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Chapter 8: Problem 21

In an experiment to assess the strength of the hunger drive in rats, 30 previously trained animals were deprived of food for 24 hours. At the end of the 24 -hour period, each animal was put into a cage where food was dispensed if the animal pressed a lever. The length of time the animal continued pressing the bar (although receiving no food) was recorded for each animal. If the data yielded a sample mean of 19.3 minutes with a standard deviation of 5.2 minutes, estimate the true mean time and calculate the margin of error.

Short Answer

Expert verified
Answer: The estimated true mean time is between 17.36 minutes and 21.24 minutes.

Step by step solution

01

1. Calculate the estimated standard error

We can calculate the estimated standard error by dividing the standard deviation by the square root of the sample size: Estimated Standard Error = \(\frac{Standard Deviation}{\sqrt{Sample Size}}\) Estimated Standard Error = \(\frac{5.2}{\sqrt{30}}\) = \(0.95\) minutes.
02

2. Determine the degrees of freedom

The degrees of freedom in this case would be the sample size minus 1: Degrees of Freedom = \(Sample Size - 1\) Degrees of Freedom = \(30 - 1 = 29\).
03

3. Find the critical T-score

We will use a 95% confidence interval, which means we are looking for the T-score that corresponds to the center 95% of the T-distribution. The T-score depends on the degrees of freedom, in this case, 29. Using a T-distribution table or calculator, we find the critical T-score: T-score = \(2.045\).
04

4. Calculate the margin of error

Now, we will calculate the margin of error by multiplying the critical T-score by the estimated standard error: Margin of Error = \(Critical T-score 脳 Estimated Standard Error\) Margin of Error = \(2.045 脳 0.95\) = \(1.94\) minutes.
05

5. Estimate the true mean time

Finally, we will use the sample mean and the margin of error to estimate the true mean time: Lower Bound = Sample Mean - Margin of Error = \(19.3 - 1.94\) = \(17.36\) minutes. Upper Bound = Sample Mean + Margin of Error = \(19.3 + 1.94\) = \(21.24\) minutes. We can estimate that the true mean time that the rats will continue pressing the lever is between \(17.36\) minutes and \(21.24\) minutes with a 95% confidence interval.

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Most popular questions from this chapter

Refer to Exercise \(8.18 .\) The means and standard deviations for 50 billing statements from each of the computer databases of each of the three hotel chains are given in the table: $$\begin{array}{lccc} & \text { Marriott } & \text { Radisson } & \text { Wyndham } \\\\\hline \text { Sample Average } & \$ 170 & \$ 145 & \$ 160 \\\\\text { Sample Standard Deviation } & 17.5 & 10 & 16.5\end{array}$$ a. Find a \(95 \%\) confidence interval for the difference in the average room rates for the Marriott and the Wyndham hotel chains. b. Find a \(99 \%\) confidence interval for the difference in the average room rates for the Radisson and the Wyndham hotel chains. c. Do the intervals in parts a and b contain the value \(\left(\mu_{1}-\mu_{2}\right)=0 ?\) Why is this of interest to the researcher? d. Do the data indicate a difference in the average room rates between the Marriott and the Wyndham chains? Between the Radisson and the Wyndham chains?

You want to estimate the difference in grade point averages between two groups of college students accurate to within .2 grade point, with probability approximately equal to \(.95 .\) If the standard deviation of the grade point measurements is approximately equal to .6, how many students must be included in each group? (Assume that the groups will be of equal size.)

A sample survey is designed to estimate the proportion of sports utility vehicles being driven in the state of California. A random sample of 500 registrations are selected from a Department of Motor Vehicles database, and 68 are classified as sports utility vehicles. a. Use a \(95 \%\) confidence interval to estimate the proportion of sports utility vehicles in California b. How can you estimate the proportion of sports utility vehicles in California with a higher degree of accuracy? (HINT: There are two answers.)

Exercise 8.106 presented statistics from a study of fast starts by ice hockey skaters. The mean and standard deviation of the 69 individual average acceleration measurements over the 6 -meter distance were 2.962 and .529 meters per second, respectively. a. Find a \(95 \%\) confidence interval for this population mean. Interpret the interval. b. Suppose you were dissatisfied with the width of this confidence interval and wanted to cut the interval in half by increasing the sample size. How many skaters (total) would have to be included in the study?

A small amount of the trace element selenium, \(50-200\) micrograms \((\mu \mathrm{g})\) per day, is considered essential to good health. Suppose that random samples of \(n_{1}=n_{2}=30\) adults were selected from two regions of the United States and that a day's intake of selenium, from both liquids and solids, was recorded for each person. The mean and standard deviation of the selenium daily intakes for the 30 adults from region 1 were \(\bar{x}_{1}=167.1\) and \(s_{1}=24.3 \mu \mathrm{g}\) respectively. The corresponding statistics for the 30 adults from region 2 were \(\bar{x}_{2}=140.9\) and \(s_{2}=\) 17.6. Find a \(95 \%\) confidence interval for the difference in the mean selenium intakes for the two regions. Interpret this interval.
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