Math Problem Statement

How many computers? In a simple random sample of 200 households, the sample mean number of personal computers was 1.45. Assume the population standard deviation is σ = 0.85. (a) Construct a 95% confidence interval for the mean number of personal computers. (b) If the sample size were 180 rather than 200, would the margin of error be larger or smaller than the result in part (a)? (c) If the confidence level were 99.5% rather than 95%, would the margin of error be larger or smaller than the result in part (a)? (d) Based on the confidence interval constructed in part (a), is it likely that the mean number of personal computers is less than 1.17?

Solution

Let's go through each part of the problem step-by-step:

Part 1

For part (a), the confidence interval calculation is based on the sample mean, sample size, population standard deviation, and the specified confidence level (95%).

Given:

  • Sample mean (xˉ\bar{x}) = 1.45
  • Population standard deviation (σ\sigma) = 0.85
  • Sample size (nn) = 200
  • Confidence level = 95%

Using the formula for a confidence interval: xˉ±zα/2×σn\bar{x} \pm z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}} For a 95% confidence level, zα/21.96z_{\alpha/2} \approx 1.96.

Calculating the margin of error: Margin of Error=1.96×0.852000.12\text{Margin of Error} = 1.96 \times \frac{0.85}{\sqrt{200}} \approx 0.12 So, the confidence interval is approximately: 1.45±0.12(1.33,1.57)1.45 \pm 0.12 \Rightarrow (1.33, 1.57)

Part 2

In part (b), if the sample size were 180 instead of 200, we would have a smaller sample size, leading to a larger margin of error because: Standard Error=σn\text{Standard Error} = \frac{\sigma}{\sqrt{n}} A decrease in nn increases the standard error, which increases the margin of error. Thus, the correct choice is "larger" margin of error.

Part 3

For part (c), if the confidence level were increased to 99.5%, the zz-value would increase (99.5% confidence level corresponds to a zz-value around 2.807). Higher confidence levels result in larger margins of error because the confidence interval becomes wider to ensure higher certainty. Therefore, the correct answer is "larger" margin of error.

Part 4

In part (d), to determine if it is likely that the mean number of personal computers is less than 1.17, we check the confidence interval from part (a), which is (1.33, 1.57). Since 1.17 is not within this range, it is unlikely that the mean number of personal computers is less than 1.17. So, the answer is "It is not likely".

Would you like further details on any of these calculations, or do you have additional questions?

Additional Related Questions

  1. How does changing the sample size impact the width of a confidence interval?
  2. What are the implications of using the population standard deviation instead of the sample standard deviation?
  3. How would the confidence interval change if the population standard deviation were unknown?
  4. Why does a higher confidence level result in a larger margin of error?
  5. What are the practical implications of a confidence interval that does not include certain values?

Tip

When interpreting confidence intervals, remember that a 95% confidence interval does not mean there's a 95% probability that the population mean lies within the interval; rather, it means that if we were to take many samples and calculate confidence intervals for each, approximately 95% of them would contain the true population mean.

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Math Problem Analysis

Mathematical Concepts

Confidence Intervals
Margin of Error
Standard Error
Hypothesis Testing

Formulas

Confidence Interval: \(\bar{x} \pm z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}}\)
Standard Error: \(\frac{\sigma}{\sqrt{n}}\)
Margin of Error: \(z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}}\)

Theorems

Central Limit Theorem

Suitable Grade Level

Undergraduate Introductory Statistics