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.

Diastolic Blood Pressure Diastolic blood pressure levels of 60 mm Hg or lower are considered to be too low. For the 300 diastolic blood pressure levels listed in Data Set 1 “Body Data” from Appendix B, the mean is 70.75333 mm Hg and the standard deviation is 11.61618 mm Hg. Use a 0.01 significance level to test the claim that the sample is from a population with a mean greater than 60 mm Hg.

Accepted Answer

Welcome back, everyone. In this problem, a nutritionist claims that the average daily calcium intake for adults is more than 900 mg. A simple random sample of 250 adults yields a mean daily calcium intake of 980.531 mg with a standard deviation of 120.883 mg. Use a 0.01 significance level to test the claim that the population mean daily calcium intake is greater than 900 mg. Now, what are we doing here? We're testing a claim, OK, that the population mean daily calcium intake is greater than 900 mg where our population standard deviation is not known. However, so far we know that this is a simple random sample. We also know that the sample size is 250, and since it's 250, then we can assume that this follows a normal distribution. So if we look at the information that we're given here. OK, in other words, we're given, um, we're given uh an a significance level alpha of 0.01. OK? We also know that. X bar or sample mean is 980.531 mg, OK? Me population mean is 900 mg. And our sample standard deviation S is 120.883 mg. So for us to test this claim, or another hypothesis, OK, first is that mu is less than or equal to 900. That is our default assumption because we're testing if it's greater than 900. And thus our alternate hypothesis is that mu is greater than 900. So for us to test this claim, we can use a test or, or sorry, we can use a right-tailed tea test. In other words, from our alternate hypothesis, we are saying that if, OK, if our test statistic. is greater than our critical value, then we can reject our null hypothesis, OK? Visually, what we, what that means is that if we have our distribution, OK, and here we have our critical value. Then if our test statistic falls in this region, we're going to reject the null hypothesis. That's basically what we're looking for. So from that, it tells us that we need the test statistic T and the critical T value. Let's start with the test statistic, OK? Recall that the test statistic T is going to be equal to the sample mean minus the population mean divided by the sample standard deviation divided by the square root of the sample size. So in that case then, this means T is going to be equal to, actually, let me put that in red. T will be equal to. 980.531 minus 9, 900, OK, divided by 120.883 divided by the square root of the sample size 250. Now when you go ahead and solve here, then we get our test statistic to be 10.533. No, approximately that is. No, we need to figure out if this T value is greater than our critical T value. No 4 in this case 4 or 1-tailed test. OK, so for this one tailed test. At a significance level of 0.01 and the degrees of freedom being equal to 1 less than the sample size, so that's 250 minus 1 or 249, we can go to the table for critical T values to find the critical value. But in this case, if we go to that table, you'll probably know that it does not have our T value. 4 degrees of freedom, 249, because when you go, you might notice that it's gonna be between 200 and 300. And here this represents the area in one tail, OK. So the significance of the area and one tail, you should see values 2.345 and 2.34, sorry, 339, OK? For a significance level of 0.01. So now we'll need to do linear interpolation to find the criticality value and by interpolation. OK. Then that criticality value. for a significant 0.01 and 249 degrees of freedom minus 2.345, the criticalt value for uh 200 degrees of freedom divided by 249 minus 200 is going to be equal to. 2.339, the criticality value for 300 degrees of freedom minus 2.345 divided by 300 minus 200. In other words, both these values must be the same. And if we go ahead and substitute here, then that the value, then, OK. Criticality value should be equal to. Uh, 2.339. -2.345. Divided by 300 minus 200, 100, OK. Multiplied by 249 minus 2,49, OK, plus 2.345. When we go ahead and solve here, we should get it to be approximately 2.342. OK? So this critical value is 2.342. Now, do you remember what we said earlier? We would reject or not hypothesis if the T value is greater than the critical T value. And since our T value is 10.553, OK, let me put that in red. Since our T value is approximately. 10.533, and that is greater than our critical value of 2.342. OK. That means we reject the null hypothesis. If we are rejected another hypothesis, what are we saying? Well, I had it. Let me just put it here. What we're really saying then, OK, is that there is sufficient evidence at the 0.01 significance level to support the claim that the population mean daily calcium for adults is greater than 900 mg. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Hypothesis Testing

P-value

Significance Level

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