Example 1. Test the hypothesis using the classical approach and the P-value approach. H 0 : p = 0.45 versus H 1 : p < n = 150 x = 62 = 0.

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1 Example 1 Test the hypothesis using the classical approach and the P-value approach. H 0 : p = 0.45 versus H 1 : p < 0.45 n = 150 x = 62 = 0.05 Perform the classical approach. Verify the requirements to perform the hypothesis test, that is, the sample must be a simple random sample, np 0 (1- p 0 ) > 10, and the sample size cannot be more than 5% of the population size (for independence). Is this true? 150(0.45) (1 0.45) > > 10 yes! Determine if this is a left-tailed, right-tailed, or two-tailed test. If a hypothesis test has an equal hypothesis versus a not equal hypothesis, then it s a two-tailed test. If it has an equal hypothesis versus a less than hypothesis, then it s a left-tailed test. If it has an equal hypothesis versus a greater than hypothesis, then it s a right-tailed test. This is an equal hypothesis versus a less than hypothesis, so we need to use a left-tailed test. Calculate = x/n = 62/150 = Or on the calculator!!! z 0 = ( ) / square root( (0.45(1 0.45))/150 ) = -.037/square root( ) = (rounded to two places) z = (this is 0.01 away from what you get by hand, but since you are just comparing this number with the critical value it won t make a difference!!)

2 Since = 0.05 we can find the critical value, z, by using the table in the book on page 395. Or determine the critical value using the calculator by finding invnorm(0.95) = **we use 0.95 instead of 0.05 because we always give the calculator the area to the LEFT and since we have 0.05 to the right, we have = 0.95 to the left. Compare the critical value with the negative of the test statistic, -z. z 0 = z = For a left-tailed test, if z 0 < - z we reject the null hypothesis (see summary in blue box below) > our critical value is GREATER than our test statistic, so we do not reject the null hypothesis. Summary of what we just did (look at left side for classical approach ): Use your calculator (invnorm) to find the critical value. Use your calculator (invnorm) to find the critical value.

3 Perform the P-value approach. Use your calculator to determine the P-value. STAT then TESTS, #5: 1-PropZTest (these values come from the above information scroll up to see where p 0, etc. came from) Our P-value is Compare the P-value with the level of significance,. If the P-value < we reject the null hypothesis. P-value > 0.05 level of significance, therefor there is not sufficient evidence at the 0.05 level of significance to reject the null hypothesis. So, do not reject!! Example 2 In a clinical trial, 26 out of 650 patients taking a prescription drug complained of flulike symptoms. Suppose that it is know that 2.4% of patients taking competing drugs complain of flulike symptoms. Is there sufficient evidence to conclude that more than 2.4% of this drug s users experience flulike symptoms as a side effect at the = 0.1 level of significance? Verify the requirements to perform the hypothesis test, that is, the sample must be a simple random sample, np0(1- p0) > 10, and the sample size cannot be more than 5% of the population size (for independence). 650 (0.024) ( ) > > 10 yes! What are the null and alternative hypotheses? The question is whether or not there is MORE than 2.4% H 0 : p = H 1 : p > Next, find the P-value: The P-value = (rounded to three places)

4 Since the P-value <, REJECT the null hypothesis and conclude that there is sufficient evidence that more than 2.4% of users experience flulike symptoms. Example 3 Previously, 29% of employed adults were satisfied with their chances for promotion. A manager wants to determine if this percentage has changed significantly since then. She randomly selects 200 employed adults and finds that 78 of them are completely satisfied with their chances for promotion. Is there sufficient evidence to conclude that the percentage of employed adults satisfied with their chances for promotion is significantly different from the previous percentage at the = 0.05 level of significance? Verify the requirements to perform the hypothesis test, that is, the sample must be a simple random sample, np0(1- p0) > 10, and the sample size cannot be more than 5% of the population size (for independence). (200)(0.29) (1-0.29) > > 10 yes! H 0 : p = 029 H 1 : p 0.29 (we use not equal to because it just said significantly different not greater than or less than) Find the p-value P-value = (rounded to three places) Since the P-value <, we REJECT the null hypothesis. So, YES there is sufficient evidence that the percentage of adults satisfied with their chances of promotion is significantly different than 2.4%. Example 4 Previously, 5% of mothers smoked more than 21 cigarettes during their pregnancy. An obstetrician believes that the percentage of mothers who smoke 21 cigarettes or more is less than 5% today. She randomly selects 140 pregnant mothers and finds that 2 of them smoked 21 or more cigarettes during pregnancy. Test the researcher s statement at the = 0.05 level of significance. Verify the requirements to perform the hypothesis test, that is, the sample must be a simple random sample, np0(1- p0) > 10, and the sample size cannot be more than 5% of the population size (for independence). Is this true? 140(0.05) (1 0.05) > > 10 is FALSE!!

5 She believes (wants to test) whether or not the percetage is now LESS than 5%. So the null and alternative hypotheses are: H0: p = 0.05 and H1 = p < 0.05 Since the P-value is the probability of obtaining x or fewer successes, the P-value is P(X<2). (Because 2 people in the sample said they smoked 21 or more cigarettes). So we complete P(x<2) using our Ti-84 and round to three decimal places. We find the cumulative binomial probability for x < 2 success in 140 trials if the probability of success in each trial is binomcdf(140,0.05, 2) = round to three places to get If the probability of getting a sample proportion as extreme as or more extreme than the one obtained is small under the assumption the statement in the null hypothesis is true, reject the null hypothesis. Otherwise, do not reject the null hypothesis. The P-value is less than alpha (0.05) < 0.05 Use this comparison to determine the correct conclusion regarding the researcher's statement Because the P-value is less than alpha there is sufficient evidence to conclude that the percentage of mothers who smoke 21 or more cigarettes during pregnancy is less than 5%, meaning we reject the null hypothesis.