[
\text{Lower bound} = \bar{x} - ME = 80 - \left(1.96 \times \frac{15}{\sqrt{60}}\right)
\][/tex]
[tex]\[
\text{Upper bound} = \bar{x} + ME = 80 + \left(1.96 \times \frac{15}{\sqrt{60}}\right)
\][/tex]
6. Round the answers to one decimal place:
[tex]\[
\text{Confidence Interval} = (76.2, 83.8)
\][/tex]
### Part (b): Compute the 95% confidence interval for the population mean with a sample of 120 items.
1. Given Information:
- Sample size ([tex]\( n \)[/tex]) = 120
- Same sample mean ([tex]\( \bar{x} \)[/tex]) = 80
- Population standard deviation ([tex]\( \sigma \)[/tex]) = 15
- Confidence level = 0.95
2. Determine the z-score for a 95% confidence level:
The z-score remains 1.96 for a 95% confidence level.
3. Calculate the standard error (SE):
[tex]\[
SE = \frac{\sigma}{\sqrt{n}} = \frac{15}{\sqrt{120}}
\][/tex]
4. Calculate the margin of error (ME):
[tex]\[
ME = z \times SE = 1.96 \times \frac{15}{\sqrt{120}}
\][/tex]
5. Determine the confidence interval:
[tex]\[
\text{Lower bound} = \bar{x} - ME = 80 - \left(1.96 \times \frac{15}{\sqrt{120}}\right)
\][/tex]
[tex]\[
\text{Upper bound} = \bar{x} + ME = 80 + \left(1.96 \times \frac{15}{\sqrt{120}}\right)
\][/tex]
6. Round the answers to two decimal places:
[tex]\[
\text{Confidence Interval} = (77.32, 82.68)
\][/tex]
### Part (c): What is the effect of a larger sample size on the interval estimate?
A larger sample size reduces the standard error because the denominator (the square root of the sample size) increases. Consequently, the margin of error, which is the product of the z-score and the standard error, decreases. Thus, a larger sample size provides a smaller margin of error, leading to a more precise confidence interval.
Answer:
Larger sample provides a smaller margin of error.
