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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
a) Sample means are 9 and 7. b) Standard deviations are 2.28 and 1.79. c) Point estimate is 2. d) 90% CI is (0.055, 3.945).

Step by step solution

01

Calculate the Sample Means

To find the mean of each sample, sum up all the values in each sample and divide by the number of observations.For Sample 1:\[\bar{x}_1 = \frac{10 + 7 + 13 + 7 + 9 + 8}{6} = \frac{54}{6} = 9\]For Sample 2:\[\bar{x}_2 = \frac{8 + 7 + 8 + 4 + 6 + 9}{6} = \frac{42}{6} = 7\]
02

Calculate the Sample Standard Deviations

The standard deviation is calculated using the formula:\[s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n - 1}}\]For Sample 1:- Calculate deviations: \(\{1, -2, 4, -2, 0, -1\}\)- Sum of squares: \(1^2 + (-2)^2 + 4^2 + (-2)^2 + 0^2 + (-1)^2 = 26\)- Standard deviation: \[s_1 = \sqrt{\frac{26}{5}} = \sqrt{5.2} \approx 2.28\]For Sample 2:- Calculate deviations: \(\{1, 0, 1, -3, -1, 2\}\)- Sum of squares: \(1^2 + 0^2 + 1^2 + (-3)^2 + (-1)^2 + 2^2 = 16\)- Standard deviation: \[s_2 = \sqrt{\frac{16}{5}} = \sqrt{3.2} \approx 1.79\]
03

Compute the Point Estimate of the Difference Between Means

The point estimate of the difference between the two population means is simply the difference between the two sample means.\[\bar{x}_1 - \bar{x}_2 = 9 - 7 = 2\]
04

Calculate the 90% Confidence Interval for the Difference in Population Means

For two independent samples, the formula for the confidence interval of the difference between means is:\[(\bar{x}_1 - \bar{x}_2) \pm z\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}\]- Here, \(z\) is the critical value for a 90% confidence interval, typically \(z = 1.645\) for a normal distribution.- Substitute the values:\[2 \pm 1.645 \sqrt{\frac{2.28^2}{6} + \frac{1.79^2}{6}}\]- Simplify:\[2 \pm 1.645 \sqrt{\frac{5.1984}{6} + \frac{3.2041}{6}} = 2 \pm 1.645 \sqrt{\frac{8.4025}{6}} = 2 \pm 1.645 \times 1.18385 \]- Calculate the margin of error:\[1.945 \]- Confidence interval:\[0.055, 3.945\]

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Key Concepts

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

Sample Means
When dealing with sample data, calculating the sample mean is a common task. The sample mean is a measure of central tendency, which shows the average value of a data set. To compute the sample mean, sum all the data values in the sample and divide by the number of observations.
For example, in Sample 1, which contains values {10, 7, 13, 7, 9, 8}, you add them up to get 54, and since there are 6 values in the sample, the sample mean is calculated as \( \bar{x}_1 = \frac{54}{6} = 9 \). Similarly, for Sample 2, the sum of the data values {8, 7, 8, 4, 6, 9} is 42, and the mean is \( \bar{x}_2 = \frac{42}{6} = 7 \).
Sample means provide an estimate of the population mean and help in performing further statistical analyses.
Sample Standard Deviation
The sample standard deviation quantifies the amount of variation or dispersion in a sample data set. It indicates how much individual data points tend to differ from the sample mean. The formula for sample standard deviation involves subtracting the sample mean from each data point, squaring each result, summing these squares, dividing by \( n-1 \) (where \( n \) is the sample size), and finally taking the square root of that quotient.
In Sample 1, the deviations from the mean \( \ar{x}_1 = 9 \) are {1, -2, 4, -2, 0, -1}, which results in a sum of squares of 26. Dividing by 5 yields a variance of 5.2, and its square root, the standard deviation \( s_1 = \sqrt{5.2} \approx 2.28 \).
Similarly for Sample 2, comparing each point against \( \ar{x}_2 = 7 \) gives deviations {1, 0, 1, -3, -1, 2}, with a variance of 3.2, resulting in a standard deviation \( s_2 = \sqrt{3.2} \approx 1.79 \).
This measure is crucial for computing confidence intervals and hypothesis testing.
Point Estimate
In statistical analysis, a point estimate is a single value used to estimate a population parameter. In this exercise, the point estimate is for the difference between the means of two populations. This is achieved by simply subtracting one sample mean from the other.
The formula used is \( \bar{x}_1 - \bar{x}_2 \). For instance, with Sample 1 having a mean of 9 and Sample 2 having a mean of 7, the point estimate of the difference is \( 9 - 7 = 2 \).
This point estimate is a straightforward, yet insightful calculation that provides a simple comparison between the average outcomes of two groups. It acts as the foundation for more complex inferential statistics, like computing confidence intervals.
Normal Distribution
Understanding the normal distribution is crucial for statistical analysis, especially when estimating population parameters like means. It is a continuous probability distribution that is symmetric around the mean, showing that data near the mean are more frequent in occurrence than data far from the mean.
Many statistical tests and estimates, including confidence intervals and point estimates, rely on the assumption of normal distribution. This is partly because of its properties, such as the Empirical Rule, which states that approximately 68% of the data falls within one standard deviation of the mean, 95% within two, and 99.7% within three.
In this exercise, since the samples are taken from normal populations, we use the normal distribution for calculating confidence intervals. For instance, the critical value used (\( z = 1.645 \)) for a 90% confidence interval comes from the standard normal distribution. These characteristics make it vital for making predictions and understanding variability in data sets.

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Most popular questions from this chapter

Four different paints are advertised as having the same drying time. To check the manufacturer's claims, five samples were tested for each of the paints. The time in minutes until the paint was dry enough for a second coat to be applied was recorded. The following data were obtained. $$\begin{array}{cccc} \text { Paint 1 } & \text { Paint 2 } & \text { Paint 3 } & \text { Paint 4 } \\\ 128 & 144 & 133 & 150 \\ 137 & 133 & 143 & 142 \\ 135 & 142 & 137 & 135 \\ 124 & 146 & 136 & 140 \\ 141 & 130 & 131 & 153 \end{array}$$ At the \(\alpha=.05\) level of significance, test to see whether the mean drying time is the same for each type of paint.

Consider the following hypothesis test. $$\begin{array}{l} H_{0}: \mu_{1}-\mu_{2}=0 \\ H_{\mathrm{a}}: \mu_{1}-\mu_{2} \neq 0 \end{array}$$ The following results are for two independent samples taken from the two populations. $$\begin{array}{ll} \text { Sample 1 } & \text { Sample 2 } \\ n_{1}=80 & n_{2}=70 \\ \bar{x}_{1}=104 & \bar{x}_{2}=106 \\ \sigma_{1}=8.4 & \sigma_{2}=7.6 \end{array}$$ a. What is the value of the test statistic? b. What is the \(p\) -value? c. With \(\alpha=.05,\) what is your hypothesis testing conclusion?

The U.S. Department of Transportation provides the number of miles that residents of the 75 largest metropolitan areas travel per day in a car. Suppose that for a simple random sample of 50 Buffalo residents the mean is 22.5 miles a day and the standard deviation is 8.4 miles a day, and for an independent simple random sample of 40 Boston residents the mean is 18.6 miles a day and the standard deviation is 7.4 miles a day. a. What is the point estimate of the difference between the mean number of miles that Buffalo residents travel per day and the mean number of miles that Boston residents travel per day? b. What is the \(95 \%\) confidence interval for the difference between the two population means?

A well-known automotive magazine took three top-of-the-line midsize automobiles manufactured in the United States, test-drove them, and compared them on a variety of criteria. In the area of gasoline mileage performance, five automobiles of each brand were each test-driven 500 miles; the miles per gallon data obtained follow. Use \(\alpha=.05\) to test whether there is a significant difference in the mean number of miles per gallon for the three types of automobiles. $$\begin{array}{ccc} & \text { Automobile } & \\ \mathbf{A} & \mathbf{B} & \mathbf{C} \\ 19 & 19 & 24 \\ 21 & 20 & 26 \\ 20 & 22 & 23 \\ 19 & 21 & 25 \\ 21 & 23 & 27 \end{array}$$

Safegate Foods, Inc., is redesigning the checkout lanes in its supermarkets throughout the country and is considering two designs. Tests on customer checkout times conducted at two stores where the two new systems have been installed result in the following summary of the data. $$\begin{array}{cl} \text { System A } & \text { System B } \\ n_{1}=120 & n_{2}=100 \\ \bar{x}_{1}=4.1 \text { minutes } & \bar{x}_{2}=3.4 \text { minutes } \\ \sigma_{1}=2.2 \text { minutes } & \sigma_{2}=1.5 \text { minutes } \end{array}$$ Test at the .05 level of significance to determine whether the population mean checkout times of the two systems differ. Which system is preferred?

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