PSY 230: Psychological Statistics and Measurement University of Arizona Fall 201
ID: 2929403 • Letter: P
Question
PSY 230: Psychological Statistics and Measurement University of Arizona Fall 2017 Written Assignment #1: Hypothesis Testing ***due on Wednesday, October 18 at 2:00 pm*** Overview Imagine that you are a psychologist who has invented a new medication ("Wildcatozine") to treat peopl 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 ofn- 30 college students who have been diagnosed with depression, and 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 th scale, with a standard deviation (o) 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, 5,7, 10, 6,3,4,9,2,7,5, 6, 4,8,4Explanation / Answer
We have data set of depression scores after Wildcatozine medication along with descriptive statistics in given below tables. Descriptive statistics has been calculated using data analysis package in excel.
Depression Scores after Wildcatozine Medication
4
7
6
10
6
3
8
1
5
3
8
2
4
9
3
9
5
7
10
6
3
4
9
2
7
5
6
4
8
4
Depression Scores after Wildcatozine Medication
Statistic
Mean
5.6
Standard Error
0.461382231
Median
5.5
Mode
4
Standard Deviation
2.527094556
Sample Variance
6.386206897
Range
9
Minimum
1
Maximum
10
Count
30
Sum
168
1.1 Descriptive Statistics:
Mean= Sum of sample observations/Number of observations
=(4+7+6+………+4)/30=5.6
Median: We sort the sample data in ascending or descending order, and the in case of even number of observations, we take take mean of middle two observations. In case of thirty observations in a sample,
Median= (15th observation+16th observation)/2=5.5
(Provided, first we sort data first)
Standard error=
sd= Standard Deviation=2.527094556
n: Number of observations.
Formulla for calculating Standard Deviation=
is the ith observation in a sample data.
is the mean of the sample data.
Sample variance is the square of Standard deviation.
Range= Maximum observation- Minimum observation=10-1=9
1.2 Z-Score Information
Z-Scores calculation of the sample data is given in the below table:
Depression Scores after Wildcatozine Medication
Z-Score
4.00
-0.633138161
7.00
0.553995891
6.00
0.15828454
10.00
1.741129943
6.00
0.15828454
3.00
-1.028849512
8.00
0.949707241
1.00
-1.820272213
5.00
-0.23742681
3.00
-1.028849512
8.00
0.949707241
2.00
-1.424560862
4.00
-0.633138161
9.00
1.345418592
3.00
-1.028849512
9.00
1.345418592
5.00
-0.23742681
7.00
0.553995891
10.00
1.741129943
6.00
0.15828454
3.00
-1.028849512
4.00
-0.633138161
9.00
1.345418592
2.00
-1.424560862
7.00
0.553995891
5.00
-0.23742681
6.00
0.15828454
4.00
-0.633138161
8.00
0.949707241
4.00
-0.633138161
(i)
Z-score is simply standard normal score with the formula given as
Z-score=
sd= Standard Deviation=2.527094556
is the ith observation in a sample data.
is the mean of the sample data.
(ii)
Z-score is the number of standard deviations from the mean a sample data points. In general and technically it is a measure of how many sd’s (standard deviations) below or above the mean of the population raw-score is. A Z-score is also known as a standard score and it can be placed on a normal model curve.
(iii)
We have 50% sample observations below mean and 50% observations above mean. Standard normal distribution is a symmetrical distribution about mean. It has mean zero and standard deviation one. So above mean, it covers 50% area and mellow mean it covers 50% area. When we compare the standardized scores (Z-scores) with mean of the standata normal distribution (which is zero), we find 50% observations below zero and 50% observations above zero.
Z-Score below zero
Z-Score above zero
-1.820272213
0.15828454
-1.424560862
0.15828454
-1.424560862
0.15828454
-1.028849512
0.15828454
-1.028849512
0.553995891
-1.028849512
0.553995891
-1.028849512
0.553995891
-0.633138161
0.949707241
-0.633138161
0.949707241
-0.633138161
0.949707241
-0.633138161
1.345418592
-0.633138161
1.345418592
-0.23742681
1.345418592
-0.23742681
1.741129943
-0.23742681
1.741129943
1.3 Hypothesis testing information
The null hypothesis in case of our data is
H0: Depressed individuals who take Wildcatozine medication will experience same depression symptoms as depressed individuals who do not take medication.
The alternative hypothesis in case of our data is
H1: Depressed individuals who take Wildcatozine medication will experience fever depression symptoms as depressed individuals who do not take medication.
It is left tailed test. Or we can say it is directional test because we want to check whether depressed individuals who take Wildcatozine medication will experience fever depression symptoms as depressed individuals who do not take medication. So in this regard, we have to conduct one tailed Z-test.
One Sample z-test
Depression Scores after Wildcatozine Medication
z = -3.0344, n = 30, sd = 2.52710,
Standard deviation of the sample mean = 0.46138, p-value = 0.001205
Alternative hypothesis: true mean is less than 7
95 percent confidence interval: { -Inf 6.358906}
Since P-value is less than 0.05, we reject the null hypothesis and accept alternative hypothesis that depressed individuals who take Wildcatozine medication will experience fever depression symptoms as depressed individuals who do not take medication.
Sample z scores exceed the cutoff for p<0.01 but do not exceed the cutoff for p<0.001 as the calculated Z-test p-value is 0.001205.
We can reject the null hypothesis at 0.01 level of significance as well.
To be on safer side from Type-I error, we set the level of significance as level of significance is the Maximum probability of committing a Type I Error.
Depression Scores after Wildcatozine Medication
4
7
6
10
6
3
8
1
5
3
8
2
4
9
3
9
5
7
10
6
3
4
9
2
7
5
6
4
8
4
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