In Exercises 13–18, test the claim about the population mean μ...

Question

In Exercises 13–18, test the claim about the population mean μ at the level of significance α. Assume the population is normally distributed.

Claim: μ≠52,200; α=0.05. Sample statistics: x_bar=53,220, s=2700, n=34

Accepted Answer

Hello everyone. Let's take a look at this question together. A car manufacturer's competitor claims that the mean fuel efficiency of their new model is different from 35 MPG or MPG. A sample of 40 cars is tested. The sample has a mean fuel efficiency of 36.7 MPG and a standard deviation of 2.9 MPG. Assumed fuel efficiency is normally distributed. Test the competitor's claim at the 0.05 significance level. So, in order to solve this question, we have to test the competitor's claim at the 0.05 significance level that the mean fuel efficiency of their new model is different from 35 MPG. And we also know we have a sample of 40 cars with a mean fuel efficiency of 36.7 MPG, and a standard deviation of 2.9 MPG. And we also have to assume fuel efficiency is normally distributed. And based on that information, since the population standard deviation is unknown, the population is normally distributed, and our sample size N is greater than 30, we can use the T test. And the first step in solving this problem is stating the null and alternative hypotheses, and from the provided information, the competitor claims that the mean fuel efficiency of the manufacturer's new model is different from 35 MPG, which this statement can be expressed as me is not equal to 35. And since this is a statement of strict inequality, it is the alternative hypothesis. So our alternative hypothesis is mute is not equal to 35, and the complement of the alternative hypothesis is the null hypothesis. So the null hypothesis is mute is equal to 35. Therefore, the null and alternative hypotheses are the null hypothesis is mu is equal to 35, and the alternative hypothesis is mu is not equal to 35, which is the claim. And the given level of significance for this test, which is alpha, is alpha equals 0.05, and we have to identify the degrees of freedom, which we know the degrees of freedom are calculated as N minus 1. So, 40 minus 1 is equal to 39, so our degrees of freedom is 39, and now we can determine the critical values, which we know since our alternative hypothesis contains the not equal to symbol, the hypothesis test is a two-tailed test. And so we use the table for T distribution and look for the intersection of our degrees of freedom, which is 39, and an alpha of 0.05, as this is a two-tailed test, and from the table, we can identify the critical value is equal to 2.023. However, we need to consider both the negative critical value and the positive critical value, since this is a two-tailed test. Therefore, our critical values include the negative critical value of -2.023 and the positive critical value of 2.023. And then we have to determine the rejection regions, which since the test. is a two-tailed test, and the critical values are -2.023 and 2.023. The rejection regions are given as T is less than -2.023, and T is greater than 2.023. And now we find the standardized test statistic T. Which is given by the following formula of T is equal to our sample mean, subtracted by the population mean divided by the sample standard deviation, which is divided by the square root of the sample size. So then we substitute the values into the. Formula giving us 36.7, subtracted by 35 divided by 2.9, which is divided by the square root of 40. And when simplified, our standardized test statistic T is equal to 3.707. And using the standardized test statistic and the rejection regions, we can make our decision, which is since T is greater than our critical value in a two-tailed test, T is in the rejection region. Therefore, we reject the null hypothesis. And from this information, since the claim is our alternative hypothesis, and we reject the null hypothesis, the interpretation for this decision is that there is sufficient evidence to support the claim that the mean fuel efficiency is different from 35 MPG. Therefore, the The result of testing the competitors claim at the 0.05 significance level is that there is sufficient evidence to support the claim that the mean fuel efficiency is different from 35 MPG, which makes answer choice A the correct answer, and all other answer choices are incorrect. I hope you found this video to be helpful. Thank you and goodbye.

In Exercises 13–18, test the claim about the population mean μ at the level of significance α. Assume the population is normally distributed.
Claim: μ≠52,200; α=0.05. Sample statistics: x_bar=53,220, s=2700, n=34

Key Concepts

Hypothesis Testing

Significance Level (α)

Confidence Interval

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