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Chapter 9: Problem 4

Find the \(p\) -value for the following large-sample \(z\) tests: a. A right-tailed test with observed \(z=1.15\) b. A two-tailed test with observed \(z=-2.78\) c. A left-tailed test with observed \(z=-1.81\)

Short Answer

Expert verified
Answer: a. Right-tailed test with observed z=1.15: p-value = 0.1251 b. Two-tailed test with observed z=-2.78: p-value = 0.0054 c. Left-tailed test with observed z=-1.81: p-value = 0.0351

Step by step solution

01

Determine the tail(s) of the test

We are given the tail(s) of the test for each scenario: a. Right-tailed b. Two-tailed c. Left-tailed
02

Find the p-value using z-table for right-tailed test

For scenario (a), we have a right-tailed test with observed z=1.15. We need to look up the area to the left of z=1.15 in the z-table. The area to the left of z=1.15 is 0.8749. Since it's a right-tailed test, the p-value is the area to the right of z=1.15. So, p-value for (a) = 1 - 0.8749 = 0.1251.
03

Find the p-value using z-table for two-tailed test

For scenario (b), we have a two-tailed test with observed z=-2.78. First, we need to look up the area to the left of z=-2.78 in the z-table. The area to the left of z=-2.78 is 0.0027. Since it's a two-tailed test, we need to find the area in both tails. The area in the left tail is 0.0027, and the area in the right tail is the same. So, p-value for (b) = 2 * 0.0027 = 0.0054.
04

Find the p-value using z-table for left-tailed test

For scenario (c), we have a left-tailed test with observed z=-1.81. We need to look up the area to the left of z=-1.81 in the z-table. The area to the left of z=-1.81 is 0.0351. Since it's a left-tailed test, the p-value is the area to the left of z=-1.81. So, p-value for (c) = 0.0351.
05

Final Answer

The p-values for the given scenarios are: a. Right-tailed test with observed z=1.15: p-value = 0.1251 b. Two-tailed test with observed z=-2.78: p-value = 0.0054 c. Left-tailed test with observed z=-1.81: p-value = 0.0351

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Most popular questions from this chapter

Critical Value Approach Fill in the blanks in the table below. $$\begin{array}{|l|l|l|l|l|l|}\hline \begin{array}{l}\text { Test } \\\\\text { Statistic }\end{array} & \begin{array}{l}\text { Significance } \\\\\text { Level }\end{array} &\begin{array}{l}\text { One or } \\\\\text { Two-Tailed Test? }\end{array} & \text { Critical Value } & \begin{array}{l}\text { Rejection } \\\\\text { Region }\end{array} & \text { Conclusion } \\\\\hline z=0.88 & \alpha=.05 & \text { Two-tailed } & & & \\\\\hline z=-2.67 & \alpha=.05 & \text { 0ne-tailed (lower) } & & & \\\\\hline z=5.05 & \alpha=.01 & \text { Two-tailed } & & & \\\\\hline z=-1.22 & \alpha=.01 & \text { One- tailed (lower) } & & & \\\\\hline\end{array}$$

A random sample of \(n=1400\) observations from a binomial population produced \(x=529\) a. If your research hypothesis is that \(p\) differs from .4 what hypotheses should you test? b. Calculate the test statistic and its \(p\) -value. Use the \(p\) -value to evaluate the statistical significance of the results at the \(1 \%\) level. c. Do the data provide sufficient evidence to indicate that \(p\) is different from \(.4 ?\)

Suppose you wish to detect a difference between \(\mu_{1}\) and \(\mu_{2}\) (either \(\mu_{1}>\mu_{2}\) or \(\mu_{1}<\mu_{2}\) ) and, instead of running a two-tailed test using \(\alpha=.05,\) you use the following test procedure. You wait until you have collected the sample data and have calculated \(\bar{x}_{1}\) and \(\bar{x}_{2}\) If \(\bar{x}_{1}\) is larger than \(\bar{x}_{2},\) you choose the alternative hypothesis \(H_{\mathrm{a}}: \mu_{1}>\mu_{2}\) and run a one-tailed test placing \(\alpha_{1}=.05\) in the upper tail of the \(z\) distribution. If, on the other hand, \(\bar{x}_{2}\) is larger than \(\bar{x}_{1},\) you reverse the procedure and run a one-tailed test, placing \(\alpha_{2}=.05\) in the lower tail of the \(z\) distribution. If you use this procedure and if \(\mu_{1}\) actually equals \(\mu_{2},\) what is the probability \(\alpha\) that you will conclude that \(\mu_{1}\) is not equal to \(\mu_{2}\) (i.e., what is the probability \(\alpha\) that you will incorrectly reject \(H_{0}\) when \(H_{0}\) is true)? This exercise demonstrates why statistical tests should be formulated prior to observing the data.

What's Normal? What is normal, when it comes to people's body temperatures? A random sample of 130 human body temperatures, provided by Allen Shoemaker \(^{3}\) in the Journal of Statistical Education, had a mean of 98.25 degrees and a standard deviation of 0.73 degrees. Does the data indicate that the average body temperature for healthy humans is different from 98.6 degrees, the usual average temperature cited by physicians and others? Test using both methods given in this section. a. Use the \(p\) -value approach with \(\alpha=.05\). b. Use the critical value approach with \(\alpha=.05 .\) c. Compare the conclusions from parts a and b. Are they the same? d. The 98.6 standard was derived by a German doctor in \(1868,\) who claimed to have recorded 1 million temperatures in the course of his research. \({ }^{4}\) What conclusions can you draw about his research in light of your conclusions in parts a and \(b\) ?

A Maze Experiment In a maze running study, a rat is run in a T maze and the result of each run recorded. A reward in the form of food is always placed at the right exit. If learning is taking place, the rat will choose the right exit more often than the left. If no learning is taking place, the rat should randomly choose either exit. Suppose that the rat is given \(n=100\) runs in the maze and that he chooses the right exit \(x=64\) times. Would you conclude that learning is taking place? Use the \(p\) -value approach, and make a decision based on this \(p\) -value.
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