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The eating habits of 12 bats were examined in the article "Foraging Behavior of the Indian False Vampire Bat" (Biotropica \([1991]: 63-67) .\) These bats consume insects and frogs. For these 12 bats, the mean time to consume a frog was \(\bar{x}=21.9 \mathrm{~min}\). Suppose that the standard deviation was \(s=7.7 \mathrm{~min} .\) Construct and interpret a \(90 \%\) confidence interval for the mean suppertime of a vampire bat whose meal consists of a frog. What assumptions must be reasonable for the one-sample \(t\) interval to be appropriate?

Short Answer

Expert verified
The 90% confidence interval for the mean time it takes for a vampire bat to consume a frog is approximately (18.0 minutes, 25.8 minutes). We are 90% confident that the true mean time lies within this range. The assumptions necessary for this interval to be valid include a random sample and an approximately normal distribution or a large enough sample size for the Central Limit Theorem to apply.

Step by step solution

01

Identify the given data

From the exercise, it's given that the sample mean, denoted as \(\bar{x}\), is 21.9 minutes, the standard deviation, denoted as \(s\), is 7.7 minutes, and the sample size, denoted as \(n\), is 12 bats. The confidence level is 90%.
02

Find the critical value

The critical value can be found using a t-distribution table. Given that the confidence level is 90% and the degrees of freedom is \(n-1 = 12 - 1 = 11\), the critical value (t*) for a two-tailed test is approximately 1.796.
03

Compute the Confidence Interval

The confidence interval is computed using the formula \(\bar{x} \pm t^* \cdot \frac{s}{\sqrt{n}}\). Substituting the given values, the confidence interval is \(21.9 \pm 1.796 \cdot \frac{7.7}{\sqrt{12}}\), which approximately equals \(21.9 \pm 3.91\), hence, the interval is (18.0, 25.8).
04

Interpret the result

The interpretation of the confidence interval is as follows: We are 90% confident that the mean time it takes for a vampire bat to consume a frog lies between 18.0 and 25.8 minutes.
05

Discuss Assumptions

The assumptions that need to be made for this one-sample t interval to be appropriate are: 1. The sample is random. 2. The distribution of the eating time is approximately normal or the sample size is large enough so that the Central Limit Theorem applies. Under these conditions, the calculation will provide a valid confidence interval.

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Most popular questions from this chapter

The interval from \(-2.33\) to \(1.75\) captures an area of \(.95\) under the \(z\) curve. This implies that another largesample \(95 \%\) confidence interval for \(\mu\) has lower limit \(\bar{x}-2.33 \frac{\sigma}{\sqrt{n}}\) and upper limit \(\bar{x}+1.75 \frac{\sigma}{\sqrt{n}} .\) Would you recommend using this \(95 \%\) interval over the \(95 \%\) interval \(\bar{x} \pm 1.96 \frac{\sigma}{\sqrt{n}}\) discussed in the text? Explain. (Hint: Look at the width of each interval.)

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