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Chapter 10: Problem 18

The following information was obtained from two independent samples selected from two normally distributed populations with unknown but equal standard deviations. \(\begin{array}{llllllll}\text { Sample 1: } & 2.18 & 2.23 & 1.96 & 2.24 & 2.72 & 1.87 & 2.68 \\ & 2.15 & 2.49 & 2.05 & & & & \\ \text { Sample 2: } & 1.82 & 1.26 & 2.00 & 1.89 & 1.73 & 2.03 & 1.43 \\ & 2.05 & 1.54 & 2.50 & 1.99 & 2.13 & & \end{array}\) 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 a \(2.5 \%\) significance level if \(\mu_{1}\) is lower than \(\mu_{2}\).

Short Answer

Expert verified
The short answer to this problem would consist of specific numerical values for the point estimate of \(\mu_{1}-\mu_{2}\), the 99% confidence interval for \(\mu_{1}-\mu_{2}\), and the conclusion from the hypothesis testing (whether \(\mu_{1}\) is lower than \(\mu_{2}\) or not with the given significance level). However, without the exact calculations, these exact numerical results cannot be provided.

Step by step solution

01

Calculation of point estimate

Firstly, the point estimate of \(\mu_{1}-\mu_{2}\) is the difference between the sample averages, as this is an unbiased estimator for the difference in population means. For sample 1, the sample mean \(\overline{x}_{1}\) will be calculated by adding all the sample 1 values and dividing it by the number of observations. The same will be done for sample 2, producing \(\overline{x}_{2}.\) The point estimate of \(\mu_{1}-\mu_{2}\) is then \(\overline{x}_{1} - \overline{x}_{2}.\)
02

Calculation of 99% confidence interval

To construct a 99% confidence interval for \(\mu_{1}-\mu_{2},\) calculate the standard error (SE) of the difference of the sample means. The standard error is calculated by the formula, \( SE=\sqrt{(s_{1}^{2}/n_{1})+(s_{2}^{2}/n_{2})},\) where \(s_{1}^{2}\) and \(s_{2}^{2}\) are the sample variances for sample 1 and 2 respectively and \(n_{1}\) and \(n_{2}\) are the respective sample sizes. The 99% confidence interval is then, \(\overline{x}_{1}-\overline{x}_{2} \pm (t_{\alpha/2,n_{1}+n_{2}-2} \times SE),\) where \(t_{\alpha/2,n_{1}+n_{2}-2}\) is the t-value for a 99% confidence interval with \(n_{1}+n_{2}-2\) degrees of freedom.
03

Hypothesis Testing

Next, conduct a hypothesis test at a 2.5% significance level for \(H_{0}:\mu_{1} \geq \mu_{2}\) against \(H_{1}:\mu_{1} < \mu_{2}.\) The test statistic is: \(t = (\overline{x}_{1} - \overline{x}_{2})/SE.\) This value is compared to the critical t-value for a 2.5% one-tailed test with \(n_{1}+n_{2}-2\) degrees of freedom. If the calculated t-statistic is less than the critical t-value, reject the null hypothesis. Otherwise, do not reject the null hypothesis.

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Key Concepts

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

Confidence Interval
A confidence interval is a range of values that is used to estimate an unknown population parameter. In this case, it is used to estimate the difference between two population means, \(\mu_1\) and \(\mu_2\).
To construct a confidence interval, one must know the point estimate (the sample mean difference) and the standard error.
  • The confidence interval provides an estimated range believed to encompass the true population parameter 99% of the time.
  • To find this interval, you calculate two bounds: the lower and upper limits.
  • Here, you add and subtract a calculated margin of error from the point estimate to get these bounds.
The formula for the confidence interval in this scenario is:\[ \overline{x}_{1} - \overline{x}_{2} \pm \left( t_{\alpha/2, n_1+n_2-2} \times \text{SE} \right) \].
The \(t_{\alpha/2, n_1+n_2-2}\) is a value found using the t-distribution for your desired confidence level, in this situation, it corresponds to the 99% confidence level.
Point Estimate
A point estimate in statistics is a single value given as an estimate of the population parameter. It is derived from sample data, making it essential for the overall hypothesis testing process.
  • The point estimate in this case is the difference between the sample means of the two groups \((\overline{x}_1 - \overline{x}_2)\).
  • It serves as a representation of the difference of means \((\mu_1 - \mu_2)\) from the population.
  • Consider it as a snapshot, which provides the best guess value of a parameter without including an estimate of variability.
Finding this estimate involves calculating the average of each sample and then subtracting the sample 2 average from the sample 1 average.
Standard Error
The concept of standard error (SE) is crucial in hypothesis testing.
Essentially, it measures the variability, or dispersion, of sample statistics over different samples from the same population. In simpler terms, it tells us how much the sample mean is likely to differ from the true population mean.
  • The smaller the standard error, the more representative the sample is of the overall population.
  • The standard error for a difference of means is calculated using the formula \(SE = \sqrt{(s_1^2/n_1) + (s_2^2/n_2)}\).
  • It incorporates sample variances and sizes of both groups, making it a balanced estimate of how reliable your sample means difference is.
This value is vital since it affects both the confidence interval and the t-test results.
T-Distribution
The t-distribution, often called Student's t-distribution, is a type of probability distribution that is symmetric and bell-shaped.
It is particularly useful when dealing with small sample sizes or when the population standard deviation is not known.
  • It helps in determining the critical values for confidence intervals and hypothesis tests.
  • Unlike a normal distribution, it has heavier tails, which means there is a greater likelihood of extreme values.
  • The shape of the t-distribution is determined by the degrees of freedom, which are directly related to the sample size.
In the context of hypothesis testing and confidence intervals, the t-distribution offers a way to estimate population parameters when only sample data is available. It allows us to use the computed standard error along with the point estimate to calculate the confidence interval and assess the hypothesis test results.

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Most popular questions from this chapter

According to an estimate, the average earnings of female workers who are not union members are \(\$ 1120\) per week and those of female workers who are union members are \(\$ 1244\) per week. Suppose that these average earnings are calculated based on random samples of 1500 female workers who are not union members and 2000 female workers who are union members. Further assume that the standard deviations for the two corresponding populations are \(\$ 70\) and \(\$ 90\), respectively. a. Construct a \(95 \%\) confidence interval for the difference between the two population means. b. Test at a \(2.5 \%\) significance level whether the mean weekly earnings of female workers who are not union members is less than that of female workers who are union members.

A local college cafeteria has a self-service soft ice cream machine. The cafeteria provides bowls that can hold up to 16 ounces of ice cream. The food service manager is interested in comparing the average amount of ice cream dispensed by male students to the average amount dispensed by female students. A measurement device was placed on the ice cream machine to determine the amounts dispensed. Random samples of 85 male and 78 female students who got ice cream were selected. The sample averages were \(7.23\) and \(6.49\) ounces for the male and female students, respectively. Assume that the population standard deviations are \(1.22\) and \(1.17\) ounces, respectively. a. Let \(\mu_{1}\) and \(\mu_{2}\) be the population means of ice cream amounts dispensed by all male and all female students at this college, respectively. What is the point estimate of \(\mu_{1}-\mu_{2} ?\) b. Construct a \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). c. Using a \(1 \%\) significance level, can you conclude that the average amount of ice cream dispensed by all male college students is larger than the average amount dispensed by all female college students? Use both approaches to make this test.

A state that requires periodic emission tests of cars operates two emission test stations, \(\mathrm{A}\) and \(\mathrm{B}\), in one of its towns. Car owners have complained of lack of uniformity of procedures at the two stations, resulting in different failure rates. A sample of 400 cars at Station A showed that 53 of those failed the test; a sample of 470 cars at Station B found that 51 of those failed the test. a. What is the point estimate of the difference between the two population proportions? b. Construct a \(95 \%\) confidence interval for the difference between the two population proportions. c. Testing at a \(5 \%\) significance level, can you conclude that the two population proportions are different? Use both the critical-value and the \(p\) -value approaches.

Manufacturers of two competing automobile models, Gofer and Diplomat, each claim to have the lowest mean fuel consumption. Let \(\mu_{1}\) be the mean fuel consumption in miles per gallon (mpg) for the Gofer and \(\mu_{2}\) the mean fuel consumption in mpg for the Diplomat. The two manufacturers have agreed to a test in which several cars of each model will be driven on a 100 -mile test run. The fuel consumption, in \(\mathrm{mpg}\), will be calculated for each test run. The average mpg for all 100 -mile test runs for each model gives the corresponding mean. Assume that for each model the gas mileages for the test runs are normally distributed with \(\sigma=2 \mathrm{mpg}\). Note that each car is driven for one and only one 100 -mile test run. a. How many cars (i.e., sample size) for each model are required to estimate \(\mu_{1}-\mu_{2}\) with a \(90 \%\) confidence level and with a margin of error of estimate of \(1.5 \mathrm{mpg}\) ? Use the same number of cars (i.e., sample size) for each model. b. If \(\mu_{1}\) is actually \(33 \mathrm{mpg}\) and \(\mu_{2}\) is actually \(30 \mathrm{mpg}\), what is the probability that five cars for each model would yield \(\bar{x}_{1} \geq \bar{x}_{2}\) ?

A sample of 1000 observations taken from the first population gave \(x_{1}=290\). Another sample of 1200 observations taken from the second population gave \(x_{2}=396\). a. Find the point estimate of \(p_{1}-p_{2}\). b. Make a \(98 \%\) 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}
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