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Chapter 8: Problem 44

Independent random samples of size \(n_{1}=n_{2}=\) 500 were selected from each of two populations. The mean and standard deviations for the two samples were \(\bar{x}_{1}=125.2, \bar{x}_{2}=123.7, s_{1}=5.6,\) and \(s_{2}=6.8\) a. Find a point estimate for the difference in the two population means. Calculate the margin of error. b. Based on the results in part a, can you conclude that there is a difference in the two population means? Explain.

Short Answer

Expert verified
Based on the results obtained from the given data, we found that the point estimate for the difference in the two population means was 1.5, and the margin of error was approximately 0.772 with a 95% confidence level. As the confidence interval of (0.728, 2.272) does not contain zero, we can conclude that there is a significant difference in the two population means with 95% confidence.

Step by step solution

01

Calculate the point estimate for the difference in the two population means

To find the point estimate for the difference in the two population means, we'll subtract the sample mean of population two from the sample mean of population one: \(\bar{x}_{1} - \bar{x}_{2}\). Using the given sample means, this is: \(125.2 - 123.7 = 1.5\)
02

Calculate the standard error

The standard error for the difference in means is given by the following formula: \(\text{Standard Error} (SE) = \sqrt{\frac{s_{1}^2}{n_{1}} + \frac{s_{2}^2}{n_{2}}}\) Plugging in our given values: \(SE = \sqrt{\frac{5.6^2}{500} + \frac{6.8^2}{500}} = \sqrt{\frac{31.36}{500} + \frac{46.24}{500}} = \sqrt{ 0.06272 + 0.09248 } = \sqrt{0.1552} \approx 0.394\)
03

Calculate the margin of error

To calculate the margin of error, we will use the standard error and the critical value from the standard normal distribution. Since the sample sizes are relatively large, we can approximate the critical value with a \(95\%\) confidence level which is equal to \(1.96\): \(\text{Margin of Error} = \text{Critical Value} \times \text{Standard Error} = 1.96 \times 0.394 \approx 0.772\)
04

Determine if there is a difference in the two population means based on the margin of error

The point estimate of the difference in the population means is \(1.5\) and the margin of error is \(0.772\). We can construct a \(95\%\) confidence interval for the difference in population means by subtracting and adding the margin of error to the point estimate: \(\text{Confidence Interval} = (1.5 - 0.772, 1.5 + 0.772) = (0.728, 2.272)\) Since the confidence interval does not contain zero, we can conclude that there is a difference in the two population means at a \(95\%\) confidence level.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Point Estimate
When you want to understand the difference between two population means, a point estimate provides a valuable starting point. In this exercise, the point estimate is calculated by finding the difference between the two sample means. The formula used is simple:
  • Subtract the sample mean of the second population from the sample mean of the first population: \( \bar{x}_{1} - \bar{x}_{2} \).
  • In our example, it's computed as \(125.2 - 123.7 = 1.5 \).
This value, 1.5, is the best guess or estimate of what the actual difference in the population means would be based on the samples we have.
Margin of Error
The margin of error helps quantify the uncertainty in our point estimate. It's calculated using the standard error of the difference and a critical value from the normal distribution.
For a 95% confidence level, the critical value is commonly 1.96.
  • First, we calculate the standard error (SE), a measure of how much the sample means vary: \[\text{SE} = \sqrt{\frac{s_{1}^2}{n_{1}} + \frac{s_{2}^2}{n_{2}}} = \sqrt{0.1552} \approx 0.394\].
  • Then, the margin of error is found by multiplying the critical value by the standard error: \[\text{Margin of Error} = 1.96 \times 0.394 \approx 0.772\].
This margin of error indicates that our point estimate could vary by 卤0.772 due to random sampling variability.
Confidence Interval
A confidence interval presents a range of plausible values for the difference between population means. It's derived by adding and subtracting the margin of error from the point estimate.
  • In our example, the confidence interval is calculated as: \[(1.5 - 0.772, 1.5 + 0.772) = (0.728, 2.272)\].
  • This interval means we're 95% confident that the true difference in population means lies somewhere between 0.728 and 2.272.
A key aspect of this interval is that it does not include zero, suggesting a significant difference between the populations at the 95% confidence level.
Population Mean Difference
Understanding the difference in population means allows us to conclude whether one population has a higher mean than the other.
  • With a point estimate of 1.5 and a confidence interval of (0.728, 2.272), we infer that the difference in means is positive.
  • This suggests that, on average, the first population's mean is greater than the second population's mean.
  • Because the confidence interval does not span zero, we have evidence against the null hypothesis (no difference in means) at a 95% confidence level.
This result provides a clear indication that there is a discernible difference between the two population means.

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Most popular questions from this chapter

Refer to Exercise \(8.7 .\) What effect does increasing the sample size have on the margin of error?

A random sample of \(n=64\) observations has a mean \(\bar{x}=29.1\) and a standard deviation \(s=3.9 .\) a. Give the point estimate of the population mean \(\mu\) and find the margin of error for your estimate. b. Find a \(90 \%\) confidence interval for \(\mu\). What does "90\% confident" mean? c. Find a \(90 \%\) lower confidence bound for the population mean \(\mu\). Why is this bound different from the lower confidence limit in part b? d. How many observations do you need to estimate \(\mu\) to within .5, with probability equal to \(.95 ?\)

Consumer Confidence An increase in the rate of consumer savings is frequently tied to a lack of confidence in the economy and is said to be an indicator of a recessional tendency in the economy. A random sampling of \(n=200\) savings accounts in a local community showed a mean increase in savings account values of \(7.2 \%\) over the past 12 months, with a standard deviation of \(5.6 \% .\) Estimate the mean percent increase in savings account values over the past 12 months for depositors in the community. Find the margin of error for your estimate.

A random sample of \(n=50\) observations from a quantitative population produced \(\bar{x}=56.4\) and \(s^{2}=2.6\). Give the best point estimate for the population mean \(\mu\), and calculate the margin of error.

An experimenter fed different rations, \(A\) and \(B\), to two groups of 100 chicks each. Assume that all factors other than rations are the same for both groups. Of the chicks fed ration \(\mathrm{A}, 13\) died, and of the chicks fed ration \(\mathrm{B}, 6\) died. a. Construct a \(98 \%\) confidence interval for the true difference in mortality rates for the two rations. b. Can you conclude that there is a difference in the mortality rates for the two rations?
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