Hypothesis Testing Using Rejection Regions In Exercises 23–30,...

Question

Hypothesis Testing Using Rejection Regions In Exercises 23–30, (a) identify the claim and state H0 and Ha, (b) find the critical value(s) and identify the rejection region(s), (c) find the standardized test statistic X^2, (d) decide whether to reject or fail to reject the null hypothesis, and (e) interpret the decision in the context of the original claim. Assume the population is normally distributed.

Salaries The annual salaries (in dollars) of 12 randomly chosen nursing supervisors are shown in the table at the left. At α=0.10, is there enough evidence to reject the claim that the standard deviation of the annual salaries is \$18,630?

Salaries The annual salaries (in dollars) of 12 randomly chosen nursing supervisors are shown in the table at the left. At α=0.10, is there enough evidence to reject the claim that the standard deviation of the annual salaries is $18,630?

tab2

Accepted Answer

Hello everyone. Let's take a look at this question together. The monthly electricity bills in dollars for 12 randomly selected households in a city are listed below, and here we have the data values for the monthly electricity bills in dollars. At the alpha equals 0.10 level of significance, is there sufficient evidence to reject the claim that the standard deviation of monthly electricity bills is $5. Is it answer choice A at alpha equals 0.10 significance level, there is sufficient evidence to reject the claim that the standard deviation of monthly electricity bills is $5. Answer choice B at alpha equals 0.10 significance level, there is no sufficient evidence to reject the claim that the standard deviation of monthly electricity bills is $5 or answer choice C, not enough information. So, in order to solve this question, we have to determine at the alpha equals 0.10 level of significance, is there sufficient evidence to reject the claim that the standard deviation of monthly electricity bills is $5 given that the monthly electricity bills in dollars for 12 randomly selected. households are given by the following data values. And so the first step in determining if there is sufficient evidence is to identify the information that we are given, which we know includes a sample size of N equals 12, a claimed standard deviation of 5. A significance level of alpha equals 0.10. Our degrees of freedom is calculated as N minus 1, so 12 minus 1 is equal to 11, and then the data values for the monthly electricity bills in dollars for the 12 randomly selected households. And then using the given information, we must identify the claim and state the hypotheses, which we know the claim is that the standard deviation is $5. Therefore, our null hypothesis is the standard deviation is equal to 5, and our alternative hypothesis is the standard deviation is not equal to 5. Therefore, this is a two-tailed test. And so now we must calculate the test statistic, starting with the sample mean, which is equal to the sum of the X values divided by the sample size N. So plugging in the values for the data of the monthly electricity bills divided by our sample size and the value for our mean is equal to 153.67. And then we calculate the sum of squared deviations, which is equal to 308.67, and using these values, we can calculate our sample variants, which is the sample variant. is equal to the sum of our squared deviations divided by N minus 1, which substituting the values into the formula, our sample variance is equal to 28.06. And now we can determine our test statistic, which is calculated by 2 is equal to and minus 1 in parenthesis multiplied by our sample variants divided by the population variance. Which substituting the values into the formula for the test statistic, our test statistic is approximately 12.35. And now we determine our critical values, which for alpha equals 0.10, and this is a two-tailed test, and our degrees of freedom, which is equal to 11, the left critical value. is approximately 4.575, and the right critical value is approximately 19.675. Therefore, our rejection region is given as we reject the null hypothesis if our test statistic is less than 4.575 or greater than 19.675. So now we make our decision, which we know that since 12.35, is between 4.575 and 19.675, we know that the test statistic does not fall in the rejection region. Therefore, we fail to reject the null hypothesis. So in conclusion, at the 0.10 significance level, there is no sufficient evidence to reject the claim that the standard deviation of monthly electricity bills is $5 which looking at our answer choices, we can identify answer choice B is the correct answer. As we know that at the 0.10 significance level, there is no sufficient evidence to reject the claim that the standard deviation of monthly electricity bills is $5 and all other answer choices are incorrect. I hope you found this video to be helpful. Thank you and goodbye.

Key Concepts

Hypothesis Testing

Critical Value and Rejection Region

Standard Deviation and Chi-Square Test

Watch next