In Exercises 15–22, test the claim about the population varian...

Question

In Exercises 15–22, test the claim about the population variance or standard deviation at the level of significance Assume the population is normally distributed.

Claim: σ^2=63, α=0.01 . Sample statistics: s^2=58, n=29

Accepted Answer

All right, hello, everyone. So this question says, a quality control manager claims that the variance in the life spans of a certain type of light bulb is exactly 36 hours. To test this claim, a random sample of 18 bulbs is selected, and the sample variance is found to be 42 hours squared. At the alpha equals 0.05 significance level, test the manager's claim. Assume the lifespans are normally distributed. And here we've got 4 different entry choices labeled A through D. All right, so first, what information do we actually have? First, we have the claim that the variance that sigma squared is equal to 36. We also know the sample variance that's Squad is equal to 42. Now The sample size that's N is equal to 18, and alpha, the significance level is 0.05. So first, let's identify the hypothesis in this case. We're testing if the true population variance is equal to 36 based on a sample. This means that the null hypothesis H0. Is where sigma squared is indeed equal to 36. So the alternative hypothesis, H1. Would be Where sigma squared is not equal to 36. And for the record, this is a two-tailed test. So first, we have to find the test statistic using RI square formula. So let me actually open up some space here. And recall that Kay Square. Is equal to. And subtracted by 1 multiplied by a squared. And then divided by sigma squared. So plugging in the information we already have, chi square is equal to. 18 subtracted by 1. Multiplied by 42. And divided by 36. Which equals approximately 19.83. From here you would find the critical values. So, as mentioned before, this is a two-tailed guy square test with In this case, 17 degrees of freedom. Recall that the degrees of freedom is equal to N, the sample size subtracted by 1. And, as mentioned before, alpha is equal to 0.05. So We can split the significance level. Starting off with the left tail critical value. That's at alpha divided by 2. Which equals 0.025? So Chi squared At 0.025, and with 17 degrees of freedom. Equals approximately 30.191. So now for the right side For the right side That critical value is going to be at 1 subtracted by alpha divided by 2. So here, that's one subtracted by 0.025. Which gives you 0.975. Therefore, Chi squared at 0.975 and 17 degrees of freedom. is equal to approximately 7.564. So now we're left to make a decision by comparing the test statistic to the critical values that we've just obtained. And here, We can see that our test statistic is between or it lies between the two critical values that we've obtained. Therefore, we fail to reject the null hypothesis. Because to reject the null hypothesis, recall that I square would have to be. Either less than 7.564 or greater than 30.191. So, because we failed to reject the null hypothesis, there is not enough evidence to reject the manager's claim that the population variance is exactly 36 hours squared. Which now leads you to your correct answer. And based on the information that we've just obtained, the correct answer is going to be option C in the multiple choice. Which reads, there is no sufficient evidence to reject the manager's claim that the population variance is exactly 36 hours squared. And there you have it. So with that being said, thank you so very much for watching and I hope you found this helpful.

In Exercises 15–22, test the claim about the population variance or standard deviation at the level of significance Assume the population is normally distributed.
Claim: σ^2=63, α=0.01 . Sample statistics: s^2=58, n=29

Key Concepts

Population Variance

Hypothesis Testing

Significance Level (α)

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