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Chapter 10: Problem 24

Assuming that the two populations are normally distributed with unequal and unknown population standard deviations, construct a \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) for the following. $$ \begin{array}{lll} n_{1}=14 & \bar{x}_{1}=109.43 & s_{1}=2.26 \\ n_{2}=15 & \bar{x}_{2}=113.88 & s_{2}=5.84 \end{array} $$

Short Answer

Expert verified
The 95% confidence interval for \( \mu_{1} - \mu_{2} \) is calculated using the above steps which involves first getting the degree of freedom via the Welch-Satterthwaite equation, then the standard error and finally using these in the interval formula. The final confidence interval obtained is dependent on the t value obtained which is looked up from the t-distribution table using the computed degrees of freedom.

Step by step solution

01

Compute the Difference of Two Means

First, compute the difference of the two sample means. Given \(\bar{x}_{1}=109.43\) and \(\bar{x}_{2}=113.88\), then the difference is: \( \mu = \bar{x}_{1} - \bar{x}_{2} = 109.43 - 113.88 = -4.45 \)
02

Setup the Welch-Satterthwaite Equation

The degrees of freedom are approximated via the Welch-Satterthwaite equation: \(df = \frac{{(s_{1}^2/n_{1} + s_{2}^2/n_{2})^2}}{{(s_{1}^4/(n_{1}^2 (n_{1} - 1)) + (s_{2}^4/(n_{2}^2 (n_{2} - 1))}}\). Substituting the given values into this equation \(df = \frac{{(2.262/14 + 5.842/15)^2}}{{(2.26^4/(14^2(14 - 1)) + (5.84^4/(15^2 (15 - 1))}}\) can compute the degree of freedom.
03

Compute the Standard Error

Next, calculate the standard error using the formula: \(SE = \sqrt{(s_{1}^2/n_{1} + s_{2}^2/n_{2})}\) Substituting the values, we get: \(SE = \sqrt{(2.26^2/14 + 5.84^2/15)}\)
04

Construct the Confidence Interval

To construct the confidence interval, use the formula: \((\mu - t * SE, \mu + t * SE)\) where \(t\) is the t-score for the appropriate degree of freedom and confidence level (95%). Find the t-value from the t-distribution table using the degree of freedom calculated in step 2 and then substitute into the interval formula.

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Key Concepts

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

Welch's t-test
Welch's t-test is a statistical test used to determine if there is a significant difference between the means of two groups, especially when the groups have unequal variances and sample sizes. Unlike the traditional t-test, Welch's t-test is designed to handle situations where the assumption of equal variances doesn't hold, making it a more reliable choice in many real-world scenarios.

In the context of the exercise, Welch's t-test is used to construct a confidence interval for the difference between two independent sample means. This test is ideal when comparing two groups with different standard deviations and sample sizes. The flexibility of handling different group variances without the assumption of equal variances makes it widely applicable.
  • Welch's t-test is robust against breaking the assumption of equal variances.
  • It uses a modified formula to calculate degrees of freedom, making it more versatile.
  • Applicable for comparing means when standard deviations differ significantly.
Difference of Means
The difference of means measures how far apart the average values of two separate groups are from each other. It is calculated by simply taking one sample mean and subtracting it from another. This calculation tells us the average difference in outcomes between the two groups being studied.

In the provided exercise, the difference of means equals -4.45, which indicates that, on average, the second population has a higher mean than the first by 4.45 units. A negative difference suggests that the mean of the second sample (\(\bar{x}_{2} = 113.88\)) is greater than the mean of the first sample (\(\bar{x}_{1} = 109.43\)).
  • Simple subtraction: \(\mu = \bar{x}_{1} - \bar{x}_{2}\).
  • Negative difference: suggests the second mean is higher than the first.
  • Important for interpreting group variations.
Degrees of Freedom
Degrees of freedom are a critical component of many statistical calculations. They reflect the number of independent values that can vary in an analysis without violating any imposed constraints. In the context of the Welch's t-test, degrees of freedom help determine the appropriate distribution to use for calculating t-scores.

The Welch-Satterthwaite formula is employed in this situation because it accounts for unequal sample variances. It approximates the degrees of freedom by combining the variances and sample sizes of both groups. Although the formula appears complex, its purpose is straightforward: adjusting for unequal variability to make inferences about the population.
  • Degrees of freedom play a vital role in hypothesis testing and confidence intervals.
  • The larger the degrees of freedom, the closer the t-distribution is to the normal distribution.
  • In irregular or small sample sizes, this adjustment is essential for accuracy.
Standard Error
The standard error (SE) is a measure of how much the sample mean is expected to vary from the true population mean. It essentially gives us an understanding of the precision of our sample mean estimate. The smaller the standard error, the more accurate our sample's mean is likely to be a reflection of the true population mean.

In our exercise, the standard error is calculated using the formula combining the variances and the sample sizes of both groups. This accounts for how each group's data might vary from their means and provides a combined measure of variability for the difference between the means.
  • Standard Error formula: \(SE = \sqrt{(s_1^2/n_1 + s_2^2/n_2)}\).
  • The smaller the SE, the more reliable the mean of the sample is.
  • Used to construct confidence intervals and conduct hypothesis tests.

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Most popular questions from this chapter

Manufacturers of two competing automobile models, Gofer and Diplomat, each claim to have the lowest mean fuel consumption. Let \(\mu_{1}\) be the mean fuel consumption in miles per gallon (mpg) for the Gofer and \(\mu_{2}\) the mean fuel consumption in mpg for the Diplomat. The two manufacturers have agreed to a test in which several cars of each model will be driven on a 100 -mile test run. The fuel consumption, in \(\mathrm{mpg}\), will be calculated for each test run. The average mpg for all 100 -mile test runs for each model gives the corresponding mean. Assume that for each model the gas mileages for the test runs are normally distributed with \(\sigma=2 \mathrm{mpg}\). Note that each car is driven for one and only one 100 -mile test run. a. How many cars (i.e., sample size) for each model are required to estimate \(\mu_{1}-\mu_{2}\) with a \(90 \%\) confidence level and with a margin of error of estimate of \(1.5 \mathrm{mpg}\) ? Use the same number of cars (i.e., sample size) for each model. b. If \(\mu_{1}\) is actually \(33 \mathrm{mpg}\) and \(\mu_{2}\) is actually \(30 \mathrm{mpg}\), what is the probability that five cars for each model would yield \(\bar{x}_{1} \geq \bar{x}_{2}\) ?

Briefly explain the meaning of independent and dependent samples. Give one example of each.

In a random sample of 800 men aged 25 to 35 years, \(24 \%\) said they live with one or both parents. In another sample of 850 women of the same age group, \(18 \%\) said that they live with one or both parents. a. Construct a \(95 \%\) confidence interval for the difference between the proportions of all men and all women aged 25 to 35 years who live with one or both parents. b. Test at a \(2 \%\) significance level whether the two population proportions are different. c. Repeat the test of part b using the \(p\) -value approach.

The following information was obtained from two independent samples selected from two normally distributed populations with unequal and unknown population standard deviations. $$ \begin{array}{lll} n_{1}=14 & \bar{x}_{1}=.109 .43 & s_{1}=2.26 \\ n_{2}=15 & \bar{x}_{2}=.113 .88 & s_{2}=5.84 \end{array} $$ Test at a \(1 \%\) significance level if \(\mu_{1}\) is less than \(\mu_{2}\).

An economist was interested in studying the impact of the recession of a few years ago on dining out, including drive-through meals at fast-food restaurants. A random sample of 48 families of four with discretionary incomes between \(\$ 300\) and \(\$ 400\) per week indicated that they reduced their spending on dining out by an average of \(\$ 31.47\) per week, with a sample standard deviation of \(\$ 10.95 .\) Another random sample of 42 families of five with discretionary incomes between \(\$ 300\) and \(\$ 400\) per week reduced their spending on dining out by an average of \(\$ 35.28\) per week, with a sample standard deviation of \(\$ 12.37\). (Note that the two groups of families are differentiated by the number of family members.) Assume that the distributions of reductions in weekly dining-out spendings for the two groups have unknown and unequal population standard deviations. a. Construct a \(90 \%\) confidence interval for the difference in the mean weekly reduction in dining out spending levels for the two populations. b. Using a \(5 \%\) significance level, can you conclude that the average weekly spending reduction for all families of four with discretionary incomes between \(\$ 300\) and \(\$ 400\) per week is less than the average weekly spending reduction for all families of five with discretionary incomes between \(\$ 300\) and \(\$ 400\) per week?
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