" Over several years and many thousands of students, 85% of the high school students in a large city have passed the competency test that is one of the requirements for a diploma. Now reformers claim that a new mathematics curriculum will increase the percentage who pass. A random sample of 1000 students follow the new curriculum. The school board wants to see an improvement that is statistically significant at the 5% level before it will adopt the new program for all students. If p is the proportion of all students who would pass the exam if they followed the new curriculum, we must test H0:p=0.85 Ha:p>0.85 A)Suppose that 868 of the 1000 students in the sample pass the test. Show that this is not significant at the 5% level. (Follow the method of Example 3, page 521, in Chapter 22.) B)Suppose that 869 of the 1000 students pass. Show that this is significant at the 5% level. C)Is there a practical difference between 868 successes in 1000 tries and 869 successes? What can you conclude about the importance of a fixed significance level?" (For part c, skip the question "What can you conclude ...?")

Avea to the left of 2- Score Table P Percentiles of the " ormal distributions Standard Perc tile Standard Percentile Standard Percentile score score score -3.4 0.03 -1.1 13.57 1.2 88.49 -3.3 0.05 -1.0 15.87 1.3 90.32 -3.2 0.07 -0.9 18.41 1.4 91.92 -3.1 0.10 -0.8 21.19 1.5 93.32 -3.0 0.13 -0.7 24.20 1.6 94.52 -2.9 0.19 -0.6 27.42 1.7 95.54 -2.8 0.26 -0.5 30.85 1.8 96.41 -2.7 0.35 -0.4 34.46 1.9 97.13 -2.6 0.47 -0.3 38.21 2.0 97.73 -2.5 0.62 -0.2 42.07 2.1 98.21 -2.4 0.82 -0.1 46.02 2.2 98.61 -2.3 1.07 0.0 50.00 2.3 98.93 -2.2 1.39 0.1 53.98 2.4 99.18 -2.1 1.79 0.2 57.93 2.5 99.38 -2.0 2.27 0.3 61.79 2.6 99.53 -1.9 2.87 0.4 65.54 2.7 99.65 -1.8 3.59 0.5 69.15 2.8 99.74 -1.7 4.46 0.6 72.58 2.9 99.81 -1.6 5.48 0.7 75.80 3.0 99.87 -1.5 6.68 0.8 78.81 3.1 99.90 -1.4 8.08 0.9 81.59 3.2 99.93 -1.3 9.68 1.0 84.13 3.3 99.95 -1.2 11.51 1.1 86.43 3.4 99.97Ho : p = 0.5 Ha : p > 0.5 Here, , 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 p follows a Normal distribution with mean 0.5 and standard deviation 0.0707. The data. A sample of 50 people found that 36 preferred fresh coffee. The sample proportion is p = 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 p = 0.72 isThe data. A sample of 50 people found that 36 preferred fresh coffee. The sample proportion is p = 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 p = 0.72 is observation - mean standard score = standard error 0.72 - 0.5 = = 3.1 0.0707 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