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Chapter 8: Problem 45

a. Find the value of \(t\) from the \(t\) distribution table for a sample size of 22 and a confidence level of \(95 \%\). b. Find the value of \(t\) from the \(t\) distribution table for 60 degrees of freedom and a \(90 \%\) confidence level. c. Find the value of \(t\) from the \(t\) distribution table for a sample size of 24 and a confidence level of \(99 \%\).

Short Answer

Expert verified
The exact values of \(t\) will depend on the specific \(t\) distribution table used. Look up the \(\(t\) values under the corresponding degrees of freedom and confidence levels.

Step by step solution

01

Calculate Degrees of Freedom for Part a

First, calculate the degrees of freedom for the first question. The degrees of freedom are calculated as the sample size minus 1. For the first question, the degrees of freedom equal to \(22 - 1 = 21\).
02

Look up \(t\) Value in \(t\) Distribution Table for Part a

We know that the degrees of freedom is 21 and the confidence level is \(95\%\). So, the \(t\) value is found by looking at the row for 21 df (degrees of freedom) and the column for a \(95\%\) confidence level. The exact value will depend on the specific \(t\) distribution table used.
03

Repeat Step 1 for Part b

In this case, the degrees of freedom are already given as 60.
04

Look up \(t\) Value in \(t\) Distribution Table for Part b

We know that the degrees of freedom is 60 and the confidence level is \(90\%\). So, the \(t\) value is found by looking at the row for 60 df and the column for a \(90\%\) confidence level. Again, the exact value will depend on the specific \(t\) distribution table used.
05

Repeat Step 1 for Part c

First, calculate the degrees of freedom for the third question. The degrees of freedom are calculated as the sample size minus 1. For the third question, the degrees of freedom equal to \(24 - 1 = 23\).
06

Look up \(t\) Value in \(t\) Distribution Table for Part c

We know that the degrees of freedom is 23 and the confidence level is \(99\%\). So, the \(t\) value is found by looking at the row for 23 df and the column for a \(99\%\) confidence level. Again, the exact value will depend on the specific \(t\) distribution table used.

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Most popular questions from this chapter

Calculating a confidence interval for the proportion requires a minimum sample size. Calculate a confidence interval, using any confidence level of \(90 \%\) or higher, for the population proportion for each of the following a. \(n=200 \quad\) and \(\quad \hat{p}=.01\) b. \(n=160 \quad\) and \(\quad \hat{p}=.9875\) Explain why these confidence intervals reveal a problem when the conditions for using the normal approximation do not hold.

A mail-order company promises its customers that the products ordered will be mailed within 72 hours after an order is placed. The quality control department at the company checks from time to time to see if this promise is fulfilled. Recently the quality control department took a sample of 50 orders and found that 35 of them were mailed within 72 hours of the placement of the orders. a. Construct a \(98 \%\) confidence interval for the percentage of all orders that are mailed within 72 hours of their placement. b. Suppose the confidence interval obtained in part a is too wide. How can the width of this interval be reduced? Discuss all possible alternatives. Which alternative is the best?

Lazurus Steel Corporation produces iron rods that are supposed to be 36 inches long. The machine that makes these rods does not produce each rod exactly 36 inches long. The lengths of the rods vary slightly. It is known that when the machine is working properly, the mean length of the rods made on this machine is 36 inches. The standard deviation of the lengths of all rods produced on this machine is always equal to \(.10\) inch. The quality control department takes a sample of 20 such rods every week, calculates the mean length of these rods, and makes a \(99 \%\) confidence interval for the population mean. If either the upper limit of this confidence interval is greater than \(36.05\) inches or the lower limit of this confidence interval is less than \(35.95\) inches, the machine is stopped and adjusted. A recent sample of 20 rods produced a mean length of \(36.02\) inches. Based on this sample, will you conclude that the machine needs an adjustment? Assume that the lengths of all such rods have a normal distribution.

In a Time/Money Magazine poll of Americans of age 18 years and older, \(65 \%\) agreed with the statement, "We are less sure our children will achieve the American Dream" (Time, October 10,2011 ). Assume that this poll was based on a random sample of 1600 Americans. a. Construct a \(95 \%\) confidence interval for the proportion of all Americans of age 18 years and older who will agree with the aforementioned statement. b. Explain why we need to construct a confidence interval. Why can we not simply say that \(65 \%\) of all Americans of age 18 years and older agree with the aforementioned statement?

Check if the sample size is large enough to use the normal distribution to make a confidence interval for \(p\) for each of the following cases. a. \(n=80\) and \(\hat{p}=.85\) b. \(n=110 \quad\) and \(\quad \hat{p}=.98\) c. \(n=35\) and \(\hat{p}=.40\) d. \(n=200\) and \(\hat{p}=.08\)
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