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Chapter 10: Problem 3

The following information is obtained from two independent samples selected from two normally distributed populations. $$ \begin{array}{lll} n_{1}=18 & \bar{x}_{1}=7.82 & \sigma_{1}=2.35 \\ n_{2}=15 & \bar{x}_{2}=5.99 & \sigma_{2}=3.17 \end{array} $$ a. What is the point estimate of \(\mu_{1}-\mu_{2} ?\) b. Construct a \(99 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). Find the margin of error for this estimate.

Short Answer

Expert verified
a. The point estimate of \(\mu_{1}-\mu_{2}\) is 1.83. b. The 99% confidence interval for \(\mu_{1}-\mu_{2}\) is \(-1.395, 4.065\). The margin of error for this estimate is 3.225.

Step by step solution

01

Point Estimate Calculation

The point estimate for \(\mu_{1}-\mu_{2}\) is given by \(\bar{x}_{1} - \bar{x}_{2}\). Substitute the given sample means to get the point estimate: \(7.82 - 5.99 = 1.83\)
02

Standard Error Calculation

To construct a confidence interval, the standard error of the difference in means is required. The formula for the standard error is \(\sqrt{ \left(\frac{\sigma_{1}^2}{n_{1}}\right) + \left(\frac{\sigma_{2}^2}{n_{2}}\right)}\). Substituting the values from the exercise, we get the standard error as \(\sqrt{ \left(\frac{2.35^2}{18}\right) + \left(\frac{3.17^2}{15}\right)} = 1.25\)
03

Z-Score Calculation

Since the exercise demands a 99% confidence interval, we need the Z-score for a two-tailed test at this confidence level. From the standard normal distribution table, the z-score is roughly 2.58.
04

Margin of Error Calculation

Margin of error can be found by multiplying the standard error by the Z-score. In this case it is \(1.25 \times 2.58 = 3.225\).
05

Confidence Interval Calculation

The 99% confidence interval is calculated as (point estimate - margin of error, point estimate + margin of error). That is, \(1.83 - 3.225\) and \(1.83 + 3.225\) which yields the interval \(-1.395, 4.065\).

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Key Concepts

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

Point Estimate
In statistics, the point estimate is a single value that serves as a best guess or estimation of an unknown population parameter. For this exercise, we are interested in the point estimate of the difference between two population means, \(\mu_1 - \mu_2\). To find this, we simply subtract the average of the second sample from the average of the first one.
For example:
  • Mean of Sample 1 (\(\bar{x}_1\) = 7.82
  • Mean of Sample 2 (\(\bar{x}_2\) = 5.99
  • Point Estimate = \(\bar{x}_1 - \bar{x}_2 = 1.83\)
This value indicates the estimated average difference between the populations from which your samples are drawn.
Standard Error
The standard error gives us an idea of how much the sample mean might fluctuate from the actual population mean. It鈥檚 crucial for constructing confidence intervals because it reflects the variability in our estimate.
The formula used is:\[ SE = \sqrt{ \left(\frac{\sigma_1^2}{n_1}\right) + \left(\frac{\sigma_2^2}{n_2}\right) } \]Here:
  • \(\sigma_1, \sigma_2\) are the standard deviations of the populations
  • \(n_1, n_2\) are the sample sizes
When you plug in the given values, you'll get:
\( SE = 1.25 \).

This standard error indicates the average difference we expect due to random sampling.
Z-Score
The Z-score helps to determine how far away our sample mean is from the population mean, measured in terms of standard deviations. When constructing confidence intervals, we use the Z-score to define the confidence level we're interested in.
For a 99% confidence interval, the Z-score is about 2.58.
Why 2.58? Because it covers the middle 99% of the distribution in a standard normal distribution.
This score guides us in understanding how extreme our observed result is relative to the benchmark or null hypothesis, especially when constructing the interval.
Margin of Error
The margin of error quantifies the extent of possible error in the point estimate. It shows the range within which we expect the true population parameter to lie, given some level of confidence.
It's calculated by multiplying the standard error by the Z-score:\[ \text{Margin of Error} = SE \times Z \text{-score} \]In this case:
\( 1.25 \times 2.58 = 3.225 \)
This margin provides a buffer around our point estimate, creating a span that forms the confidence interval.
With this margin, we express our confidence that the true difference sits within the interval we're calculating.

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Most popular questions from this chapter

A local college cafeteria has a self-service soft ice cream machine. The cafeteria provides bowls that can hold up to 16 ounces of ice cream. The food service manager is interested in comparing the average amount of ice cream dispensed by male students to the average amount dispensed by female students. A measurement device was placed on the ice cream machine to determine the amounts dispensed. Random samples of 85 male and 78 female students who got ice cream were selected. The sample averages were \(7.23\) and \(6.49\) ounces for the male and female students, respectively. Assume that the population standard deviations are \(1.22\) and \(1.17\) ounces, respectively. a. Let \(\mu_{1}\) and \(\mu_{2}\) be the population means of ice cream amounts dispensed by all male and all female students at this college, respectively. What is the point estimate of \(\mu_{1}-\mu_{2} ?\) b. Construct a \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). c. Using a \(1 \%\) significance level, can you conclude that the average amount of ice cream dispensed by all male college students is larger than the average amount dispensed by all female college students? Use both approaches to make this test.

The following information was obtained from two independent samples selected from two populations with unknown but equal standard deviations. $$ \begin{array}{lll} n_{1}=55 & \bar{x}_{1}=90.40 & s_{1}=11.60 \\ n_{2}=50 & \bar{x}_{1}=86.30 & s_{2}=10.25 \end{array} $$ Test at a \(5 \%\) significance level if \(\mu_{1}\) is greater than \(\mu_{2}\).

A car magazine is comparing the total repair costs incurred during the first three years on two sports cars, the T-999 and the XPY. Random samples of 45 T-999s and 51 XPYs are taken. All 96 cars are 3 years old and have similar mileages. The mean of repair costs for the 45 T-999 cars is \(\$ 3300\) for the first 3 years. For the 51 XPY cars, this mean is \(\$ 3850\). Assume that the standard deviations for the two populations are \(\$ 800\) and \(\$ 1000\), respectively. a. Construct a 99\% confidence interval for the difference between the two population means. b. Using a \(1 \%\) significance level, can you conclude that such mean repair costs are different for these two types of cars? c. What would your decision be in part b if the probability of making a Type I error were zero? Explain.

Sample of 500 observations taken from the first population gave \(x_{1}=305\). Another sample of 600 observations taken from the second population gave \(x_{2}=348\). a. Find the point estimate of \(p_{1}-p_{2}\). b. Make a \(97 \%\) confidence interval for \(p_{1}-p_{2}\). c. Show the rejection and nonrejection regions on the sampling distribution of \(\hat{p}_{1}-\hat{p}_{2}\) for \(H_{0}: p_{1}=p_{2}\) versus \(H_{1}: p_{1}>p_{2}\). Use a significance level of \(2.5 \%\). d. Find the value of the test statistic \(z\) for the test of part \(\mathrm{c}\). e. Will you reject the null hypothesis mentioned in part \(\mathrm{c}\) at \(\mathrm{a}\) significance level of \(2.5 \% ?\)

Maine Mountain Dairy claims that its 8-ounce low-fat yogurt cups contain, on average, fewer calories than the 8 -ounce low-fat yogurt cups produced by a competitor. A consumer agency wanted to check this claim. A sample of 27 such yogurt cups produced by this company showed that they contained an average of 141 calories per cup. A sample of 25 such yogurt cups produced by its competitor showed that they contained an average of 144 calories per cup. Assume that the two populations are approximately normally distributed with population standard deviations of \(5.5\) and \(6.4\) calories, repectively. a. Make a \(98 \%\) confidence interval for the difference between the mean number of calories in the 8 -ounce low-fat yogurt cups produced by the two companies. b. Test at a \(1 \%\) significance level whether Maine Mountain Dairy's claim is true. c. Calculate the \(p\) -value for the test of part b. Based on this \(p\) -value, would you reject the null hypothesis if \(\alpha=.05 ?\) What if \(\alpha=.025 ?\)
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