To solve these problems, we need to construct confidence intervals based on the given data. The formula for constructing a confidence interval is:

$CI = \bar{X} \pm Z \times \frac{S}{\sqrt{n}}$

Where:

• $\bar{X}$ is the sample mean,
• $Z$ is the z-value associated with the confidence level,
• $S$ is the sample standard deviation,
• $n$ is the sample size,
• $\frac{S}{\sqrt{n}}$ is the standard error (SE).

Let's solve each part of the question.

Problem 1: Heights of Guyanese Men

Given:

• Sample size $n = 100$,
• Sample mean $\bar{X} = 71$ inches,
• Sample standard deviation $S = 2.8$ inches.

a. 90% Confidence Interval

For 90% confidence, $Z = 1.64$.

The standard error $SE$ is: $SE = \frac{S}{\sqrt{n}} = \frac{2.8}{\sqrt{100}} = 0.28$

Now, compute the confidence interval: $CI = 71 \pm 1.64 \times 0.28$ $CI = 71 \pm 0.4592$ $CI = (70.5408, 71.4592)$

So, the 90% confidence interval is approximately $(70.54, 71.46)$.

b. 95% Confidence Interval

For 95% confidence, $Z = 1.96$.

Now, compute the confidence interval: $CI = 71 \pm 1.96 \times 0.28$ $CI = 71 \pm 0.5488$ $CI = (70.4512, 71.5488)$

So, the 95% confidence interval is approximately $(70.45, 71.55)$.

c. 99% Confidence Interval

For 99% confidence, $Z = 2.58$.

Now, compute the confidence interval: $CI = 71 \pm 2.58 \times 0.28$ $CI = 71 \pm 0.7224$ $CI = (70.2776, 71.7224)$

So, the 99% confidence interval is approximately $(70.28, 71.72)$.

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Problem 2: Time to Complete Tax Forms

Given:

• Sample size $n = 100$,
• Sample mean $\bar{X} = 23.6$ hours,
• Sample standard deviation $S = 7.0$ hours.

a. 90% Confidence Interval

For 90% confidence, $Z = 1.64$.

The standard error $SE$ is: $SE = \frac{S}{\sqrt{n}} = \frac{7.0}{\sqrt{100}} = 0.7$

Now, compute the confidence interval: $CI = 23.6 \pm 1.64 \times 0.7$ $CI = 23.6 \pm 1.148$ $CI = (22.452, 24.748)$

So, the 90% confidence interval is approximately $(22.45, 24.75)$.

b. 95% Confidence Interval

For 95% confidence, $Z = 1.96$.

Now, compute the confidence interval: $CI = 23.6 \pm 1.96 \times 0.7$ $CI = 23.6 \pm 1.372$ $CI = (22.228, 24.972)$

So, the 95% confidence interval is approximately $(22.23, 24.97)$.

c. 99% Confidence Interval

For 99% confidence, $Z = 2.58$.

Now, compute the confidence interval: $CI = 23.6 \pm 2.58 \times 0.7$ $CI = 23.6 \pm 1.806$ $CI = (21.794, 25.406)$

So, the 99% confidence interval is approximately $(21.79, 25.41)$.

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Summary:

1. 90% CI for heights: $(70.54, 71.46)$
2. 95% CI for heights: $(70.45, 71.55)$
3. 99% CI for heights: $(70.28, 71.72)$
4. 90% CI for tax form time: $(22.45, 24.75)$
5. 95% CI for tax form time: $(22.23, 24.97)$
6. 99% CI for tax form time: $(21.79, 25.41)$

Would you like further explanation on any of these calculations?

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Follow-up Questions:

1. What does a confidence interval tell us about the population?
2. How does increasing the confidence level affect the width of the interval?
3. How does sample size influence the standard error and confidence interval?
4. What assumptions are necessary for constructing these confidence intervals?
5. Why do we use the z-value and how is it related to the normal distribution?

Tip: Increasing the confidence level widens the interval, because we need to account for more variability to be more certain.