/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 24 The following information was ob... [FREE SOLUTION] | 91Ó°ÊÓ

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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 a \(2.5 \%\) significance level if \(\mu_{1}\) is lower than \(\mu_{2}\).

Short Answer

Expert verified
The point estimate of \( \mu_{1}-\mu_{2} \) can be obtained by subtracting the mean of sample 2 from sample 1. The 99% confidence interval for \( \mu_{1}-\mu_{2} \) can be calculated using the formula \( \mu_{1}-\mu_{2} \pm Z \frac{s}{\sqrt{n}} \). A hypothesis test can be conducted to determine if \( \mu_{1} \) is lower than \( \mu_{2} \) using a one-tailed t-test.

Step by step solution

01

Find means of both populations

We need to compute the means \( \mu_{1} \) and \( \mu_{2} \) of both populations. The mean of a sample is given by sum of all values divided by the number of values. So, calculate the sum of all values in sample 1 and divide by the total number of values in sample 1. Do the same for sample 2.
02

Compute point estimate

The point estimate of \( \mu_{1} - \mu_{2} \) is simply \( \mu_{1} - \mu_{2} \), where \( \mu_{1} \) and \( \mu_{2} \) are the means of populations 1 and 2 calculated in the previous step.
03

Construct confidence interval

A 99% confidence interval for \( \mu_{1}-\mu_{2} \) can be computed using the formula: \( \mu_{1}-\mu_{2} \pm Z \frac{s}{\sqrt{n}} \), where Z is the z-value from the standard normal distribution corresponding to 99% confidence level, s is the standard deviation of the sample difference and n is the sample size. Find these values and substitute in the formula to find the interval.
04

Conduct Hypothesis Test

Here we want to test the hypothesis that \( \mu_{1} \) is lower than \( \mu_{2} \) at a 2.5% significance level. This is a one-tailed test. The Null Hypothesis H0: \( \mu_{1} - \mu_{2} \geq 0 \) and Alternative Hypothesis H1: \( \mu_{1} - \mu_{2} < 0 \). Conduct this test using a t statistic, which is given by \( t = \frac{(\mu_{1}-\mu_{2}) - D}{s_{d}/\sqrt{n}} \), where D=0 (difference under H0), \( s_{d} \) is standard deviation of the difference and n is the number of pairs. If calculated t score is less than critical t value at 2.5% significance level, we 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 provides a range of values, bounded by an upper and lower limit, which likely includes the true value of an unknown population parameter, with a given level of confidence. For example, a 99% confidence interval indicates that if we were to take 100 different samples and compute the interval for each, about 99 of those intervals will include the population parameter we're trying to estimate.

It's important to remember that while a confidence interval gives us a range, the precise value of the population parameter is still unknown. In the context of hypothesis testing, constructing a confidence interval is a way to understand the precision of our point estimate. It also helps us infer if a population parameter is statistically significant enough to be considered different from another parameter. When calculating a confidence interval for the difference of means, as in this exercise, one needs the calculated means, sample size, and the correct Z-value for the desired confidence level.
Point Estimate
A point estimate is a single value used to estimate a population parameter. In hypothesis testing, we often use the sample mean as a point estimate for the true population mean. In this exercise, the point estimate is of the difference between two population means, \( \mu_{1} - \mu_{2} \), which is simply computed by finding the respective means of two independent samples and subtracting one from the other.

The point estimate provides a specific, constant number derived from our sample data that serves as our best guess for the population parameter. While the point estimate is a concise way to represent an unknown parameter, it doesn’t account for the variability in estimates that would arise from different samples. Thus, it is often supplemented by a confidence interval to provide a fuller picture.
Z-Value
The Z-value is a key component of many statistics calculations, particularly in hypothesis testing and constructing confidence intervals. It represents the number of standard deviations a data point is from the mean in a standard normal distribution. In hypothesis testing, a Z-value can be used to determine the critical region by linking the significance level with the probability distribution.

For constructing a confidence interval, particularly with a large sample size or known population standard deviation, we often look up the Z-value that corresponds to our desired level of confidence (such as 99%). This Z-value is then used along with the standard deviation and sample size to calculate the confidence interval. It is crucial for understanding how far away our sample mean is from the population mean under the assumption that the null hypothesis is true.
T Statistic
The t statistic is vital when dealing with smaller sample sizes or when the population standard deviation is unknown. Unlike the Z-distribution, the t-distribution is thicker in the tails, which helps account for increased uncertainty with smaller samples. When performing hypothesis tests, particularly those concerning means, the t statistic helps compare the sample data to the null hypothesis.

To calculate the t statistic in this context, we determine how far the observed difference between sample means is from the hypothesized population difference, measured in standard errors. The formula is t = ((mean1 - mean2) - (population difference)) / (standard deviation of difference / sqrt(n)).
  • The t statistic provides insight into whether the sample means are significantly different.
  • Based on the calculated t value and the critical t value from t-distribution tables, we make conclusions about our hypothesis.
It’s particularly helpful in real-world scenarios where population parameters are largely unknown.

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Most popular questions from this chapter

The global recession has led more and more people to move in with relatives, which has resulted in a large number of multigenerational households. An October 2011 Pew Research Center poll showed that \(11.5 \%\) of people living in multigenerational households were living below the poverty level, and 14.6\% of people living in other types of households were living below the poverty level (www. pewsocialtrends.org/201 1/10/03/fighting-poverty-in-a-bad- cconomy-americans-move-in-with-relatives/? sre-pre-headline). Suppose that these results were based on samples of 1000 people living in multigenerational households and 2000 people living in other types of households. a. Let \(p_{1}\) be the proportion of all people in multigenerational households who live below the poverty level and \(p_{2}\) be the proportion of all people in other types of households who live below the poverty level. Construct a 98\% confidence interval for \(p_{1}-p_{2}\) - b. Using a \(2.5 \%\) significance level, can you conclude that \(p_{1}\) is less than \(p_{2}\) ? Use the critical-value approach. c. Repeat part b using the \(p\) -value approach.

Gamma Corporation is considering the installation of govemors on cars driven by its sales staff. These devices would limit the car speeds to a preset level, which is expected to improve fuel economy. The company is planning to test several cars for fuel consumption without governors for 1 week. Then governors would be installed in the same cars, and fuel consumption will be monitored for another week. Gamma Corporation wants to estimate the mean difference in fuel consumption with a margin of error of estimate of 2 mpg with a 90 \% confidence level. Assume that the differences in fuel consumption are normally distributed and that previous studies suggest that an estimate of \(s_{d}=3 \mathrm{mpg}\) is reasonable. How many cars should be tested? (Note that the critical value of \(t\) will depend on \(n\), so it will be necessary to use trial and error.)

The owner of a mosquito-infested fishing camp in Alaska wants to test the effectiveness of two rival brands of mosquito repellents, \(\mathrm{X}\) and \(\mathrm{Y}\). During the first month of the season, eight people are chosen at random from those guests who agree to take part in the experiment. For each of these guests, Brand \(\bar{X}\) is randomly applied to one arm and Brand \(\mathrm{Y}\) is applied to the other arm. These guests fish for 4 hours, then the owner counts the number of bites on each arm. The table below shows the number of bites on the arm with Brand \(X\) and those on the arm with Brand \(Y\) for each guest. \begin{tabular}{l|rrrrrrrr} \hline Guest & A & B & C & D & E & F & G & H \\ \hline Brand X & 12 & 23 & 18 & 36 & 8 & 27 & 22 & 32 \\ \hline Brand Y & 9 & 20 & 21 & 27 & 6 & 18 & 15 & 25 \\ \hline \end{tabular} a. Construct a \(95 \%\) confidence interval for the mean \(\mu_{d}\) of population paired differences, where a paired difference is defined as the number of bites on the arm with Brand \(X\) minus the number of bites on the arm with Brand \(Y\). b. Test at a \(5 \%\) significance level whether the mean number of bites on the arm with Brand \(\mathrm{X}\) and the mean number of bites on the arm with Brand \(Y\) are different for all such guests.

According to Exercise \(10.27\), an insurance company wants to know if the average speed at which men drive cars is higher than that of women drivers. The company took a random sample of 27 cars driven by men on a highway and found the mean speed to be 72 miles per hour with a standard deviation of \(2.2\) miles per hour. Another sample of 18 cars driven by women on the same highway gave a mean speed of 68 miles per hour with a standard deviation of \(2.5\) miles per hour. Assume that the speeds at which all men and all women drive cars on this highway are both normally distributed with unequal population standard deviations. a. Construct a \(98 \%\) confidence interval for the difference between the mean speeds of cars driven by all men and all women on this highway. b. Test at a \(1 \%\) significance level whether the mean speed of cars driven by all men drivers on this highway is higher than that of cars driven by all women drivers. c. Suppose that the sample standard deviations were \(1.9\) and \(3.4\) miles per hour, respectively. Redo parts a and b. Discuss any changes in the results.

Employees of a large corporation are concerned about the declining quality of medical services provided by their group health insurance. A random sample of 100 office visits by employees of this corporation to primary care physicians during 2004 found that the doctors spent an average of 19 minutes with each patient. This year a random sample of 108 such visits showed that doctors spent an average of \(15.5\) minutes with each patient. Assume that the standard deviations for the two populations are \(2.7\) and \(2.1\) minutes, respectively. a. Construct a \(95 \%\) confidence interval for the difference between the two population means for these two years. b. Using a \(2.5 \%\) level of significance, can you conclude that the mean time spent by doctors with each patient is lower for this year than for \(2004 ?\) c. What would your decision be in part \(\mathrm{b}\) if the probability of making a Type I error were zero? Explain.

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