Testing Hypotheses
In Exercises 13–24, assume that a sim...

Question

Testing Hypotheses

In Exercises 13–24, assume that a simple random sample has been selected and test the given claim. Unless specified by your instructor, use either the P-value method or the critical value method for testing hypotheses. Identify the null and alternative hypotheses, test statistic, P-value (or range of P-values), or critical value(s), and state the final conclusion that addresses the original claim.

Systolic Blood Pressure Systolic blood pressure levels above 120 mm Hg are considered to be high. For the 300 systolic blood pressure levels listed in Data Set 1 “Body Data” from Appendix B, the mean is 122.96000 mm Hg and the standard deviation is 15.85169 mm Hg. Use a 0.01 significance level to test the claim that the sample is from a population with a mean greater than 120 mm Hg.

Accepted Answer

Welcome back, everyone. In this problem, a company claims that the average time to resolve a customer complaint is less than 48 hours. A simple random sample of 60 complaints shows a mean resolution time of 45.2 hours with a standard deviation of 6.5 hours. At the 0.01 significance level, test the claim that the mean resolution time is less than 48 hours. Now what do we know so far? Well, for starters, we are testing a claim about a population mean with a population standard deviation unknown. However, we know that this is a simple random sample, OK? In other words, a simple random sample was taken, and it has a sample size of 60, which is greater than 30. So we can apply the central limit theorem and treat this as a normal distribution. No, since the population standard deviation is known, but we know the standard deviation of the sample, then that means we can use the T test, OK? So first, we'll need to form our hypotheses. No, our null hypothesis is that our mean is greater than or equal to 48. Remember, we're testing if it's less than 48. So our default default assumption is that it is greater than or equal to 48 hours, which means then that our null hypothesis is that the mean resolution time is less than 48 hours, OK? And so far. We know that we're testing this at a significance level alpha of 0.01. The sample size N is 60, or sample mean is 45.2 hours. And the standard deviation S is 6.5. Now, since we're testing here that it's less than 48 hours, then we're going to use a left tailed T test, OK, and see if the test statistic is less than the critical T value. Because if it's less, if I were to just do a quick diagram here of what we're searching for, for this left-tailed test, if. The test statistic is less than the critical T value, OK? That means we can reject the null hypothesis. But if it's greater than the critical T value, OK, that is towards the right of the graph, then there's not enough evidence for us to reject that hypothesis. So let's find our test statistic and our critical T value to compare. Now, starting with the test statistic, recall that the T value will be equal to the sample mean, OK, minus mu knot. In this case it's 48 divided by the standard deviation S, the sample standard deviation divided by the square root of the sample size. So in that case, the NRT value is going to be 45.2 minus 48 divided by 6.5 divided by the square root of 60. And when we go ahead and solve here, we should get that to be approximately equal to -3.337. So that is our test statistic. Next, what is our criticality value going to be? Well, for a left tailed test, for a left tailed tea test, OK. With N minus 1 degree of freedom, remember N is 60, so 1 less is 59, OK. So with the degrees of freedom equal to 59. And our significance level of 0.01, we can turn to our table for critical values to help us. However, when we refer to the table, we should you, you, you'll probably notice that there is no entry with 59 degrees of freedom. That poses a problem for us. However, we can use the idea of linear interpolation to help us to figure out what that what those degrees of freedom are, OK? Because on in our table is going to show 50 and then 6040 degrees of freedom and 59 is between 50 and 60. So since we're considering the left tail, we need to negate the values and then by interpolation. OK, by interpolation. We can infer that the T value, the critical value for a 0.01 alpha level and 59 degrees of freedom, minus the T value, OK, for where it's 50. OK. Alright, the degrees of freedom is 50, still 0.01, so I should probably add that here. 0.0 and 50, divided by 59 minus 50 is going to be equal to the T value for that same level, and 60 minus the T value for it. And 50 divided by 60 minus 50. Now, when we use what we know, OK, let me make some space here. Then that means T 0.0159, OK. Minus -2.403. Remember we're negating our values for our left tail test divided by 59 minus 50 will be equal to -2.390 minus -2.403 divided by 60 minus 50. I know if we do some transposition here. Then that means, OK. Tea or tea valley that we're looking for. OK, it's going to be equal to. Negative 2.390 plus 2.403 divided by 10, OK, multiplied by 59 minus 50, which is 9, OK, minus 2.403. And when we work that out and we solve here, then we should get that T value to be approximately equal to -2.391. So now we have our critical T value. Remember earlier we said in this left tail test we reject our another hypothesis if the T value is less than our critical T value. And in this case, we can tell that the T value we found earlier, -3.337 is less than -2.391. Thus we reject our null hypothesis, and we can make a conclusion. So since we reject the null hypothesis, OK. I reject the null hypothesis. No, for thus. We can say that at the 0.01 significance level. There is sufficient evidence. There is sufficient evidence. To support The company's claim. OK. That the mean resolution time. Mean resolution time. is less than. 48 hours. That's a conclusion that we can come to. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Hypothesis Testing

P-value

Significance Level

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