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Briefly explain the procedure used to calculate the \(p\) -value for a two- tailed and for a one-tailed test. respectively.

Short Answer

Expert verified
The p-value is calculated by obtaining a test statistic, using a statistical table or software. For a two-tailed test, this value checks for extreme values on both ends and hence, is double the p-value of a one-tailed test, which checks for extreme values only in the direction of the alternative hypothesis.

Step by step solution

01

Definition of P-value

The p-value is a statistical measure that helps scientists determine whether their hypotheses are right or wrong. It is a probability that measures the evidence against the null hypothesis. Lower p-values suggest that the null hypothesis is less likely to be true.
02

Calculation of a P-value

To calculate a p-value, you first need to calculate a test statistic (such as t, z, or chi-square) using your sample data. Then, the p-value is the probability of obtaining a test statistic as extreme or more extreme than the observed data under the null hypothesis. This can be done using a statistical table or software.
03

Two-tailed Test P-value

In a two-tailed test, you are testing for the possibility of the relationship in both directions. When you calculate the p-value for a two-tailed test, the calculated probability is for the occurrence of the extreme values on both ends (tails) of the distribution curve. This means that the p-value will be twice that of a one-tailed test as it tests for the effect in both directions.
04

One-tailed Test P-value

In a one-tailed test, you're testing the statistical significance in one direction of interest. When calculating the p-value for such a test, you consider only one tail of the probability distribution. Hence, the p-value for a one-tailed test is the probability of getting a value as extreme or more extreme in the direction of your alternative hypothesis.

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Most popular questions from this chapter

According to the Magazine Publishers of America (www.magazine.org), the average visit at the magazines' Web sites during the fourth quarter of 2007 lasted \(4.145\) minutes. Forty-six randomly selected visits to magazine's Web sites during the fourth quarter of 2009 produced a sample mean visit of \(4.458\) minutes, with a standard deviation of \(1.14\) minutes. Using the \(10 \%\) significance level and the critical value approach, can you conclude that the length of an average visit to these Web sites during the fourth quarter of 2009 was longer than \(4.145\) minutes? Find the range for the \(p\) -value for this test. What will your conclusion be using this \(p\) -value range and \(\alpha=.10\) ?

A study claims that all homeowners in a town spend an average of 8 hours or more on house cleaning and gardening during a weekend. A researcher wanted to check if this claim is true. A random sample of 20 homeowners taken by this researcher showed that they spend an average of \(7.68\) hours on such chores during a weekend. The population of such times for all homeowners in this town is normally distributed with the population standard deviation of \(2.1\) hours. a. Using the \(1 \%\) significance level, can you conclude that the claim that all homeowners spend an average of 8 hours or more on such chores during a weekend is false? Use both approaches. b. Make the test of part a using a \(2.5 \%\) significance level. Is your decision different from the one in part a? Comment on the results of parts a and \(b\).

In each of the following cases, do you think the sample size is large enough to use the normal distribution to make a test of hypothesis about the population proportion? Explain why or why not. a. \(n=30\) and \(p=.65\) b. \(n=70\) and \(p=.05\) c. \(n=60\) and \(p=.06\) d. \(n=900\) and \(p=.17\)

The manufacturer of a certain brand of auto batteries claims that the mean life of these batteries is 45 months. A consumer protection agency that wants to check this claim took a random sample of 24 such batteries and found that the mean life for this sample is \(43.05\) months. The lives of all such batteries have a normal distribution with the population standard deviation of \(4.5\) months. a. Find the \(p\) -value for the test of hypothesis with the alternative hypothesis that the mean life of these batteries is less than 45 months. Will you reject the null hypothesis at \(\alpha=.025\) ? b. Test the hypothesis of part a using the critical-value approach and \(\alpha=.025\).

Brooklyn Corporation manufactures DVDs. The machine that is used to make these DVDs is known. to produce not more than \(5 \%\) defective DVDs. The quality control inspector selects a sample of 200 DVDs each week and inspects them for being good or defective. Using the sample proportion, the quality control inspector tests the null hypothesis \(p \leq .05\) against the alternative hypothesis \(p>.05\), where \(p\) is the proportion of DVDs that are defective. She always uses a \(2.5 \%\) significance level. If the null hypothesis is rejected, the production process is stopped to make any necessary adjustments. A recent sample of 200 DVDs contained 17 defective DVDs. a. Using the \(2.5 \%\) significance level, would you conclude that the production process should be stopped to make necessary adjustments? b. Perform the test of part a using a \(1 \%\) significance level. Is your decision different from the one in part a?

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