Naughty or Nice? An experiment was conducted in which 16 ten-m...

Question

Naughty or Nice? An experiment was conducted in which 16 ten-month-old babies were asked to watch a climber character attempt to ascend a hill. On two occasions, the baby witnesses the character fail to make the climb. On the third attempt, the baby witnesses either a helper toy push the character up the hill, or a hinderer toy preventing the character from making the ascent. The helper and hinderer toys were shown to each baby in a random fashion for a fixed amount of time. In Problem 41 from Section 10.2, we learned that, after watching both the helper and hinderer toy in action, 14 of 16 ten-month-old babies preferred to play with the helper toy when given a choice as to which toy to play with. A second part of this experiment showed the climber approach the helper toy, which is not a surprising action, and then alternatively the climber approached the hinderer toy, which is a surprising action. The amount of time the ten-month-old watched the event was recorded. The mean difference in time spent watching the climber approach the hinderer toy versus watching the climber approach the helper toy was 1.14 seconds with a standard deviation of 1.75 second. Source: J. Kiley Hamlin et al., “Social Evaluation by Preverbal Infants,” Nature, Nov. 2007.
b. Assuming the differences are normally distributed with no outliers, test if the difference in the amount of time the baby will watch the hinderer toy versus the helper toy is greater than 0 at the 0.05 level of significance.

b. Assuming the differences are normally distributed with no outliers, test if the difference in the amount of time the baby will watch the hinderer toy versus the helper toy is greater than 0 at the 0.05 level of significance.

Accepted Answer

Hello. In this video we are told that a psychologist measures the difference in reaction times for 20 participants before and after consuming caffeine. The mean difference is 0.82 seconds with a standard deviation of 1.2 seconds. Assume the differences are normally distributed with no outliers. At a 0.05 significance level, we want to test whether the mean reaction time or the mean difference in reaction times is greater than zero. So, let's go ahead and start by writing down the given information in the problem. We are going to define new to be the mean reaction times. Now in this case here, we are given a sample size of 20 participants, so n is equal to 20. We are also working with the mean difference which we will label D bar of 0.82. And we are given a standard deviation of 1.20. And we want to work with a significance level of 0.05. Now, here, we are going to test or compare the same group of participants before and after consuming caffeine. So this is going to be a paired sampled tea test. Furthermore, because we know the standard deviation, we are going to set the original claim as our alternate hypothesis. Now again, the claim here is that the mean reaction time is greater than 0. So that means that new, which is our alternate hypothesis, is going to be that new is greater than 0. And this means that our null hypothesis is going to be the complement of this claim, is that the mean reaction time is less than or equal to 0. So, let's go ahead and approach this problem by using the paired sample T test. Furthermore, we are also going to be using a right-sided T test for the group, because we are using it greater than symbol for the alternate hypothesis. So, let's go ahead and start by defining our T test statistic. Now, in a paired sample T test, our T test statistic is defined as It is defined as the mean difference in reaction times D bar divided by the standard deviation divided by the square root of the sample size. Now we do have the information we need to find the T test statistic. So our T test statistic is going to be 0.82 divided by the quantity, 1.20, divided by the square root of 20. And with the help of a calculator, this is going to give us an approximate test statistic of 3.06. Now the next thing we need to do, is we need to find the critical value. Now for the critical value, there are two things that we need. We need the degree of freedom and the and the significance level. Now, the degree of freedom in this case is going to be the sample size minus 1. And we were given a sample size of 20, so our degree of freedom is going to be 20 minus 1, which is 19. So for a critical value, we are going to be using the significance level given to us, which is 0.05, for a right-sided test. And we are using a degree of freedom of 19, and by using a tea table we get a t critical value of approximately 1.729. Now that we've calculated our T test statistic and our T critical value, we are going to compare the two values together. But as we can see, the test statistic of 3.06 is greater than the critical value of 1.729. So we know that our critical value is less than our test statistic. What this tells us is that we're going to reject the null hypothesis. Since we are rejecting the null hypothesis, this tells us that there is sufficient evidence to conclude that the mean difference in reaction time is greater than zero, indicating that the caffeine provides or improves reaction time. And with that being said, the solution to this problem is going to be option A. So I hope this video helps you in understanding how to approach this problem, and we will go ahead and see you all in the next video.

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