Imagine that you are a psychologist who has invented a new medication ("Wildcatozine") to treat people with depression. You predict that depressed individuals who take Wildcatozine will experience fewer depression symptoms than depressed individuals who do not take the medication, but now you need to test this hypothesis.

You recruit a sample of n = 30 college students who have been diagnosed with depression, and you give them all Wildcatozine. Several weeks later, you give all participants a questionnaire about their depression symptoms. This questionnaire gives each participant a depression score from 0 (not at all depressed) to 10 (extremely depressed).

You may assume that this questionnaire has been widely-used previously with other depressed students. As such, we know that the population of untreated depressed individuals has an average score () of 7.00 on this scale, with a standard deviation () of 2.50, and that this distribution is normally distributed.

Data

After several weeks of taking Wildcatozine, these are the depression scores reported by your sample:

4, 7, 6, 10, 6, 3, 8, 1, 5, 3, 8, 2, 4, 9, 3, 9, 5, 8, 10, 6, 3, 4, 9, 10, 7, 5, 3, 5, 8, 4

Descriptive Statistics

o Central Tendency: Measures of central tendency are statistics that indicate the typical or 'central' values of distributions. They are representative of the entire distribution. The 3 different ways of identifying the central tendency are by calculating the mean, median, and the mode of a distribution.

oMean: The mean is the numerical average of scores in a distribution. The mean is also one of the most common measures of central tendency that describes the average of the scores. The mean may also be referred to as the balance point of a distribution. In order to find the mean, you must take the sum of all of the scores and divide it by the number of scores. There are two different means that may be calculated, the population or sample mean. The mean of the scores is 4+ 7+ 6+ 10+ 6+ 3+ 8+ 1+ 5+ 3+ 8+ 2+ 4+ 9+ 3+ 9+ 5+ 8+ 10+ 6+ 3+ 4+ 9+ 10+ 7+ 5+ 3+ 5+ 8+ 4= 175 then, 175/30= 5.83

oMedian: The median is the middle score when all of the scores are arranged from lowest to highest. The median divides the scores into 2 groups, both of equal size. If there is an odd number of scores, the median will be the middle number. If there is an even number of scores, the median will be the average of the two middle scores (add the 2 middle scores then divide them by 2). The median may also be referred to as the midpoint of the distribution where half of the scores are on each side. When working with discrete or continuous variables, the median will be different. For discrete variables, the median is a whole number. For continuous variables, the median is a precise point where exactly 50% of the distribution is above and below. The median of the distribution is 6.

oMode: The mode is the most common or most frequent score. There also may be more than one mode. There may be a bimodal or unimodal distribution, both of which are symmetrical when looking at the shape of the distribution. The mode of the sample is 3.

oVariability: Measures of variability describe how clustered together or spread out a distribution's scores are. There are two types of variability, high and low. A high variability means that most of the scores are far from the mean. A low variability means that most of the scores are close to the mean.

oRange: The range is a measure of variance. In order to find the range, you must subtract the lowest score from the highest score. 10-1=9, the range= 9.

oVariance: Is the average of squared deviations. In order to find variance, you must take the sum of the squared deviations and divide it by the number of scores. 198.1667/29= 6.83 The sample variance is 6.83.

oStandard Deviation: The average deviation score. In order to find the standard deviation, you must take the square root of the variance. Take the square root of 6.83= 2.61, the standard deviation= 2.61.

Z-score Information

o Z-score: The z- score is the number of standard deviations that a raw score is above or below the mean. In order to find the z- score of a population you must first subtract the population mean from the raw score, then divide by the population standard deviation. Once you know the z- score, it gives context on how to interpret scores within a distribution. The z- score also provides a common standard for comparing data that is measured in different ways. The z- score is -2.54.

oIn order to find the z- score of a population you must first subtract the population mean from the raw score, then divide by the population standard deviation. The standard error is .48. In order to get this, 5.83 (sample mean) - 7 (population mean) divided by 2.50 (population standard deviation)/ 5.47 (square root of n=30).

o Among the entire population, what proportion of scores falls above your sample mean? .9945 proportion of population scores fall above the sample mean.

oWhat proportion of scores falls below your sample mean? .0055 proportion of scores fall below the sample mean.

oUnit Normal Table: To use the unit normal table you must find the corresponding numbers

Hypothesis Testing Information

o Step 1: State the hypothesis

The medication 'Wildcatozine' significantly decreases the symptoms of depression in people.

What is the null hypothesis (H0) in this study? (H₀): M= 7

What is the alternative hypothesis (H1) in this study? (H₁): M< 7.00

Is the alternative hypothesis directional or non-directional? How do you know? And based on this answer, will you be conducting a one-tailed test or a two-tailed test? The alternative hypothesis is directional and a one- tailed test will be conducted.

o Step 2: Set the criteria for a decision

If you decide to use an alpha level of .05, what is the z-score that you will use as a cutoff score for the hypothesis test? A z- score of -1.64 will be used as the cutoff score for an alpha level of .05.

If you decide to use an alpha level of .01, what is the z-score that you will use as a cutoff score for the hypothesis test? A z- score of -2.32 will be used as the cutoff score for an alpha level of .01.

If you decide to use an alpha level of .001, what is the z-score that you will use as a cutoff score for the hypothesis test? A z- score of -3.08 will be used as the cutoff score for an alpha level of .001.

o Step 3: Collect data & compute sample statistics

What is the sample mean (M)? What is the z-score that corresponds with this sample mean? (Hint: You already figured both of these statistics earlier in the paper!) M= 5.83, Z- score= -2.54.

o Step 4: Make a decision

Does the sample z-score exceed the cutoff for p < .05? Based on your answer, what conclusion do you reach about the null hypothesis (reject or fail to reject)? Does the sample z-score exceed the cutoff for p < .01? Based on your answer, what conclusion do you reach about the null hypothesis (reject or fail toreject)?  

Does the sample z-score exceed the cutoff for p < .001? Based on your answer, what conclusion do you reach about the null hypothesis (reject or fail to reject)? Fail to reject the null hypothesis because it lies beyond the critical region.

Conclusions

o Look at the three conclusions that you reached about the null hypothesis in Step 4 of your hypothesis test.

Although you would not definitively know this as the researcher, is it possible that your p < .05 conclusion is a Type I error? Is it possible that it is a Type II error? Briefly explain your answers. This is a Type I error because p< .05 falls within the critical region.

Although you would not definitively know this as the researcher, is it possible that your p < .01 conclusion is a Type I error? Is it possible that it is a Type II error? Briefly explain your answers. This is a Type I error because p< .01 falls within the critical region.

Although you would not definitively know this as the researcher, is it possible that your p < .001 conclusion is a Type I error? Is it possible that it is a Type II error? Briefly explain your answers. 

o Assuming that you did not make any Type I or Type II errors, explain the three conclusions that you reached about the null hypothesis in Step 4 of hypothesis testing. Taking all three of these results into consideration, would you say that we can make valid claims that Wildcatozine is an effective antidepressant?

Reject the null hypothesis H₀ 'Wildcatozine' medication decreases the symptoms of depression.

Your job is to analyze the data from the Wildcatozine study, and to create paper about whether or not the medication significantly decreased symptoms of depression.

Did 'Wildcatozine' significantly decrease symptoms of depression? 

The unanwered/ highlighted questions please!