DATA:
The Population Mean for Women’s Height is 65” and the Population Standard Deviation is 3.5”. A researcher tests whether a new Stretching Machine will significantly increase height in women. The researcher selects a sample of 12 women and has each use the stretching machine for 10 minutes per day for 12 months. At the end of the 12-month period the research measures each woman’s height and calculates the sample mean. The sample mean is 67”. The researcher wants to maximize the power of the test. The data are presented below.
SUBJECT |
HEIGHT |
1 |
68 |
2 |
69 |
3 |
67 |
4 |
66 |
5 |
70 |
6 |
66 |
7 |
65 |
8 |
65 |
9 |
66 |
10 |
67 |
11 |
68 |
12 |
67 |
Sum 804
Mean 67
N 12
(1) What are the Null and Alternative Hypotheses for the test of whether there is significant Skew present in the data?
Group of answer choices:
(a) Ho: Skew ≠ 0, H1: Skew = 0
(b) Ho: Skew ≤ 0, H1: Skew > 0
(c) Ho: Skew ≥ 0, H1: Skew < 0
(d) Ho: Skew < 0, H1: Skew ≥ 0
(e) Ho: Skew = 0, H1: Skew ≠ 0
(2)Calculate the Coefficient of Skew to test the Assumption of Normality. Input Answer the answer you calculate to 2 Decimal Places.
(3)What is the Absolute Value of the Critical Value for the Coefficient of Skew to test the Assumption of Normality?
1). To test the presense of Skewness:
(e) Ho: Skew = 0, H1: Skew ≠ 0
2). Skewness = 3 * ( mean - median)/standard deviation
Here, mean = 67,
sorted data :
SUBJECT | HEIGHT |
1 | 65 |
2 | 65 |
3 | 66 |
4 | 66 |
5 | 66 |
6 | 67 |
7 | 67 |
8 | 67 |
9 | 68 |
10 | 68 |
11 | 69 |
12 | 70 |
median = (67+67)/2 = 0
skewness = 3 * 0/(standard deviation) = 3*0 = 0
The data is normal.
3. The critical value for normality is 0 because less than 0 mean it is left or negatively skewed and when skewness is greater than 1 it is right or positively skewed.
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