/*! 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 49 Of the 130 people in Exercise 8.... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Of the 130 people in Exercise 8.37,65 were female and 65 were male. The means and standard deviation of their temperatures are shown below. $$\begin{array}{lrr} & \text { Men } & \text { Women } \\\\\hline \text { Sample Mean } & 98.11 & 98.39 \\\\\text { Standard Deviation } & 0.70 & 0.74\end{array}$$ Find a \(95 \%\) confidence interval for the difference in the average body temperatures for males versus females. Based on this interval, can you conclude that there is a difference in the average temperatures for males versus females? Explain.

Short Answer

Expert verified
Answer: Yes, there is a statistically significant difference in the average body temperatures between males and females. The 95% confidence interval does not include 0 and it is negative, indicating that the average body temperature of males is lower than that of females.

Step by step solution

01

Find the Point Estimate (Difference in Means)

The point estimate for the difference in average temperatures between males and females is simply the difference in sample means, which we can denote as \(\bar{X}_m - \bar{X}_f\). Based on the given data, we can calculate this value as follows: $$ \bar{X}_m - \bar{X}_f = 98.11 - 98.39 = -0.28 \hspace{10mm} (1) $$
02

Calculate the Standard Error

Next, we need to find the standard error for the difference in means, which we can calculate using the formula: $$ SE = \sqrt{\frac{(s_m^2)}{n_m} + \frac{(s_f^2)}{n_f}} $$ Where \(s_m\) and \(s_f\) are the standard deviations for males and females, and \(n_m\) and \(n_f\) are the sample sizes. Using the given data, we can calculate the standard error: $$ SE = \sqrt{\frac{(0.70^2)}{65} + \frac{(0.74^2)}{65}} = 0.1258 \hspace{10mm} (2) $$
03

Find the Confidence Interval

Now, we can find the \(95\%\) confidence interval using the point estimate, standard error, and the critical value (\(z\)) for a \(95\%\) confidence level. The critical value for \(95\%\) confidence is \(z = 1.96\). We can calculate the margin of error (ME) and then the confidence interval: $$ ME = z \times SE = 1.96 \times 0.1258 = 0.2466 $$ Now, let's find the lower limit (LL) and upper limit (UL) of the confidence interval: $$ LL = (\bar{X}_m - \bar{X}_f) - ME = -0.28 - 0.2466 = -0.5266 $$ $$ UL = (\bar{X}_m - \bar{X}_f) + ME = -0.28 + 0.2466 = -0.0334 $$ Thus, the \(95 \%\) confidence interval for the difference in average body temperatures for males versus females is \((-0.5266, -0.0334)\).
04

Conclude based on the Confidence Interval

Since the \(95\%\) confidence interval does not include \(0\), we can confidently say that there is a statistically significant difference in the average body temperatures for males and females. The negative sign of the confidence interval values indicates that the average body temperature of males is lower than that of females.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

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 as the best guess or estimate of a population parameter. In this exercise, the point estimate focuses on the difference between the average body temperatures of males and females. Specifically, we're looking at the difference between their sample means. The calculation is simple: you subtract the sample mean of men's temperatures from that of women's. In this case: 98.11 (men) - 98.39 (women) = -0.28.
This result means that, on average, men's temperatures are 0.28 degrees Fahrenheit lower than women's. It's a snapshot of the general temperature difference you'll apply inside your confidence interval calculations.
Key points about point estimates:
  • They are derived from sample data.
  • They provide a single value estimate for the population parameter.
  • They often serve as the baseline for more complex statistical analyses, like confidence intervals.
Standard Error
Standard error measures how far the sample mean of the data is expected to be from the true population mean. In essence, it helps to give an idea of the accuracy of the point estimate. For differences in means, like in this exercise, the standard error combines the variability of both groups' estimates.
Mathematically, it's calculated with the formula:
\[ SE = \sqrt{\frac{(s_m^2)}{n_m} + \frac{(s_f^2)}{n_f}} \]
where \( s_m \) and \( s_f \) are the standard deviations of men's and women's temperatures, while \( n_m \) and \( n_f \) are their respective sample sizes. In this case, the standard error was found to be 0.1258.
This value helps us assess how much variation we might expect if we took many samples. A smaller standard error implies more precise estimates of the mean difference. Remember:
  • Standard error decreases with larger sample sizes.
  • It reflects the inherent variability in the sample data.
  • It is crucial for constructing confidence intervals.
Statistical Significance
Statistical significance helps us understand whether an observed effect in the data is unlikely due to random chance. When examining the confidence interval, if it does not include 0, it suggests a statistically significant difference between groups.
In this scenario, the confidence interval calculated was \((-0.5266, -0.0334)\). Since zero is not within this range, we can confidently say there is a statistically significant difference in the average body temperatures of men and women.
This result signifies that the observed difference (with men averaging lower temperatures than women) is unlikely to occur just by random sampling variability.
Some important points on statistical significance:
  • A confidence interval not including 0 indicates significance in this context.
  • Statistical significance does not imply practical significance; even small differences can be statistically significant with large samples.
  • The context of the data and its implications should be considered alongside statistical significance.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Last year's records of auto accidents occurring on a given section of highway were classified according to whether the resulting damage was \(\$ 1000\) or more and to whether a physical injury resulted from the accident. The data follows: $$\begin{array}{lcc} & \text { Under } \$ 1000 & \$ 1000 \text { or More } \\\\\hline \text { Number of Accidents } & 32 & 41 \\\\\text { Number Involving Injuries } & 10 & 23\end{array}$$ a. Estimate the true proportion of accidents involving injuries when the damage was \(\$ 1000\) or more for similar sections of highway and find the margin of error. b. Estimate the true difference in the proportion of accidents involving injuries for accidents with damage under \(\$ 1000\) and those with damage of \(\$ 1000\) or more. Use a \(95 \%\) confidence interval.

Suppose you wish to estimate a population mean based on a random sample of \(n\) observations, and prior experience suggests that \(\sigma=12.7\). If you wish to estimate \(\mu\) correct to within 1.6 , with probability equal to .95, how many observations should be included in your sample?

A quality-control engineer wants to estimate the fraction of defectives in a large lot of film cartridges. From previous experience, he feels that the actual fraction of defectives should be somewhere around .05. How large a sample should he take if he wants to estimate the true fraction to within .01, using a \(95 \%\) confidence interval?

Do you own an iPod Nano or a Sony Walkman Bean? These and other brands of MP3 players are becoming more and more popular among younger Americans. An iPod survey reported that \(54 \%\) of 12 - to 17 -year-olds, \(30 \%\) of 18 - to 34 -year-olds, and \(13 \%\) of 35 - to 54 -year-olds own MP3 players. \({ }^{6}\) Suppose that these three estimates are based on random samples of size \(400,350,\) and \(362,\) respectively. a. Construct a \(95 \%\) confidence interval estimate for the proportion of 12 - to 17 -year-olds who own an MP3 player. b. Construct a \(95 \%\) confidence interval estimate for the proportion of 18 - to 34 -year-olds who own an MP3 player.

Do well-rounded people get fewer colds? A study on the Chronicle of Higher \(E d u\) cation was conducted by scientists at Carnegie Mellon University, the University of Pittsburgh, and the University of Virginia. They found that people who have only a few social outlets get more colds than those who are involved in a variety of social activities. \(^{14}\) Suppose that of the 276 healthy men and women tested, \(n_{1}=96\) had only a few social outlets and \(n_{2}=105\) were busy with six or more activities. When these people were exposed to a cold virus, the following results were observed: $$\begin{array}{lcc} & \text { Few Social Outlets } & \text { Many Social Outlets } \\\\\hline \text { Sample Size } & 96 & 105 \\\\\text { Percent with Colds } & 62 \% & 35 \%\end{array}$$ a. Construct a \(99 \%\) confidence interval for the difference in the two population proportions. b. Does there appear to be a difference in the population proportions for the two groups? c. You might think that coming into contact with more people would lead to more colds, but the data show the opposite effect. How can you explain this unexpected finding?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.