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Calculating P-values*

Finding the P-values we gave in Examples 1 and 2 requires doing Normal distribution calculations using Table B of Normal percentiles. That was optional reading in Chapter 13 (pages 304–306). In practice, software does the calculation for us, but here is an example that shows how to use Table B.

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EXAMPLE 3 Tasting coffee

The hypotheses. In Example 1, we want to test the hypotheses

H0: p = 0.5

Ha: p > 0.5

Here, p is the proportion of the population of all coffee drinkers who prefer fresh coffee to instant coffee.

The sampling distribution. If the null hypothesis is true, so that p = 0.5, we saw in Example 1 that ˆp follows a Normal distribution with mean 0.5 and standard deviation, or standard error, 0.0707.

The data. A sample of 50 people found that 36 preferred fresh coffee. The sample proportion is ˆp=0.72.

The P-value. The alternative hypothesis is one-sided on the high side. So, the P-value is the probability of getting an outcome at least as large as 0.72. Figure 22.1 displays this probability as an area under the Normal sampling distribution curve. To find any Normal curve probability, move to the standard scale. When we convert a sample statistic to a standard score when conducting a statistical test of significance, the standard score is commonly referred to as a test statistic. The test statistic for the outcome ˆp=0.72 is

standard score=observation

Table B says that standard score 3.1 is the 99.9 percentile of a Normal distribution. That is, the area under a Normal curve to the left of 3.1 (in the standard scale) is 0.999. The area to the right is therefore 0.001, and that is our P-value.

The conclusion. The small P-value means that these data provide very strong evidence that a majority of the population prefers fresh coffee.

Test statistic

When conducting a statistical test of significance, the standard score that is computed based on the sample data is commonly referred to as a test statistic.

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NOW IT’S YOUR TURN

Question 22.2

22.2 Coin tossing. Refer to Exercise 22.1 (page 528). We tossed a coin only 50 times and got 21 heads, so the proportion of heads is

p^=2150=0.42

This is less than one-half. Is this evidence that our coin is not balanced? For the hypotheses you formulated in Exercise 22.1, find the P-value based on the results of our 50 tosses. Are the results significant at the 0.05 level?

22.2 The data. The sample proportion is p^=0.42. The standard score for this outcome is

standardscore=observationmeanstandard deviation

=0.420.50.0707

= −1.13

The P-value. To use Table B, round the standard score to −1.1. This is the 13.57 percentile of a Normal distribution. So the area to the left of −1.1 is 0.1357. The area to the left of −1.1 and to the right of 1.1 is double this, or 0.2714. This is our approximate P-value.

Conclusion. The large P-value gives no reason to think that the true proportion of heads differs from 0.5.

*This section is optional.

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