/*! 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 23 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. \(\begin{array}{llllllllllllll}\text { Sample 1: } & 47.7 & 46.9 & 51.9 & 34.1 & 65.8 & 61.5 & 50.2 & 40.8 & 53.1 & 46.1 & 47.9 & 45.7 & 49.0 \\\ \text { Sample 2: } & 50.0 & 47.4 & 32.7 & 48.8 & 54.0 & 46.3 & 42.5 & 40.8 & 39.0 & 68.2 & 48.5 & 41.8 & \end{array}\) 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 \(98 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) c. Test at a \(1 \%\) significance level if \(\mu_{1}\) is greater than \(\mu_{2}\).

Short Answer

Expert verified
Detailed step-by-step calculations are required for precise answers to sections a, b, and c. Here, you'll find generic descriptions of what the answers should look like. a) The point estimate of \(\mu_{1}-\mu_{2}\) is the difference between the sample means. b) The 98% confidence interval for \(\mu_{1}-\mu_{2}\) will be an interval range. c) The conclusion of the hypothesis test could be either 'Reject the null hypothesis' or 'Fail to reject the null hypothesis' depend on our calculated p-value or test statistic.

Step by step solution

01

Calculate Sample Means

First, calculate the sample means \(\bar{x}_{1}\) and \(\bar{x}_{2}\) for sample 1 and sample 2 respectively. This can be done by summing up all the values in each sample and dividing by the number of observations in each sample. These sample means serve as the point estimates for the population means \(\mu_{1}\) and \(\mu_{2}\).
02

Point Estimate of \(\mu_{1}-\mu_{2}\)

The point estimate of \(\mu_{1}-\mu_{2}\) is obtained by subtracting the sample mean of population 2 from the sample mean of population 1 i.e. \(\bar{x}_{1} - \(\bar{x}_{2}\).
03

Calculate Sample Variances

Next, calculate the sample variances for sample 1 and sample 2 respectively. The variance is a measure of the spread or dispersion. It can be calculated by taking the squared difference of each observation from the sample mean, summing these squared differences, and dividing by the number of observations minus 1. Since the distributions have equal standard deviations, we also calculate the pooled variance.
04

Construct a 98% Confidence Interval for \(\mu_{1}-\mu_{2}\)

To construct a 98% confidence interval for \(\mu_{1}-\mu_{2}\), use the formula \(\bar{x}_{1} - \bar{x}_{2} \pm Z_{\alpha/2} * \sqrt{\frac{S_{p}^{2}}{n_{1}} + \frac{S_{p}^{2}}{n_{2}}}\) where \(Z_{\alpha/2}\) is the standard normal variable corresponding to the desired confidence level and \(S_{p}\) is the pooled standard deviation.
05

Hypothesis Testing

Now, test the hypothesis that \(\mu_{1}\) is greater than \(\mu_{2}\) at a 1% significance level. This is a one-sided test. The null hypothesis is \(\mu_{1} = \mu_{2}\) and the alternative hypothesis is \(\mu_{1} > \mu_{2}\). Calculate the test statistic using \(\frac{(\bar{x}_{1}-\bar{x}_{2}) - 0}{\sqrt{\frac{S_{p}^{2}}{n_{1}} + \frac{S_{p}^{2}}{n_{2}}}}\). If the test statistic falls in the critical region or p-value is less than the significance level, 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 gives a range of values within which we believe the true population parameter (like the mean) lies. Imagine you're guessing a number, and you say, "I'm 98% sure the number is between 5 and 10." Similarly, when we create a 98% confidence interval for the population mean difference \(\mu_1 - \mu_2\), we are saying that in 98% of all possible samples, the true mean difference would fall within our calculated interval.
To calculate this interval, we use formulas that factor in sample statistics such as means and variances. For example, in comparing two populations, we would use the pooled variance and the means from our samples. The standard normal variable \(Z_{\alpha/2}\) reflects the confidence level we choose. For a 98% confidence level, this \(Z\) value would be around 2.33.
Thus, the confidence interval formula for two means is: \[\bar{x}_1 - \bar{x}_2 \pm Z_{\alpha/2} \times \sqrt{\frac{S_p^2}{n_1} + \frac{S_p^2}{n_2}}\] Each term assesses how much the sample means vary and accounts for the spread in the data, helping us establish a reliable range for our estimate.
Point Estimate
A point estimate is a single value used to approximate a parameter of a population. In this exercise, the point estimate for the difference in means of two populations \(\mu_1 - \mu_2\) is derived from the corresponding sample means.
Point estimates provide a snapshot view rather than a detailed picture. This is because they don't convey any information about the accuracy or variability of the estimation.
The formula for the point estimate of the difference between two means involves subtracting the mean of one sample from the mean of another: \[\bar{x}_1 - \bar{x}_2\] This simplistic approach is easy to comprehend and quickly gives us a general idea of the difference between population means. However, since it's a single value, it carries uncertainty about its closeness to the actual parameter value deep in the population.
Sample Mean
The sample mean is a fundamental concept in statistics, representing the average value of a particular set of numbers. When we talk about the sample mean, denoted as \(\bar{x}\), we are discussing the arithmetic mean calculated from the data in a sample.
To compute the sample mean, add up all your observations (all the numbers in your sample), and divide by the number of observations. For instance, with a sample like \(47.7, 46.9, 51.9,\ldots, 49.0\), sum these figures and divide by the quantity of numbers.
This calculation reveals a central or typical value, informing us about the middle or expected value in the dataset.
  • It's efficient for summarizing data with a single figure.
  • It lays the groundwork for more complex statistical analyses, such as finding point estimates or constructing confidence intervals.
The sample mean helps provide insight into the central tendency of the population from which the sample was drawn.
Significance Level
The significance level, often denoted as \(\alpha\), is a threshold used in hypothesis testing to decide whether to reject the null hypothesis. It's a measure of how confident we want to be when making our decision. A common choice is 5% (0.05), but in exercises like this, a more stringent 1% (0.01) level is used.
The level chosen reflects the probability of falsely rejecting a true null hypothesis, commonly referred to as a Type I error. In the context of this exercise, a 1% significance level indicates that there is only a 1% risk of incorrectly concluding that \(\mu_1 > \mu_2\) when they are, in fact, equal.
The significance level plays a pivotal role in determining the critical value against which the test statistic is compared. If the test statistic exceeds this critical threshold, or if the p-value is less than \(\alpha\), the null hypothesis is rejected.
  • This balance of error risk helps maintain the integrity of statistical conclusions.
  • Adopting higher confidence or more relaxed error risk can modify these thresholds according to specific requirements.
Understanding \(\alpha\) is crucial for appropriately conducting and interpreting hypothesis tests.

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Most popular questions from this chapter

Maria and Ellen both specialize in throwing the javelin. Maria throws the javelin a mean distance of 200 feet with a standard deviation of 10 feet, whereas Ellen throws the javelin a mean distance of 210 feet with a standard deviation of 12 feet. Assume that the distances cach of these athletes throws the javelin are normally distributed with these population means and standard deviations. If Maria and Ellen each throw the javelin once, what is the probability that Maria's throw is longer than Ellen's?

Refer to Exercise \(10.32\). As mentioned in that exercise, according to the credit rating agency Equifax, credit limits on newly issued credit cards increased between January 2011 and May 2011 . Suppose that random samples of 400 new credit cards issued in January 2011 and 500 new credit cards issued in May 2011 had average credit limits of \(\$ 2635\) and \(\$ 2887\), respectively. Suppose that the sample standard deviations for these two samples were \(\$ 365\) and \(\$ 412\), respectively. Now assume that the population standard deviations for the two populations are unknown and not equal. a. Let \(\mu_{1}\) and \(\mu_{2}\) be the average credit limits on all credit cards issued in January 2011 and in May 2011 , respectively. What is the point estimate of \(\mu_{1}-\mu_{2}\) ? b. Construct a \(98 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). c. Using a \(1 \%\) significance level, can you conclude that the average credit limit for all new credit cards issued in January 2011 was lower than the corresponding average for all credit cards issued in May 2011 ? Use both the \(p\) -value and the critical-value approaches to make this test.

Several retired bicycle racers are coaching a large group of young prospects. They randomly select seven of their riders to take part in a test of the effectiveness of a new dietary supplement that is supposed to increase strength and stamina. Each of the seven riders does a time trial on the same course. Then they all take the dietary supplement for 4 weeks. All other aspects of their training program remain as they were prior to the time trial. At the end of the 4 weeks, these riders do another time trial on the same course. The times (in minutes) recorded by each rider for these trials before and after the 4 -week period are shown in the following table. \begin{tabular}{l|lllrrll} \hline Before & 103 & 97 & 111 & 95 & 102 & 96 & 108 \\ \hline After & 100 & 95 & 104 & 101 & 96 & 91 & 101 \\ \hline \end{tabular} a. Construct a \(99 \%\) confidence interval for the mean \(\mu_{d}\) of the population paired differences, where a paired difference is equal to the time taken before the dietary supplement minus the time taken after the dietary supplement. b. Test at a \(2.5 \%\) significance level whether taking this dietary supplement results in faster times in the time trials.

The standard recommendation for automobile oil changes is once every 3000 miles. A local mechanic is interested in determining whether people who drive more expensive cars are more likely to follow the recommendation. Independent random samples of 45 customers who drive luxury cars and 40 customers who drive compact lower-price cars were selected. The average distance driven between oil changes was 3187 miles for the luxury car owners and 3214 miles for the compact lower-price cars. The sample standard deviations were \(42.40\) and \(50.70\) miles for the luxury and compact groups, respectively. Assume that the population distributions of the distances between oil changes have the same standard deviation for the two populations. a. Construct a \(95 \%\) confidence interval for the difference in the mean distances 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 less for all luxury cars than that for all compact lower-price cars?

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.)

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