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The following information is obtained from two independent samples selected from two populations. $$ \begin{array}{lll} n_{1}=650 & \bar{x}_{1}=1.05 & \sigma_{1}=5.22 \\ n_{2}=675 & \bar{x}_{2}=1.54 & \sigma_{2}=6.80 \end{array} $$ ah 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.

Short Answer

Expert verified
The point estimate of \(\mu_{1}-\mu_{2}\) is -0.49. The confidence interval for this estimate at 95% confidence level can be calculated using the formula described in step 2. The margin of error can be found as outlined in step 3.

Step by step solution

01

Find the point estimate

Point estimate of \(\mu_{1}-\mu_{2}\) can be found using the sample means. It is simply \( \bar{x}_{1} - \bar{x}_{2}\). Pluging the given values, we get the equation as follows: 1.05 - 1.54 = -0.49.
02

Constructing the confidence interval

The confidence interval is calculated using the formula:\( \bar{x}_{1} - \bar{x}_{2} \pm Z \sqrt{\frac{\sigma_{1}^2}{n_{1}}+\frac{\sigma_{2}^2}{n_{2}}} \). Where Z is the z-score from the standard normal distribution for the desired confidence level. For a 95% confidence level, the z-value (Z) is 1.96. Filling out all known variables, the equation to find confidence interval will look like this: -0.49 \pm 1.96 \sqrt{(5.22)^2/650 + (6.80)^2/675 }.
03

Find the margin of error

Margin of error is the value after the \(\pm\) sign in the confidence interval formula. This simply means calculating the expression inside the square root and then multiplying it by the Z value. The margin of error is given as: 1.96\(\sqrt{(5.22)^2/650 + (6.80)^2/675 }\)

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Key Concepts

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

Point Estimate
When we talk about a point estimate, we're referring to a single value that serves as our best guess for an unknown population parameter. In this case, we're trying to estimate the difference in means between two populations, \(\mu_1 - \mu_2\). For two independent samples, the point estimate of this difference is straightforward: it's simply the difference between the two sample means.

That sounds complicated, but it boils down to a simple formula: \(\bar{x}_1 - \bar{x}_2\). This means you subtract the mean of the second sample from the mean of the first sample.

In our exercise, the problem gives us \(\bar{x}_1 = 1.05\) and \(\bar{x}_2 = 1.54\). Using the formula, we calculate the point estimate as \[1.05 - 1.54 = -0.49.\]
This negative result means that, based on our samples, the first population mean is estimated to be 0.49 less than the second population mean. Understanding this helps in interpreting the bigger picture of our analysis.
Margin of Error
The margin of error is crucial when we talk about confidence intervals because it gives us a range in which the true parameter will likely fall. Essentially, it's a measure of the uncertainty or variability in our point estimate.

It is represented as the value after the \(\pm\) sign in the confidence interval formula. It's determined by three key components:
  • The standard deviation(s) of the samples
  • The sample sizes
  • The z-score from the standard normal distribution that corresponds to our desired confidence level, usually 1.96 for a 95% confidence level.
For the given problem, the margin of error \[= 1.96 \times \sqrt{\frac{(5.22)^2}{650} + \frac{(6.80)^2}{675}}\]
This calculation assists us in forming the bounds of our confidence interval. A smaller margin of error implies more precision in our estimate, while a larger one indicates more variability and less certainty.
Independent Samples
Independent samples are essential in statistics, especially when comparing two groups or populations. But what makes them independent?

Two samples are independent if the selection of data in one sample does not influence or affect the data selection in the other. In practical terms, it often means that the data is collected from two separate groups that don't interact or influence each other.

In our exercise, we are examining two such independent samples with sizes \(n_1 = 650\) and \(n_2 = 675\). Because they are independent, we can confidently apply certain statistical methods, like the difference of means test.
  • Allows the use of certain statistical formulas and methods without bias.
  • Helps generalize findings to the broader population.
  • Reduces variability that might occur if the two samples were related or dependent.
The independence assumption is crucial to correctly applying statistical inference methods and ensuring reliable results.
Statistical Inference
Statistical inference is a powerful tool in the field of statistics that allows us to draw conclusions about a population based on a sample. This important concept uses the data from samples to make educated guesses about the broader population from which the sample is drawn.

The exercise centers on one main type of statistical inference: constructing a confidence interval for the difference of population means (\(\mu_1 - \mu_2\)). With the collected sample data, inference helps us estimate this difference and gauge how close our point estimate might be to the true population parameter.

Here's how it typically works:
  • We take our sample data and calculate a point estimate, like the mean difference.
  • Then, we use methods like the confidence interval to provide a range of plausible values for that difference.
  • Finally, this range, combined with the margin of error, helps us understand the likely accuracy of our point estimate.
By using statistical inference, we can make valid conclusions about the populations in question, guiding decisions and predictions in a scientifically sound manner.

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Most popular questions from this chapter

According to the information given in Exercise \(10.25\), a sample of 45 customers who drive luxury cars showed that their average distance driven between oil changes was 3187 miles with a sample standard deviation of \(42.40\) miles. Another sample of 40 customers who drive compact lower-price cars resulted in an average distance of 3214 miles with a standard deviation of \(50.70\) miles. Suppose that the standard deviations for the two populations are not equal. a. Construct a \(95 \%\) confidence interval for the difference in the mean distance between oil changes for all luxury cars and all compact lower-price cars. b. Using a \(1 \%\) significance level, can you conclude that the mean distance between oil changes is lower for all luxury cars than for all compact lower- price cars? c. Suppose that the sample standard deviations were \(28.9\) and \(61.4\) miles, respectively. Redo parts a and b. Discuss any changes in the results.

The following information is obtained from two independent samples selected from two normally distributed populations. $$ \begin{array}{lll} n_{1}=18 & \bar{x}_{1}=7.82 & \sigma_{1}=2.35 \\ n_{2}=15 & \bar{x}_{2}=5.99 & \sigma_{2}=3.17 \end{array} $$ a. What is the point estimate of \(\mu_{1}-\mu_{2}\) ? b. Construct a \(99 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). Find the margin of error for this estimate.

When are the samples considered large enough for the sampling distribution of the difference between two sample proportions to be (approximately) normal?

The lottery commissioner's office in a state wanted to find if the percentages of men and women who play the lottery often are different. A sample of 500 men taken by the commissioner's office showed that 160 of them play the lottery often. Another sample of 300 women showed that 66 of them play the lottery often. a. What is the point estimate of the difference between the two population proportions? b. Construct a \(99 \%\) confidence interval for the difference between the proportions of all men and all women who play the lottery often. c. Testing at a \(1 \%\) significance level, can you conclude that the proportions of all men and all women who play the lottery often are different?

According to a Randstad Global Work Monitor survey, \(52 \%\) of men and \(43 \%\) of women said that working part-time hinders their career opportunities (USA TODAY, October 6,2011 ). Suppose that these results are based on random samples of 1350 men and 1480 women. a. Let \(p_{1}\) and \(p_{2}\) be the proportions of all men and all women, respectively, who will say that working part-time hinders their career opportunities. Construct a \(95 \%\) confidence interval for \(p_{1}-p_{2}\) b. Using a \(2 \%\) significance level, can you conclude that \(p_{1}\) and \(p_{2}\) are different? Use both the critical-value and the \(p\) -value approaches.

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