In Exercises 23 and 24, (a) identify the claim and state Ho an...

Question

In Exercises 23 and 24, (a) identify the claim and state Ho and Ha , (b) find the critical value(s) and identify the rejection region(s), (c) calculate d̄ and sd, (d) find the standardized test statistic t, (e) decide whether to reject or fail to reject the null hypothesis, and (f) interpret the decision in the context of the original claim. Assume the samples are random and dependent, and the populations are normally distributed.

A physical fitness instructor claims that a weight loss supplement will help users lose weight after two weeks. The table shows the weights (in pounds) of 9 adults before using the supplement and two weeks after using the supplement. At α=0.10, is there enough evidence to support the physical fitness instructor’s claim?

Accepted Answer

All right, hello, everyone. So, this question says, a psychologist wants to determine whether a new mindfulness training program improves reaction time. The table below shows the reaction times in milliseconds for 10 participants before and after completing the program. At the alpha equals 0.05 significance level, is there sufficient evidence to support the psychologist's claim that the training program reduces reaction time? And here we have 4 different answer choices labeled A through D. All right, so first, let's state our hypotheses. Let's say that mu sub D represents the mean difference in reaction times, so it's the before subtracted by the after. The null hypothesis would state that mu sub D is equal to 0, which implies that there is no change in the reaction time. H A on the other hand, would state that mu sub D is greater than 0. Which means that the reaction time decreased after trading. Now, here, this is a one-tailed test because the psychologist is claiming a reduction. So now, our next step is to calculate the differences, the before subtracted by the after. So here The values for D, the differences in order for each participant are as follows. 10 10 10 1277. 555 and 5. Now we can find D bar, which is the mean of the differences. The mean of the differences is equal to the sum of all values of D divided by 10, which is the number of participants here. So the sum is equal to 76, so 76 divided by 10 means D bar is equal to 7.6. So now The next step is to ultimately find the squared deviations for each participant. So first, what you do is you take the difference for each data point and subtract it by the mean of the differences. And then you square that difference, which is why it's known as the square deviation. So here This allows you to find the standard deviation of the differences, or as sub D. S D is equal to the square root of The sum of all squared deviations. Divided by and subtracted by one. This simplifies 2. 64.4. Divided by 9. Which gives you approximately 2.673. So now, tea. Is equal to D bar. Divided by S sub D divided by the square root of N. Which equals 7.6. Divided by 2.673. Divided by the square root of. And this equals approximately 8.99. So now, this is our test statistic, which we would then compare to a critical value. So our T critical value. From the Table distribution, or alpha equals 0.05, and a one-tailed test with 9 degrees of freedom. Equals approximately 1.833. This now brings you to your decision. Right? Because our test statistic, 8.99 is greater than our critical value, we can reject. The null hypothesis So, with this information in mind, going back to our answer choices. The correct answer is option B, which reads there is sufficient evidence to support the claim that the training program reduces reaction time. And there you have it. So with that being said, thank you so very much for watching, and I hope you found this helpful.

Key Concepts

Hypothesis Testing

Dependent Samples

Critical Value and Rejection Region

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