Denomination Effect A trial was conducted with 75 women in Chi...

Question

Denomination Effect A trial was conducted with 75 women in China given a 100-yuan bill, while another 75 women in China were given 100 yuan in the form of smaller bills (a 50-yuan bill plus two 20-yuan bills plus two 5-yuan bills). Among those given the single bill, 60 spent some or all of the money. Among those given the smaller bills, 68 spent some or all of the money (based on data from “The Denomination Effect,” by Raghubir and Srivastava, Journal of Consumer Research, Vol. 36). We want to use a 0.05 significance level to test the claim that when given a single large bill, a smaller proportion of women in China spend some or all of the money when compared to the proportion of women in China given the same amount in smaller bills.

c. If the significance level is changed to 0.01, does the conclusion change?

c. If the significance level is changed to 0.01, does the conclusion change?

Accepted Answer

Welcome back everyone. In this problem, researchers surveyed a simple random sample of 120 college students in Brazil. 60 students were given a single 100 real note while the other 60 received the the same amount in smaller denominations, 520 real notes. Among the single note group, 48 students spent some or all of the money, and in the small denomination group, 55 students spend some or all of the money. Using a hypothesis test at the 0.01 significance level, test the claim that fewer students spend the money when given a single large note. Now what do we, what do we know here? OK? Well, if we think about it, so far we know that we have 2 sample proportions from two simple random samples, OK? Because this researchers surveyed a simple random sample of 120 college students. We also know these two samples are independent. And for each of the two samples, if we define success, OK, if success is defined as spending all or some of the money. OK. Some are all of the money. Then sample one has 48 successes while sample. 2 has 55 successes. So for each of the two samples, there are at least 5 successes and at least 5 failures. So now, if we count sample one or if we see sample one. As the one with the large note, OK, the single large note, then we can say that N1, the sample size equals 60. X1, the number of successes equals 48, which means the sample proportion will be equal to 48 divided by 60, which is 0.8. On the other hand, if we view sample two as those with the, as the one with the smaller notes, OK. Then and 2 would be 60. X2 is 55, which means that here, P2, should have written P1 up here, but P2 is gonna be equal to 55 divided by 60, which is approximately 0.9167. So now that we have our successes defined and we know our data, or we have a few, um, we could draw a few conclusions based on the data that we're given. Then for our claim. Since we're testing that fewer students spend the money when given a single large note, then the null hypothesis H not is that P1 is greater than or equal to P2, OK. That is, the proportion of those spending with a larger note is greater than the proportion of those spending with smaller nodes, whereas our alternative hypothesis H1 is that P1 is less than P2. So now, how can we test this claim? Well, notice here that given all this information, OK, then we can compute the test statistic using the pool sample proportions, OK? Now here recall. That the pool sample proportion will be equal to X1 + X2 divided by N1 plus N2. That's gonna be 48 plus 65 divided by 120. And 48 + 65 is 103, and 103 divided by 120 is gonna be approximately equal to 0.8583. And if that's the case, then that means the pool, uh, so the pools, uh, sample failures is gonna be equal to 1 minus 103 divided by 20. That's 120, that's 17 1/20 or approximately. 0.1417. Now we needed this information to compute the test statistic because no, this means that the test statistic Z is gonna be equal to the difference between the sample proportions. OK, P1 minus P2 minus P1 minus P2 are population proportions, which the difference can be assumed to be zero because we assume in the another hypothesis that P1 equals P2. And all of this is being divided. Bye Let's let's get that value here by. P Q or pull sample proportions multiplied by Q bar divided by N1 plus P Q bar divided by N2. Let's substitute our values here. No, that's gonna be equal to 0.8 minus 0.9167. 0. All divided by the square root, OK? 0.8583 multiplied by 0.1417. Divided by 60 plus 0.8583, multiplied by 0.1417, divided by 60, OK? And now when we calculate here. We should get this to be equal to -1.833. So what does this mean? Well, now we know that our test statistic is -1.833. So here for our left tail test, we're going to reject the null hypothesis if this critical value if this test statistic value Z is less than the critical Z value. In order to do that, we need the critical Z value. So now let's see if we can figure that out. So in this case, we have a left tail test. So for a left tailed test. At a significance level of 0.01, OK. Then if we go to our table of significant values, we can try to find the corresponding critical value. And that critical value is the Z score corresponding to the cumulative area to the left of a 0.01 significant level. Now, here's the thing. If we look at our table, 0.01 is going to be between 0.0102 and 0.0099. In other words, if we go to our table, it won't have an exact value written for uh 0. alpha equals 0.01. So we'll need to perform interpolation. So, if we let Z script be the critical value, then by interpolation, that critical value minus -2.32, OK, that would be our critical value where it's at 0.1, 0.0102, OK, divided by 0.01 minus 0.0102. Should be equal to -2.33 minus -2.32 divided by 0.0099 minus 0.0102. Now if we go ahead and solve here again, we should get this critical value when we do some transposition. OK. To be equal, the critical value should be approximately equal to -2.327. Now let's compare our test statistic to this critical value. What do you notice? Well, OK, since. -1.833, our test statistic is greater than -2.327 or critical value, then we fail to reject. We fail to reject another hypothesis because the test statistic does not fall within the critical region. Therefore, we can conclude, OK, so let's write our conclusion here. OK, at the 0.01 significance level. OK. Then there is insufficient evidence. To support the claim. That a smaller proportion of students. Spend the money. When given a single large note. Compared to. 5 notes, OK, or smaller notes. Let's put smaller notes here. So there's insufficient evidence to reject non hypothesis. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Hypothesis Testing

Significance Level (α)

Proportion Comparison

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