Let’s work through each question with step-by-step solutions.

Problem 34

Given:

• Sample mean, $\bar{x} = 26$ weeks
• Sample size, $n = 50$
• Sample standard deviation, $s = 6.2$ weeks
• Confidence level = 95%

Solution:

1. Calculate the standard error (SE): $SE = \frac{s}{\sqrt{n}} = \frac{6.2}{\sqrt{50}} \approx 0.876$

2. Find the critical value for a 95% confidence level. Since the sample size is large, we can use the z-value for a 95% confidence interval, which is approximately $z = 1.96$.

3. Calculate the margin of error (ME): $ME = z \cdot SE = 1.96 \times 0.876 \approx 1.717$

4. Construct the confidence interval: $\text{CI} = \left(\bar{x} - ME, \bar{x} + ME\right) = (26 - 1.717, 26 + 1.717) = (24.283, 27.717)$

5. Conclusion: Since 28 weeks does not lie within the 95% confidence interval of (24.283, 27.717), it is not reasonable to conclude that the population mean is 28 weeks.

Problem 35

Given:

• Sample mean, $\bar{x} = 8.32$ years
• Sample size, $n = 40$
• Sample standard deviation, $s = 3.07$ years
• Confidence level = 90%

(a) What is the mean of the population?

The sample mean ($\bar{x}$) is 8.32 years. Without additional information, we estimate the population mean using the sample mean, so: $\text{Estimated population mean} = 8.32 \text{ years}$

(b) Develop a 90% confidence interval for the population mean.

1. Calculate the standard error (SE): $SE = \frac{s}{\sqrt{n}} = \frac{3.07}{\sqrt{40}} \approx 0.4853$

2. Find the critical value for a 90% confidence interval. For a 90% confidence level with $n - 1 = 39$ degrees of freedom, we use the t-distribution. The approximate t-value for a 90% confidence level and 39 degrees of freedom is $t \approx 1.685$.

3. Calculate the margin of error (ME): $ME = t \cdot SE = 1.685 \times 0.4853 \approx 0.818$

4. Construct the confidence interval: $\text{CI} = \left(\bar{x} - ME, \bar{x} + ME\right) = (8.32 - 0.818, 8.32 + 0.818) = (7.502, 9.138)$

(c) Does the sample information substantiate the claim of "almost 10 years"?

Since the 90% confidence interval for the population mean (7.502, 9.138) does not include 10 years, the sample information does not substantiate the claim that the mean length of membership is "almost 10 years."

Would you like further details on any specific calculation steps, or do you have additional questions on these problems?

Here are 5 related questions for further exploration:

1. How would the confidence interval change if the sample size were increased?
2. What is the effect of using a higher confidence level, such as 99%?
3. How would the conclusions differ if the standard deviation were smaller?
4. Can we use a z-distribution instead of a t-distribution in Problem 35? Why or why not?
5. How does the margin of error relate to the confidence level and sample size?

Tip: Remember, as sample size increases, the confidence interval becomes narrower, reflecting a more precise estimate of the population mean.