Презентация Hypothesis testing for proportions. Essential statistics онлайн

Содержание слайда: Essential Statistics

Содержание слайда: In this section, you will learn how to test a population proportion, p. If np ≥ 10 and n(1-p) ≥ 10 for a binomial distribution, then the sampling distribution for is normal with and In this section, you will learn how to test a population proportion, p. If np ≥ 10 and n(1-p) ≥ 10 for a binomial distribution, then the sampling distribution for is normal with and

Содержание слайда: Assumptions Assumptions Write hypotheses & define parameter Calculate the test statistic & p-value Write a statement in the context of the problem.

Содержание слайда: The P-Value is the probability of obtaining a test statistic that is at least as extreme as the one that was actually observed, assuming the null is true. The P-Value is the probability of obtaining a test statistic that is at least as extreme as the one that was actually observed, assuming the null is true. p-value < (>) , I reject (fail to reject) the H0.

Содержание слайда: Under Stat – Tests Under Stat – Tests Select 1 Prop Z-test Input p, x, and n P is claim proportion X is number of sampling matching claim N is number sampled Select correct Alternate Hypothesis Calculate

Содержание слайда: Provides you with the z score Provides you with the z score P-Value Sample proportion Interpret the p-value based off of your Confidence interval

Содержание слайда: 1) right-tail test z = 1.6 1) right-tail test z = 1.6 2) two-tail test z = 2.3

Содержание слайда: Represents the remaining percentage of our confidence interval. 95% confidence interval has a 5% alpha. Represents the remaining percentage of our confidence interval. 95% confidence interval has a 5% alpha.

Содержание слайда: A medical researcher claims that less than 20% of American adults are allergic to a medication. In a random sample of 100 adults, 15% say they have such an allergy. Test the researcher’s claim at  = 0.01. A medical researcher claims that less than 20% of American adults are allergic to a medication. In a random sample of 100 adults, 15% say they have such an allergy. Test the researcher’s claim at  = 0.01.

Содержание слайда: The products np = 100(0.20)= 20 and nq = 100(0.80) = 80 are both greater than 10. So, you can use the z-test. The claim is “less than 20% are allergic to a medication.” So the null and alternative hypothesis are: The products np = 100(0.20)= 20 and nq = 100(0.80) = 80 are both greater than 10. So, you can use the z-test. The claim is “less than 20% are allergic to a medication.” So the null and alternative hypothesis are: Ho: p = 0.2 and Ha: p < 0.2 (Claim)

Содержание слайда: Because the test is a left-tailed test and the level of significance is  = 0.01, the critical value is zo = -2.33 and the rejection region is z < -2.33. Using the z-test, the standardized test statistic is: Because the test is a left-tailed test and the level of significance is  = 0.01, the critical value is zo = -2.33 and the rejection region is z < -2.33. Using the z-test, the standardized test statistic is:

Содержание слайда: The graph shows the location of the rejection region and the standardized test statistic, z. Because z is not in the rejection region, you should decide not to reject the null hypothesis. In other words, there is not enough evidence to support the claim that less than 20% of Americans are allergic to the medication. The graph shows the location of the rejection region and the standardized test statistic, z. Because z is not in the rejection region, you should decide not to reject the null hypothesis. In other words, there is not enough evidence to support the claim that less than 20% of Americans are allergic to the medication.

Содержание слайда:

Содержание слайда: Since the .1056 > , I fail to reject the H0 There is not sufficient evidence to suggest that 20% of adults are allergic to medication. Since the .1056 > , I fail to reject the H0 There is not sufficient evidence to suggest that 20% of adults are allergic to medication.

Содержание слайда: Harper’s Index claims that 23% of Americans are in favor of outlawing cigarettes. You decide to test this claim and ask a random sample of 200 Americas whether they are in favor outlawing cigarettes. Of the 200 Americans, 27% are in favor. At  = 0.05, is there enough evidence to reject the claim? Harper’s Index claims that 23% of Americans are in favor of outlawing cigarettes. You decide to test this claim and ask a random sample of 200 Americas whether they are in favor outlawing cigarettes. Of the 200 Americans, 27% are in favor. At  = 0.05, is there enough evidence to reject the claim?

Содержание слайда: The products np = 200(0.23) = 45 and nq = 200(0.77) = 154 are both greater than 5. So you can use a z-test. The claim is “23% of Americans are in favor of outlawing cigarettes.” So, the null and alternative hypotheses are: The products np = 200(0.23) = 45 and nq = 200(0.77) = 154 are both greater than 5. So you can use a z-test. The claim is “23% of Americans are in favor of outlawing cigarettes.” So, the null and alternative hypotheses are: Ho: p = 0.23 (Claim) and Ha: p  0.23

Содержание слайда: Because the test is a two-tailed test, and the level of significance is  = 0.05. Because the test is a two-tailed test, and the level of significance is  = 0.05. Z = 1.344 P = .179 Since the .179 > , I fail to reject the H0 There is not sufficient evidence to suggest that more or less than 23% of Americans are in favor of outlawing cigarette’s.

Содержание слайда: The graph shows the location of the rejection regions and the standardized test statistic, z. The graph shows the location of the rejection regions and the standardized test statistic, z. Because z is not in the rejection region, you should fail to reject the null hypothesis. At the 5% level of significance, there is not enough evidence to reject the claim that 23% of Americans are in favor of outlawing cigarettes.

Содержание слайда: The Pew Research Center claims that more than 55% of American adults regularly watch a network news broadcast. You decide to test this claim and ask a random sample of 425 Americans whether they regularly watch a network news broadcast. Of the 425 Americans, 255 responded yes. At  = 0.05, is there enough evidence to support the claim? The Pew Research Center claims that more than 55% of American adults regularly watch a network news broadcast. You decide to test this claim and ask a random sample of 425 Americans whether they regularly watch a network news broadcast. Of the 425 Americans, 255 responded yes. At  = 0.05, is there enough evidence to support the claim?

Содержание слайда: The products np = 425(0.55) = 235 and nq = 425(0.45) = 191 are both greater than 5. So you can use a z-test. The claim is “more than 55% of Americans watch a network news broadcast.” So, the null and alternative hypotheses are: The products np = 425(0.55) = 235 and nq = 425(0.45) = 191 are both greater than 5. So you can use a z-test. The claim is “more than 55% of Americans watch a network news broadcast.” So, the null and alternative hypotheses are: Ho: p = 0.55 and Ha: p > 0.55 (Claim)

Содержание слайда: Because the test is a right-tailed test, and the level of significance is  = 0.05. Because the test is a right-tailed test, and the level of significance is  = 0.05. Z = 2.072 P-value = .019 Since the 0.019 < , I reject the H0. There is sufficient evidence to suggest that 20% of adults are allergic to medication.

Содержание слайда: The graph shows the location of the rejection region and the standardized test statistic, z. Because z is in the rejection region, you should decide to There is enough evidence at the 5% level of significance, to support the claim that 55% of American adults regularly watch a network news broadcast. The graph shows the location of the rejection region and the standardized test statistic, z. Because z is in the rejection region, you should decide to There is enough evidence at the 5% level of significance, to support the claim that 55% of American adults regularly watch a network news broadcast.