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Chapter 6: Problem 96

Data 4.1 on page 258 introduces a study in which mice that had a dim light on at night (rather than complete darkness) ate most of their calories when they should have been resting. These mice gained a significant amount of weight, despite eating the same number of calories as mice kept in total darkness. The time of eating seemed to have a significant effect. We look here at the effect after 8 weeks. There were 10 mice in the group with dim light at night and they gained an average of \(7.9 \mathrm{~g}\) with a standard deviation of \(3.0 .\) We see in Figure 6.9 that the data are not heavily skewed and do not have extreme outliers. Use the t-distribution to find and interpret a \(90 \%\) confidence interval for weight gain. As always, define the parameter being estimated.

Short Answer

Expert verified
The 90% confidence interval for the average weight gain for mice exposed to dim light at night is approximately between 5.75g to 10.05g.

Step by step solution

01

Identify the Parameter

The parameter to be estimated is the population mean weight gain \(\mu\) for mice who were exposed to dim light at night.
02

Compute the Sample Mean and Standard Deviation

From the exercise, it was given that the sample mean (x̄) is 7.9g and the sample standard deviation (s) is 3g. The number of mice (n) in the sample is 10.
03

Determine the Correct T Value

For a 90% confidence level and degree of freedom \(df = n-1 = 10-1 = 9\), the t-value (t*) can be found from the t-table, which is approximately 1.833.
04

Calculate the Confidence Interval

Using the formula \(CI = \bar{x} \pm t^* \cdot \frac{s}{\sqrt{n}}\), where \(CI\) is Confidence Interval, \(\bar{x}\) is the sample mean, \(t^*\) is the t-value, \(s\) is the standard deviation and \(n\) is the sample size. Substituting the given values, we find the confidence interval to be \(7.9 \pm 1.833 \cdot \frac{3}{\sqrt{10}}\).
05

Interpret the Confidence Interval

The resulting values become the lower and upper boundaries of the confidence interval. This interval is where we expect the true population mean weight gain to fall, with 90% confidence.

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Most popular questions from this chapter

Quiz Timing A young statistics professor decided to give a quiz in class every week. He was not sure if the quiz should occur at the beginning of class when the students are fresh or at the end of class when they've gotten warmed up with some statistical thinking. Since he was teaching two sections of the same course that performed equally well on past quizzes, he decided to do an experiment. He randomly chose the first class to take the quiz during the second half of the class period (Late) and the other class took the same quiz at the beginning of their hour (Early). He put all of the grades into a data table and ran an analysis to give the results shown below. Use the information from the computer output to give the details of a test to see whether the mean grade depends on the timing of the quiz. (You should not do any computations. State the hypotheses based on the output, read the p-value off the output, and state the conclusion in context.) Two-Sample T-Test and Cl \begin{tabular}{lrrrr} Sample & \(\mathrm{N}\) & Mean & StDev & SE Mean \\ Late & 32 & 22.56 & 5.13 & 0.91 \\ Early & 30 & 19.73 & 6.61 & 1.2 \\ \multicolumn{3}{c} { Difference } & \(=\mathrm{mu}(\) Late \()\) & \(-\mathrm{mu}\) (Early) \end{tabular} Estimate for difference: 2.83 $$ \begin{aligned} &95 \% \mathrm{Cl} \text { for difference: }(-0.20,5.86)\\\ &\text { T-Test of difference }=0(\text { vs } \operatorname{not}=): \text { T-Value }=1.87\\\ &\text { P-Value }=0.066 \quad \mathrm{DF}=54 \end{aligned} $$

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