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–
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-
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.
NOW IT’S YOUR TURN
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
This is less than one-
22.2 The data. The sample proportion is . The standard score for this outcome is
= −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.