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Chapter 10: Problem 24

The following information was obtained from two independent samples selected from two normally distributed populations with unknown but equal standard deviations. $$ \begin{array}{lllllllllllll} \text { Sample 1: } & 2.18 & 2.23 & 1.96 & 2.24 & 2.72 & 1.87 & 2.68 & 2.15 & 2.49 & 2.05 & & \\ \text { Sample 2: } & 1.82 & 1.26 & 2.00 & 1.89 & 1.73 & 2.03 & 1.43 & 2.05 & 1.54 & 2.50 & 1.99 & 2.13 \end{array} $$ a. Let \(\mu_{1}\) be the mean of population 1 and \(\mu_{2}\) be the mean of population \(2 .\) What is the point estimate of \(\mu_{1}-\mu_{2} ?\) b. Construct a \(99 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). c. Test at the \(2.5 \%\) significance level if \(\mu_{1}\) is lower than \(\mu_{2}\).

Short Answer

Expert verified
a) The point estimate of \(\mu_{1}-\mu_{2}\) is obtained by subtracting the mean of the second population from the first. b) The 99% confidence interval for \(\mu_{1}-\mu_{2}\) is calculated from the point estimate \(\pm\) Z-value of 99% confidence interval * standard error of difference. c) To test the hypothesis that \(\mu_{1}\) is lower than \(\mu_{2}\) at the 2.5% significance level, we compare the p-value obtained from the test statistic with the significance level.

Step by step solution

01

Point Estimate

To calculate the point estimate of \(\mu_{1}-\mu_{2}\), find the means of the two samples and subtract the mean of Sample 2 from the mean of Sample 1. Use the formula for finding the mean which is sum of all datapoints divided by the number of datapoints.
02

Confidence Interval

To obtain the 99% confidence interval for \(\mu_{1}-\mu_{2}\), we use the formula: Point estimate \(\pm\) (Z-value of 99% confidence interval * standard error of difference). The standard error of difference can be calculated using the pooled standard deviation formula (also known as standard error).
03

Hypothesis Testing

Firstly, define the null and alternate hypothesis. The null hypothesis is \(H_0: \mu_1-\mu_2 = 0\) and the alternate hypothesis is \(H_a: \mu_1 - \mu_2 < 0\). Next, calculate the test statistic using the formula: \(Z = \frac{(\mu_1 - \mu_2) - 0}{\text{Standard error of difference}}\). Finally, use a Standard Normal Table or a similar tool to find the p-value associated with this test statistic, and compare this with the provided significance level of 2.5%.

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Key Concepts

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

Point Estimation
Point estimation is all about finding a single best guess or approximation for an unknown population parameter. In our given problem, we are estimating the difference between the means of two populations, denoted as \(\mu_{1} - \mu_{2}\). A point estimate is calculated by using the means of the sample data provided.
It's like using the sample means as the best possible figures we have from the samples available. To calculate the point estimate in step one, you will:
  • Find the mean of each sample.
  • Subtract the mean of Sample 2 from the mean of Sample 1.

This gives you a direct estimate of the difference between the two population means. Even though this estimate gives a precise figure, always keep in mind it's a best guess based on the sample data, not the entire population data.
Confidence Interval
Once you have a point estimate, the next step is often to calculate a confidence interval. A confidence interval gives a range within which you believe a population parameter lies, based on your sample data. Unlike a point estimate, it provides a margin for error, which adds a layer of certainty to the estimation.
For the problem at hand, constructing a 99% confidence interval for \(\mu_{1}-\mu_{2}\) means you're 99% certain that the true population mean difference falls within this calculated range. This involves:
  • Adding and subtracting the margin of error to the point estimate.
  • The margin of error is derived from multiplying the Z-value for 99% confidence by the standard error of the difference.

The standard error of difference considers variability within the samples and requires using the pooled standard deviation, assuming the populations have equal deviations. The result is an interval showing the possible values for the true parameter difference.
Hypothesis Testing
Hypothesis testing is like a decision-making process to determine whether there's enough statistical evidence in a sample to infer that a certain condition is true for the entire population. In this problem, you're testing whether the mean of population 1 is lower than that of population 2.
The process starts with setting up the null and alternate hypotheses. Here:
  • The null hypothesis \(H_0\) assumes no difference \((\mu_1 - \mu_2 = 0)\).
  • The alternate hypothesis \(H_a\) suggests that \(\mu_1\) is less than \(\mu_2\) \((\mu_1 - \mu_2 < 0)\).

After defining the hypotheses, the next step is calculating a test statistic using your data. This involves:
  • Computing the standard error of the difference, as in the confidence interval calculation.
  • Calculating the Z-value for the observed data difference from the expected value under the null hypothesis (which is 0 in this case).

Lastly, compare the Z-value to a critical value, or use it to find a p-value. If the p-value is smaller than the given significance level (2.5% in this case), you reject the null hypothesis. Otherwise, you don't reject it, supporting the theory that the means could indeed be equal or that \(\mu_1\) is not significantly lower than \(\mu_2\).

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Most popular questions from this chapter

A consumer organization tested two paper shredders, the Piranha and the Crocodile, designed for home use. Each of 10 randomly selected volunteers shredded 100 sheets of paper with the Piranha, and then another sample of 10 randomly selected volunteers each shredded 100 sheets with the Crocodile. The Piranha took an average of 203 seconds to shred 100 sheets with a standard deviation of 6 seconds. The Crocodile took an average of 187 seconds to shred 100 sheets with a standard deviation of 5 seconds. Assume that the shredding times for both machines are normally distributed with equal but unknown standard deviations. a. Construct a \(99 \%\) confidence interval for the difference between the population means. b. Using the \(1 \%\) significance level, can you conclude that the mean time taken by the Piranha to shred 100 sheets is greater than that for the Crocodile? c. What would your decision be in part \(\mathrm{b}\) if the probability of making a Type \(\mathrm{I}\) error were zero? Explain.

A town that recently started a single-stream recycling program provided 60-gallon recycling bins to 25 randomly selected households and 75-gallon recycling bins to 22 randomly selected households. The total volume of recycling over a 10-week period was measured for each of the households. The average total volumes were 382 and 415 gallons for the households with the 60 - and 75 -gallon bins, respectively. The sample standard deviations were \(52.5\) and \(43.8\) gallons, respectively. Assume that the 10 -week total volumes of recycling are approximately normally distributed for both groups and that the population standard deviations are equal. a. Construct a \(98 \%\) confidence interval for the difference in the mean volumes of 10 -week recycling for the households with the 60 - and 75 -gallon bins. b. Using the \(2 \%\) significance level, can you conclude that the average 10 -week recycling volume of all households having 60 -gallon containers is different from the average volume of all households that have 75 -gallon containers?

Find the following confidence intervals for \(\mu_{d}\), assuming that the populations of paired differences are normally distributed. a. \(n=12, \bar{d}=17.5, \quad s_{d}=6.3, \quad\) confidence level \(=99 \%\) b. \(n=27, \quad \bar{d}=55.9, \quad s_{d}=14.7\), confidence level \(=95 \%\) c. \(n=16, \bar{d}=29.3, \quad s_{d}=8.3, \quad\) confidence level \(=90 \%\)

A sample of 500 observations taken from the first population gave \(x_{1}=305\). Another sample of 600 observations taken from the second population gave \(x_{2}=348\). a. Find the point estimate of \(p_{1}-p_{2}\). b. Make a \(97 \%\) confidence interval for \(p_{1}-p_{2}\). c. Show the rejection and nonrejection regions on the sampling distribution of \(\hat{p}_{1}-\hat{p}_{2}\) for \(H_{0}: p_{1}=p_{2}\) versus \(H_{1}: p_{1}>p_{2} .\) Use a significance level of \(2.5 \% .\) d. Find the value of the test statistic \(z\) for the test of part \(\mathrm{c}\). e. Will you reject the null hypothesis mentioned in part \(\mathrm{c}\) at a significance level of \(2.5 \%\) ?

Construct a \(99 \%\) confidence interval for \(p_{1}-p_{2}\) for the following. $$ n_{1}=300 \quad \hat{p}_{1}=.55 \quad n_{2}=200 \quad \hat{p}_{2}=.62 $$
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