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.

Taxi Fares For the first 40 taxi fares (dollars) listed in Data Set 32 “Taxis” from Appendix B, the mean is $12.035 and the standard deviation is $8.361. Use a 0.05 significance level to test the claim that the mean cost of a taxicab ride in New York City is less than $15.00.

Accepted Answer

Welcome back, everyone. In this problem, a simple random sample of 40 grocery receipts from a supermarket shows a mean of $54.825 and a standard deviation of $15.605. Tests the claim at the 0.05 significance level that the average grocery bill is less than $60. Now what are we trying to figure out here? Well, we're testing a claim about a population mean with a population standard deviation not known. So far we know that the sample is a simple random sample and it has a sample size of 40. Since it's greater than 30, then we can assume this follows a normal sampling distribution and thus we can try to test our claim using tests that apply to normal distributions. Now, since we know the sta sample standard deviation but not the population standard deviation, that means we can use the T test. So let's take our hypotheses and figure out which tail test we're going to use. Now, since we're testing the claim that the average grocery bill is less than $60 then our non hypothesis, the default assumption is that the average bill mal equals $60. Which means the alternate hypothesis, H1, is that mu is less than 60, that the average grocery bill is less than 60. So based on these hypotheses, if we were to put this on a normal distribution, OK. Then, for us to find the where it would be less than 60, where the average grocery bill would be less than 60, would correspond to a T value on the left side of our distribution, OK? That's less than the critical value that we find, OK? So in other words, if our test statistic is going to be less than our critical value, then we are going to reject the null hypothesis. Now what do we know that can help us to find the test statistic and the critical value? Well, from our problem, like we said, we know the sample size N equals 40. The sample mean is $54.825. And the. Sample standard deviation S is 15.605. Recall that our test statistic T equals the sample mean minus the population mean divided by the sample standard deviation divided by the square root of the sample size. So if we apply that, that means our test statistic T will be equal to 54.825 minus 60. Divided by 15.605 divided by the square root of 40. And when we solve, we should get our test statistic to be approximately 2.097. No, we just need to find our critical value. So far, OK, a left tailed test. And for a left tail test. With A significance level alpha of 0.05 and the degrees of freedom one less than the sample size. So with a sample size of 40, that means we'd have 39 degrees of freedom. If we go to our tea table, then we should get a critical value, OK. We should get a critical value that would be approximately equal to -1.685. It would be equal to 1.685 in the test in, in the table, but because it's a left tail test, then we negate the T value. Now let's compare our criticality value to our test statistic. What do you notice? Well, you probably note, OK, that our calculated T value, which is -2.097 is less than the critical T value of -1.685. And since it's less than the critical value then. It falls in the rejection region. In other words, we can reject our null hypothesis. So since we reject another hypothesis, what can we conclude? Well, OK, let's let's type the sort. It might be a little easier to type. No, we can say that there is sufficient evidence. Add the 0.05 significance. Let me write that properly here. 0.05 significance level. To support the claim. that the average grocery bill. is less than $60 OK? So since we reject the null hypothesis, then this is what we can conclude. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Hypothesis Testing

P-value

Significance Level

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