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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

A company that has many department stores in the southern states wanted to find at two such stores the percentage of sales for which at least one of the items was returned. A sample of 800 sales randomly selected from Store A showed that for 280 of them at least one item was returned. Another sample of 900 sales randomly selected from Store B showed that for 279 of them at least one item was returned. a. Construct a 98\% confidence interval for the difference between the proportions of all sales at the two stores for which at least one item is returned. b. Using a \(1 \%\) significance level, can you conclude that the proportions of all sales for which at least one item is returned is higher for Store A than for Store \(\mathrm{B}\) ?

Using data from the U.S. Census Bureau and other sources, www.nerdwallet.com estimated that considering only the households with credit card debts, the average credit card debt for U.S. households was \(\$ 15,523\) in 2014 and \(\$ 15,242\) in 2013 . Suppose that these estimates were based on random samples of 600 households with credit card debts in 2014 and 700 households with credit card debts in 2013 . Suppose that the sample standard deviations for these two samples were \(\$ 3870\) and \(\$ 3764\), respectively. Assume that the standard deviations for the two populations are unknown but equal. a. Let \(\mu_{1}\) and \(\mu_{2}\) be the average credit card debts for all such households for the years 2014 and 2013 , respectively. What is the point estimate of \(\mu_{1}-\mu_{2}\) ? b. Construct a \(98 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). c. Using a \(1 \%\) significance level, can you conclude that the average credit card debt for such households was higher in 2014 than in \(2013 ?\) Use both the \(p\) -value and the critical-value approaches to make this test.

Describe the sampling distribution of \(\bar{x}_{1}-\bar{x}_{2}\) for two independent samples when \(\sigma_{1}\) and \(\sigma_{2}\) are known and either both sample sizes are large or both populations are normally distributed. What are the mean and standard deviation of this sampling distribution?

The following information is obtained from two independent samples selected from two populations. $$ \begin{array}{lll} n_{1}=650 & \bar{x}_{1}=1.05 & \sigma_{1}=5.22 \\ n_{2}=675 & \bar{x}_{2}=1.54 & \sigma_{2}=6.80 \end{array} $$ a. What is the point estimate of \(\mu_{1}-\mu_{2} ?\) b. Construct a \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). Find the margin of error for this estimate.

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