Statistics in the Media A headline read, “More Than Half of Am...

Question

Statistics in the Media A headline read, “More Than Half of Americans Say Federal Taxes Too High.” The headline was based on a random sample of 1026 adult Americans in which 534 stated the amount of federal tax they have to pay is too high. Is this an accurate headline?

Accepted Answer

Hello. In this video, we are told that a local haunted house claims that at least 60% of its visitors report being scared during the tour. A skeptical blogger suspects the true proportion is less than 60% and collects a random sample of 50 visitors, of whom 25 said they were scared. Use a randomization procedure to test the blogger's claim at the 0.05 significance level. So, let's go ahead and start this problem by writing out our hypothesis. Now, the haunted house claims that at least 60% of its visitors report being scared during the tour. So, the proportion of visitors that are being scared is reportedly greater than or equal to 60%, and the alternate hypothesis, which is the blogger's suspension, is that this is less than 60%. So, if we represent the 60% as a decimal, we are testing the hypothesis that the proportion of people who are scared is greater than 0.60 or is less than 0.60. And in this case here, we're going to use a left-tailed test. So, let's go ahead and start by writing out some given information. First, we are told that we are given a random sample of 50 visitors. And we are also told that out of these 50 visitors, 25 said that they were scared. Now, we want to take the point estimate of the people who are scared and visited the attraction. That point estimate is going to be 25 divided by 50, which is equal to 0.5. No, Our, our original claim is that 60% of people, or 0.60. They are scared people. And using this, we are going to find the standard error and a Z score to test their hypothesis. So let's go ahead and start by finding our standard error. Now, our standard error is going to be defined as the original proportion, multiplied by its complement, divided by the sample size of people who visited the house. That is going to be the square root of 0.60, multiplied by 1 minus 0.60, divided by 50, and this is going to give us a standard error of 0.0693. Now we will go ahead and find our test C score. Now, this is going to be defined as our point estimate minus the minus the original proportion divided by the standard error. This Z test statistic is going to be 0.50 minus 0.60 divided by 0.0693. And with the help of a calculator, we get the Z statistic that this is approximately equal to -1.44. Now, finally, we want to find a P value. For the test, and the P value is the probability that Z is less than -1.44. With the help of the appendix, we get that this probability is 0.075 and recall that we are working with the significance level of 0.05. Comparing the p value to the significance level, notice that the p value of 0.075 is greater than 0.05. With that being said, we fail to reject our no hypothesis, and this means that there is sufficient evidence, or at least there is evidence to suggest that not that less than 60% of visitors are scared while visiting the haunted house. And with that being said, the solution to this problem is going to be B. So I hope this video helps you in understanding how to approach this problem, and we will go ahead and see you all in the next video.

Statistics in the Media A headline read, “More Than Half of Americans Say Federal Taxes Too High.” The headline was based on a random sample of 1026 adult Americans in which 534 stated the amount of federal tax they have to pay is too high. Is this an accurate headline?

Key Concepts

Sampling and Random Samples

Sample Proportion and Estimation

Margin of Error and Confidence Intervals

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