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Independent random samples of size \(n_{1}=n_{2}=\) 100 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. Construct a \(99 \%\) confidence interval for estimating the difference in the two population means. b. Does the confidence interval in part a provide sufficient evidence to conclude that there is a difference in the two population means? Explain.

Short Answer

Expert verified
Part A: Calculate the 99% confidence interval for the difference in means using the formula: $$CI = (125.2 - 123.7) \pm 2.614 \cdot SE$$ Part B: Determine if there is a difference in the population means by checking if the confidence interval calculated in Part A contains zero. If it does not contain zero, there is a significant difference between the two population means. Otherwise, we cannot conclude that there is a significant difference.

Step by step solution

01

Identify Given Information

Given information: - \(n_{1} = n_{2} = 100\) - \(\bar{x}_{1} = 125.2\) - \(\bar{x}_{2} = 123.7\) - \(s_{1} = 5.6\) - \(s_{2} = 6.8\)
02

Compute the Pooled Variance

Since \(n_1=n_2\), we can compute the pooled variance (\(s_p^2\)): $$s_p^2 = \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1 + n_2 - 2}$$ Substituting the given values: $$s_p^2 = \frac{(100-1)5.6^2 + (100-1)6.8^2}{100 + 100 - 2}$$
03

Calculate the Standard Error

Calculate the standard error for the difference of the means,(\(SE\)): $$SE = \sqrt{\frac{s_p^2}{n_1} + \frac{s_p^2}{n_2}}$$ Substituting the pooled variance: $$SE = \sqrt{\frac{s_p^2}{100} + \frac{s_p^2}{100}}$$
04

Calculate the t-Critical Value Using the t-Distribution

We need to find the t-critical value for a 99% confidence interval. - Degrees of freedom = \(n_1 + n_2 - 2 = 100 + 100 - 2 = 198\) - Confidence level = \(99\%\) - Alpha level = \(0.01\) (1 - Confidence level) Using a t-table or calculator, we find the t-critical value (\(t_{\alpha/2}\)) to be approximately \(2.614\).
05

Construct the Confidence Interval

Confidence interval formula: $$CI = (\bar{x}_{1} - \bar{x}_{2}) \pm t_{\alpha/2} \cdot SE$$ Substituting the values into the formula: $$CI = (125.2 - 123.7) \pm 2.614 \cdot SE$$
06

Interpret the Confidence Interval

Calculate the confidence interval and check if it contains zero: If the confidence interval contains zero, we cannot conclude that there is a significant difference between the two population means.
07

Part A: 99% Confidence Interval for Difference in Means

Calculate the confidence interval using the formula: $$CI = (125.2 - 123.7) \pm 2.614 \cdot SE$$
08

Part B: Determine if there is a difference in the population means

If the computed confidence interval contains zero, we cannot conclude that there is a significant difference between the two population means. If it does not contain zero, we can conclude that there is a significant difference.

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Key Concepts

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

Confidence Interval
A confidence interval provides an estimated range of values which is likely to include an unknown population parameter. In this exercise, we focus on constructing a 99% confidence interval for the difference between two population means. This means we are 99% confident that the true difference between the population means falls within this interval.

To construct the confidence interval, we need the difference of the sample means, the standard error, and the t-critical value from the t-distribution. The confidence interval is calculated using the formula:
  • Difference of means ± t-critical value × standard error
This formula indicates how much the sample means could vary to give us a true estimate of the population mean difference. If the interval does not include zero, it suggests a significant difference between the population means.

Understanding confidence intervals is crucial because it helps us make decisions based on data, showing whether differences observed in sample data might also be present in the general population.
Population Mean
The population mean is an average of a set of data points from the entire group being studied. In situations where collecting data from an entire population is impractical, we use samples to estimate the population mean.

In this exercise,
  • The sample mean of the first population ( ar{x}_{1} ) is 125.2
  • The sample mean of the second population ( ar{x}_{2} ) is 123.7
These sample means are used as our best estimates of the unknown population means. By comparing their difference, we aim to estimate the difference between the actual population means. Such comparisons are at the heart of inferential statistics, as they allow us to draw conclusions about populations based on sample data.

It’s essential to consider that sample means provide estimates which may have a margin of error. This is why the concept of confidence intervals and standard error become relevant, as they allow us to understand and account for this variability.
Pooled Variance
Pooled variance is a method used to estimate the variance of two or more populations to provide a more precise estimate when these populations' true variances are assumed equal. In this context, it's used because the two samples are independently and randomly selected, both having equal sizes (100).

To calculate the pooled variance (s_p^2), the formula is:

\[s_p^2 = \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1 + n_2 - 2}\]
This formula combines the variances of both samples, using them to provide a single estimate that accounts for the slight differences in their individual variances. The pooled variance is then used to calculate the standard error, a key component in constructing confidence intervals.

Understanding pooled variance is essential when comparing means, as it ensures more accurate results by considering data from both samples together rather than individually.
Standard Error
The standard error of the sample means is a crucial statistical value used to measure how far the sample mean of the data is likely to be from the true population mean. In this exercise, it helps in determining how different the samples' mean differences may be.
  • The formula for standard error when dealing with pooled variance is:
\[SE = \sqrt{\frac{s_p^2}{n_1} + \frac{s_p^2}{n_2}}\]
This calculation gives an idea of the variability of sample means considered together, indicating how accurately the means of two samples reflect their populations.

A smaller standard error suggests that the estimate of the difference between population means is precise and reliable, enhancing the confidence in the conclusions drawn. Hence, when constructing a confidence interval, a smaller standard error directly leads to a narrower interval, suggesting less uncertainty about the estimate.

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Most popular questions from this chapter

G. Wayne Marino investigated some of the variables related to "fast starts," the acceleration and speed of a hockey player from a stopped position. \(^{22}\) Sixty-nine hockey players, varsity and intramural, from the University of Illinois were included in the experiment. Each player was required to move as rapidly as possible from a stopped position to cover a distance of 6 meters. The means and standard deviations of some of the variables recorded for each of the 69 skaters are shown in the table: a. Give the formula that you would use to construct a \(95 \%\) confidence interval for one of the population means (e.g., mean time to skate the 6-meter distance). b. Construct a \(95 \%\) confidence interval for the mean time to skate. Interpret this interval.

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