/*! 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 124 3.124 Donating Blood to Grandma?... [FREE SOLUTION] | 91Ó°ÊÓ

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3.124 Donating Blood to Grandma? There is some evidence that "young blood" might improve the health, both physically and cognitively, of elderly people (or mice). Exercise 2.69 on page 75 introduces one study in which old mice were randomly assigned to receive transfusions of blood from either young mice or old mice. Researchers then measured the number of minutes each of the old mice was able to run on a treadmill. The data are stored in YoungBlood. We wish to estimate the difference in the mean length of time on the treadmill, between those mice getting young blood and those mice getting old blood. Use StatKey or other technology to find and interpret a \(90 \%\) confidence interval for the difference in means.

Short Answer

Expert verified
Without the actual data from the study, it's not possible to provide the numeric values of the confidence interval. However, the calculated interval will provide the range of values in which we are 90% confident the true difference of the population means resides. How this interval is interpreted depends on its values and it will show the potential difference between both treatment types for the mice.

Step by step solution

01

Gather Data

Firstly, gather the data from the YoungBlood study. You need two sets of data: first, the length of time old mice that received young blood were able to run on a treadmill. Second, the length of time old mice that received old blood were able to run on a treadmill.
02

Calculate Point Estimates

Now calculate the means of the two groups, which serve as the point estimates. Use the formula for the mean: \(\bar{X} = \frac{1}{N} \sum_{i=1}^{N} X_i\). Calculate this for both data sets.
03

Calculate Standard Errors

Next, calculate the standard errors for both groups. The standard error is calculated as \(\frac{S}{\sqrt{n}}\), where \(S\) is the standard deviation and \(n\) is the number of observations. Calculate this for both data sets.
04

Calculate Degrees of Freedom

Degrees of freedom can be calculated using the formula: \(df = n1 + n2 - 2\), where \(n1\) and \(n2\) are the sample sizes of both groups.
05

Define Confidence Interval

We're interested in a 90% confidence interval. From a t-table, find the t-value that corresponds to the 90% confidence and your degree of freedom.
06

Calculate Confidence Interval

Using your obtained t value, you calculate the confidence interval using the formula: \((\overline{x_1} - \overline{x_2}) ± t \sqrt{s_1^2/n_1 + s_2^2/n_2}\). Here, \(\overline{x_1}\) and \(\overline{x_2}\) are the means, \(s_1\) and \(s_2\) are the standard deviations, \(n_1\) and \(n_2\) are the sample sizes of two groups, and \(t\) is the t-value from the t-table.
07

Interpret Result

The resulting interval forms your 90% confidence interval for the difference in population means. If this interval does not include zero, we can conclude that the two population means are statistically different at the 90% confidence level.

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Key Concepts

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

Understanding the Difference in Means
When we talk about the difference in means, we're interested in finding how much one group's average differs from another's. In our young blood study, this means comparing the running times of old mice that received young blood versus those with old blood. This helps us understand whether the type of blood affects endurance.
  • Calculate the average running time for each group, known as the point estimate. Use the formula: \(\bar{X} = \frac{1}{N} \sum_{i=1}^{N} X_i\).
  • Subtract one mean from the other to find the difference in means.
This difference in means is crucial for determining if there's a significant effect or if the variations are just due to chance.
Calculating the Standard Error
The standard error tells us how much the sample mean is expected to vary from the true population mean. It's a measure of the sample's accuracy.
  • Calculate using \(\frac{S}{\sqrt{n}}\), where \(S\) is the standard deviation and \(n\) is the sample size.
  • The smaller the standard error, the closer our sample mean is likely to be to the true population mean.
Understanding standard error helps us gauge the reliability of our mean difference in the context of variability.
Exploring Degrees of Freedom
Degrees of freedom play a big role when determining the correct t-value for confidence intervals. They refer to the number of values in a calculation that are free to vary.
  • Calculated as \(df = n1 + n2 - 2\), where \(n1\) and \(n2\) are the sample sizes of each group.
  • The degrees of freedom affect how the distribution looks and thus influences the t-value you use.
By knowing the degrees of freedom, you ensure that you're using the correct distribution to make accurate predictions about the confidence interval.

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Most popular questions from this chapter

Are Female Rats More Compassionate Than Male Rats? Exercise 3.88 describes a study in which rats showed compassion by freeing a trapped rat. In the study, all six of the six female rats showed compassion by freeing the trapped rat while 17 of the 24 male rats did so. Use the results of this study to give a best estimate for the difference in proportion of rats showing compassion, between female rats and male rats. Then use StatKey or other technology to estimate the standard error \(^{44}\) and use it to compute a \(95 \%\) confidence interval for the difference in proportions. Use the interval to determine whether it is plausible that male and female rats are equally compassionate (i.e., that the difference in proportions is zero). The data are available in the dataset CompassionateRats.

Effect of Overeating for One Month: Average Long-Term Weight Gain Overeating for just four weeks can increase fat mass and weight over two years later, a Swedish study shows \(^{35}\) Researchers recruited 18 healthy and normal-weight people with an average age of \(26 .\) For a four-week period, participants increased calorie intake by \(70 \%\) (mostly by eating fast food) and limited daily activity to a maximum of 5000 steps per day (considered sedentary). Not surprisingly, weight and body fat of the participants went up significantly during the study and then decreased after the study ended. Participants are believed to have returned to the diet and lifestyle they had before the experiment. However, two and a half years after the experiment, the mean weight gain for participants was 6.8 lbs with a standard error of 1.2 lbs. A control group that did not binge had no change in weight. (a) What is the relevant parameter? (b) How could we find the actual exact value of the parameter? (c) Give a \(95 \%\) confidence interval for the parameter and interpret it. (d) Give the margin of error and interpret it.

Standard Deviation of NHL Penalty Minutes Exercise 3.102 describes data on the number of penalty minutes for Ottawa Senators NHL players. The sample has a fairly large standard deviation, \(s=27.3\) minutes. Use StatKey or other technology to create a bootstrap distribution, estimate the standard error, and give a \(95 \%\) confidence interval for the standard deviation of penalty minutes for NHL players. Assume that the data in OttawaSenators can be viewed as a reasonable sample of all NHL players.

SKILL BUILDER 1 In Exercises 3.41 to \(3.44,\) data from a sample is being used to estimate something about a population. In each case: (a) Give notation for the quantity that is being estimated. (b) Give notation for the quantity that gives the best estimate. A random sample of registered voters in the US is used to estimate the proportion of all US registered voters who voted in the last election.

In estimating the mean score on a fitness exam, we use an original sample of size \(n=30\) and a bootstrap distribution containing 5000 bootstrap samples to obtain a \(95 \%\) confidence interval of 67 to \(73 .\) In Exercises 3.106 to 3.111 , a change in this process is described. If all else stays the same, which of the following confidence intervals \((A, B,\) or \(C)\) is the most likely result after the change: \(\begin{array}{ll}A .66 \text { to } 74 & B .67 \text { to } 73\end{array}\) C. 67.5 to 72.5 Using 1000 bootstrap samples for the distribution.

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