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Chapter 8: Problem 43

For each of the following, find the area in the appropriate tail of the \(t\) distribution. a. \(t=-1.302\) and \(d f=42\) b. \(t=2.797\) and \(n=25\) c. \(t=1.397\) and \(n=9\) d. \(t=-2.383\) and \(d f=67\)

Short Answer

Expert verified
The areas under the t-distribution for each of the case is calculated using the Cumulative Distribution Function. The short answer differs for each of the cases.

Step by step solution

01

Step 1a: Calculating t-score and area for t=-1.302, df=42

Here, the t-value is given as -1.302 and degree of freedom as 42. Given that t < 0, we are interested in the left tail of the distribution. The area in the left tail would be computed as \(CDF = stats.t.cdf(-1.302, df=42)\).
02

Step 2a: Interpreting result for a

The calculated CDF value corresponds to the area under the t-distribution curve to the left of the t-score, which will provide the answer in percentage form.
03

Step 1b: Calculating t-score and area for t=2.797, n=25

\(t=2.797, n=25\) means t-value is 2.797 and sample size is 25, consequently having degrees of freedom as 24. Given that t > 0, we are interested in the right tail of the distribution. The area would be computed as \(1 - stats.t.cdf(2.797, df=24)\).
04

Step 2b: Interpreting result for b

The value calculated in this step is the area under the t-distribution curve to the right of the t-score.
05

Step 1c: Calculating t-score and area for t=1.397, n=9

\(t=1.397, n=9\) means t-value is 1.397 and sample size is 9, and so the degrees of freedom becomes 8. Since t > 0, we are interested in the right tail of the distribution. The area is computed as \(1 - stats.t.cdf(1.397, df=8)\).
06

Step 2c: Interpreting result for c

The result of above step is the area under the t-distribution curve to the right of the t-score.
07

Step 1d: Calculating t-score and area for t=-2.383, df=67

Here, the t-value is given as -2.383 and degree of freedom as 67. Since t < 0, we are interested in the left tail of the distribution. The area would be computed as \(CDF = stats.t.cdf(-2.383, df=67)\).
08

Step 2d: Interpreting result for d

The calculated CDF value will be the area under the t-distribution curve to the left of the t-score.

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Most popular questions from this chapter

A random sample of 20 acres gave a mean yield of wheat equal to \(41.2\) bushels per acre with a standard deviation of 3 bushels. Assuming that the yield of wheat per acre is normally distributed, construct a \(90 \%\) confidence interval for the population mean \(\mu .\)

It is said that happy and healthy workers are efficient and productive. A company that manufactures exercising machines wanted to know the percentage of large companies that provide on-site health club facilities. A sample of 240 such companies showed that 96 of them provide such facilities on site. a. What is the point estimate of the percentage of all such companies that provide such facilities on site? b. Construct a \(97 \%\) confidence interval for the percentage of all such companies that provide such facilities on site. What is the margin of error for this estimate?

For each of the following, find the area in the appropriate tail of the \(t\) distribution. a. \(t=2.467\) and \(d f=28\) b. \(t=-1.672\) and \(d f=58\) c. \(t=-2.670\) and \(n=55\) d. \(t=2.383\) and \(n=23\)

A company that produces detergents wants to estimate the mean amount of detergent in 64 -ounce jugs at a \(99 \%\) confidence level. The company knows that the standard deviation of the amounts of detergent in all such jugs is \(.20\) ounce. How large a sample should the company select so that the estimate is within \(.04\) ounce of the population mean?

You are interested in estimating the mean commuting time from home to school for all commuter students at your school. Briefly explain the procedure you will follow to conduct this study. Collect the required data from a sample of 30 or more such students and then estimate the population mean at a \(99 \%\) confidence level. Assume that the population standard deviation for such times is \(5.5\) minutes.
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