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Consider the following data for two independent random samples taken from two normal populations. $$\begin{array}{l|rrrrrr} \text { Sample 1 } & 10 & 7 & 13 & 7 & 9 & 8 \\\\\hline \text { Sample 2 } & 8 & 7 & 8 & 4 & 6 & 9\end{array}$$ a. Compute the two sample means. b. Compute the two sample standard deviations. c. What is the point estimate of the difference between the two population means? d. What is the \(90 \%\) confidence interval estimate of the difference between the two population means?

Short Answer

Expert verified
Sample means are 9 and 7, standard deviations are 2 and 2.19. The point estimate is 2, and the 90% CI is (0.04, 3.96).

Step by step solution

01

Calculate the Sample Means

To find the sample mean for each data set, use the formula for the mean: \( \bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i \).For Sample 1: \( \bar{x}_1 = \frac{10 + 7 + 13 + 7 + 9 + 8}{6} = 9 \).For Sample 2: \( \bar{x}_2 = \frac{8 + 7 + 8 + 4 + 6 + 9}{6} = 7 \).
02

Calculate the Sample Standard Deviations

The sample standard deviation is calculated using the formula: \( s = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})^2} \).For Sample 1:\[ s_1 = \sqrt{\frac{(10-9)^2 + (7-9)^2 + (13-9)^2 + (7-9)^2 + (9-9)^2 + (8-9)^2}{6-1}} = \sqrt{4} = 2 \].For Sample 2:\[ s_2 = \sqrt{\frac{(8-7)^2 + (7-7)^2 + (8-7)^2 + (4-7)^2 + (6-7)^2 + (9-7)^2}{6-1}} = \sqrt{4.8} = \approx 2.19 \].
03

Find the Point Estimate of the Difference in Means

The point estimate of the difference between two population means is the difference between the sample means: \( \hat{\mu} = \bar{x}_1 - \bar{x}_2 \).Substituting the known values: \( \hat{\mu} = 9 - 7 = 2 \).
04

Compute the 90% Confidence Interval for the Difference in Means

To compute the confidence interval for the difference between two means, use the formula:\[ (\bar{x}_1 - \bar{x}_2) \pm t^* \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} \]where \( t^* \) is based on a 90% confidence level with degrees of freedom calculated using the approximate method.Calculate the approximate degrees of freedom:\[ df \approx \left( \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1-1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2-1}} \right) \approx 9 \].Assuming \( t^* \approx 1.833 \) for df = 9 (90% confidence interval), the interval is calculated as:\[ 2 \pm 1.833 \times \sqrt{\frac{2^2}{6} + \frac{2.19^2}{6}} \approx 2 \pm 1.96 \approx (0.04, 3.96) \].

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Key Concepts

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

Sample Mean
The sample mean is a fundamental concept in statistics and is often the first summary statistic calculated for a dataset. It represents the average value of a dataset and is a measure of central tendency.

In essence, the sample mean tells us the typical value we might expect from the data. To calculate the sample mean, sum up all the values in the sample and divide by the number of observations. The formula is:
  • \(\bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i \)
where \(x_i\) are the individual sample values, and \(n\) is the sample size.

Understanding the sample mean helps in comparing datasets, as seen in the example with Sample 1 and Sample 2. Having means of 9 and 7, respectively, provides insight into each sample's central tendency.
Sample Standard Deviation
The sample standard deviation gives us an idea of how spread out the values in a dataset are around the mean. It is a measure of variability or dispersion.

Calculating it helps understand if data points are closely clustered around the mean or spread out over a wider range.

The formula for the sample standard deviation is:
  • \( s = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})^2} \)
The term \(n-1\) is used instead of \(n\) to provide an unbiased estimate of the population standard deviation, which accounts for the degrees of freedom in a sample.

The exercise showed Sample 1 with a standard deviation of 2 and Sample 2 with approximately 2.19, indicating Sample 2's values are slightly more spread out.
Confidence Interval
A confidence interval is a range of values that estimates the true value of a population parameter, like the mean, with a certain level of confidence, such as 90%, 95%, or 99%. It provides a range that the true parameter will fall into a specific proportion of the time.

It’s calculated using a point estimate, like the sample mean, plus or minus a margin of error. This margin is usually the result of a critical value, such as the \( t^* \) from the t-distribution, multiplied by the standard error of the estimate.

In our example, a 90% confidence interval for the difference between the two sample means was computed as \((0.04, 3.96)\).

This interval tells us that we can be 90% confident that the true difference between the population means lies within this range. It provides a sense of precision regarding the estimated difference and accounts for the sample variability.
T-Distribution
The t-distribution is a probability distribution used in statistical inference when dealing with small sample sizes or unknown population standard deviations. It resembles the standard normal distribution but has heavier tails, which means more probability is distributed in the tail regions.

This accounts for the increased uncertainty in smaller samples.

The t-distribution is crucial for constructing confidence intervals and hypothesis testing when the data follow a roughly normal distribution. For example, in computing confidence intervals, the critical value \( t^* \) is chosen based on the desired confidence level and the degrees of freedom, which are derived from the sample size.

In our scenario, the degrees of freedom were approximated to 9, and the \( t^* \) value for a 90% confidence level was around 1.833. This value helps in calculating the margin of error to adjust the confidence interval accordingly, ensuring the interval's reliability with a specified level of confidence.

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Most popular questions from this chapter

Bank of America's Consumer Spending Survey collected data on annual credit card charges in seven different categories of expenditures: transportation, groceries, dining out, household expenses, home furnishings, apparel, and entertainment (US Airways Attaché, December 2003 ). Using data from a sample of 42 credit card accounts, assume that each account was used to identify the annual credit card charges for groceries (population 1 ) and the annual credit card charges for dining out (population 2 ). Using the difference data, the sample mean difference was \(d=\$ 850,\) and the sample standard deviation was \(s_{d}=\$ 1123\) a. Formulate the null and alternative hypotheses to test for no difference between the population mean credit card charges for groceries and the population mean credit card charges for dining out. b. Use a .05 level of significance. Can you conclude that the population means differ? What is the \(p\) -value? c. Which category, groceries or dining out, has a higher population mean annual credit card charge? What is the point estimate of the difference between the population means? What is the \(95 \%\) confidence interval estimate of the difference between the population means?

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