/*! 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 44 An experiment was conducted to c... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An experiment was conducted to compare two diets \(\mathrm{A}\) and \(\mathrm{B}\) designed for weight reduction. Two groups of 30 overweight dieters each were randomly selected. One group was placed on diet \(\mathrm{A}\) and the other on diet \(\mathrm{B},\) and their weight losses were recorded over a 30 -day period. The means and standard deviations of the weight-loss measurements for the two groups are shown in the table. Find a \(95 \%\) confidence interval for the difference in mean weight loss for the two diets. Interpret your confidence interval. $$\begin{array}{ll}\text { Diet } A & \text { Diet B } \\\\\hline \bar{x}_{A}=21.3 & \bar{x}_{B}=13.4 \\\s_{A}=2.6 & s_{B}=1.9\end{array}$$

Short Answer

Expert verified
Answer: The 95% confidence interval for the difference in mean weight loss for diets A and B is (6.5, 9.3) pounds.

Step by step solution

01

Identify given statistics

We are provided with the following statistics: - \(\bar{x}_{A} = 21.3\), \(\bar{x}_{B} = 13.4\) (sample means) - \(s_{A} = 2.6\), \(s_{B} = 1.9\) (sample standard deviations) - \(n_A = n_B = 30\) (sample sizes) Our goal is to find the 95% confidence interval for the difference in mean weight loss, \(μ_A - μ_B\).
02

Find the standard error

The standard error for the difference in means is given by: \(SE_{\bar{X_A} - \bar{X_B}} = \sqrt{\frac{s_A^2}{n_A} + \frac{s_B^2}{n_B}}\) Now, plug in the given values: \(SE_{\bar{X_A} - \bar{X_B}} = \sqrt{\frac{2.6^2}{30} + \frac{1.9^2}{30}} \approx 0.714\)
03

Find the critical value

For a 95% confidence interval, we will use a critical value from a standard normal distribution (Z-distribution) table, since the sample size is large enough: \(z_{\frac{1-0.95}{2}} = z_{0.025} = 1.96\)
04

Calculate the margin of error

The margin of error is the product of the critical value and the standard error: \(MOE = z_{0.025} \cdot SE_{\bar{X_A} - \bar{X_B}} = 1.96 \cdot 0.714 \approx 1.40\)
05

Find the confidence interval

Finally, we can find the 95% confidence interval for the difference in mean weight loss for the two diets: \(CI_{95\%} = (\bar{x}_{A} - \bar{x}_{B}) \pm MOE\) \(CI_{95\%} = (21.3 - 13.4) \pm 1.40\) \(CI_{95\%} = 7.9 \pm 1.40\) So, the confidence interval is \((6.5, 9.3)\). In conclusion, we are 95% confident that the difference in mean weight loss for diet A and diet B lies between 6.5 and 9.3 pounds, with diet A being more effective for weight reduction.

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.

Difference in Means
When contrasting two groups, such as those on diet A and diet B, the difference in means is a crucial statistic. In this context, it refers to the average weight loss difference between the two groups. Essentially, we subtract the mean weight loss of diet B from diet A to determine how much more effective diet A might be than diet B. If we look at our experiment, this calculation is represented by \( \bar{x}_A - \bar{x}_B = 21.3 - 13.4 = 7.9 \). This means that diet A's participants, on average, lost 7.9 pounds more than those on diet B. Calculating the difference in means helps researchers understand the magnitude of increase or decrease between treatments or interventions. This difference serves as an anchor point for creating confidence intervals.
Standard Error
The standard error is a measure of the variability or dispersion of the sample mean's distribution. For our study involving weight loss on diets A and B, the standard error helps us quantify the precision of our sample mean difference. The formula to calculate the standard error for the difference in means is:
  • \( SE_{\bar{X}_A - \bar{X}_B} = \sqrt{\frac{s_A^2}{n_A} + \frac{s_B^2}{n_B}} \)
In our case, substituting our values gives \( SE_{\bar{X}_A - \bar{X}_B} \approx 0.714 \). This means the expected variation or uncertainty around our calculated mean difference is approximately 0.714 pounds, reflecting both sample sizes and standard deviations of the groups. A smaller standard error indicates a more precise estimate of the population parameter, reinforcing confidence in our findings.
Z-distribution
The Z-distribution, also known as the standard normal distribution, is symmetrical and centered at zero. It is extensively used in statistics for hypothesis testing, particularly when the sample size is large. In our experiment involving diets, the Z-distribution helps us determine critical values that define the margin of error for our confidence interval.
  • For a confidence level of 95%, we look for the Z-value representing the cutoff in the table, typically found at \( z_{0.025} = 1.96 \).
This Z-value corresponds to the middle 95% of the distribution, with 2.5% in each tail, establishing the limits for our hypothesis test. Using the Z-distribution is convenient as it assumes sample means tend to distribute normally if the sample size is reasonably large, due to the Central Limit Theorem.
Margin of Error
The margin of error expresses the range within which we expect the true difference in means to fall. It's calculated by multiplying the critical value from the Z-distribution by the standard error. In our example involving diets:
  • \( MOE = 1.96 \times 0.714 \approx 1.40 \)
This result means we can be confident that the true mean difference is within ±1.40 pounds of our observed difference. Hence, our confidence interval ranges from 6.5 to 9.3 pounds. A smaller margin of error signifies greater precision of our estimates, hence a more reliable conclusion about the comparative effectiveness of the studied diets. Always remember, a confidence interval that does not include zero reinforces the presence of a true difference.

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

A study was conducted to compare the mean numbers of police emergency calls per 8 -hour shift in two districts of a large city. Samples of 100 8-hour shifts were randomly selected from the police records for each of the two regions, and the number of emergency calls was recorded for each shift. The sample statistics are listed here: $$\begin{array}{lcc} & \multicolumn{2}{c} {\text { Region }} \\\\\cline { 2 - 3 } & 1 & 2 \\\\\hline \text { Sample Size } & 100 & 100 \\\\\text { Sample Mean } & 2.4 & 3.1 \\\\\text { Sample Variance } & 1.44 & 2.64\end{array}$$ Find a \(90 \%\) confidence interval for the difference in the mean numbers of police emergency calls per shift between the two districts of the city. Interpret the interval.

Independent random samples were selected from populations 1 and 2 . The sample sizes, means, and variances are as follows: $$\begin{array}{lcc} & \multicolumn{2}{c} {\text { Population }} \\\\\cline { 2 - 3 } & 1 & 2 \\\\\hline \text { Sample Size } & 35 & 49 \\\\\text { Sample Mean } & 12.7 & 7.4 \\\\\text { Sample Variance } & 1.38 & 4.14\end{array}$$ a. Find a \(95 \%\) confidence interval for estimating the difference in the population means \(\left(\mu_{1}-\mu_{2}\right) .\) b. Based on the confidence interval in part a, can you conclude that there is a difference in the means for the two populations? Explain.

Refer to Exercise 8.5 . What effect does an increased sample size have on the margin of error?

Independent random samples of \(n_{1}=n_{2}=n\) observations are to be selected from each of two binomial populations 1 and \(2 .\) If you wish to estimate the difference in the two population proportions correct to within . 05 , with probability equal to .98 , how large should \(n\) be? Assume that you have no prior information on the values of \(p_{1}\) and \(p_{2},\) but you want to make certain that you have an adequate number of observations in the samples.

What is normal, when it comes to people's body temperatures? A random sample of 130 human body temperatures, provided by Allen Shoemaker \(^{9}\) in the Journal of Statistical Education, had a mean of 98.25 degrees and a standard deviation of 0.73 degrees. a. Construct a \(99 \%\) confidence interval for the average body temperature of healthy people. b. Does the confidence interval constructed in part a contain the value 98.6 degrees, the usual average temperature cited by physicians and others? If not, what conclusions can you draw?

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.