In: Statistics and Probability
Report a) the appropriate statistical test or estimation procedure to use, b) the null and alternate hypotheses, c) the test statistic, d) the P value, e) whether you accept or reject the null, then f) a sentence or two about what the results mean.
Is there a difference between height of male children and female children? Show results using a parametric test and also the results using the appropriate non-parametric test. Which is the most appropriate test to use and why?
male = 1
female = 2
Height is in cm
child_sex |
child_ht |
1 |
130 |
1 |
135 |
1 |
139 |
1 |
138 |
2 |
125 |
2 |
129 |
1 |
133 |
1 |
128 |
2 |
120 |
2 |
140 |
1 |
144 |
1 |
150 |
2 |
134 |
1 |
132 |
2 |
131 |
1 |
128 |
1 |
139 |
2 |
136 |
1 |
161 |
1 |
163 |
2 |
145 |
2 |
141 |
1 |
154 |
1 |
159 |
1 |
132 |
1 |
138 |
1 |
143 |
2 |
134 |
2 |
140 |
1 |
148 |
1 |
140 |
2 |
133 |
2 |
134 |
1 |
151 |
1 |
157 |
1 |
145 |
1 |
161 |
1 |
154 |
2 |
159 |
1 |
132 |
1 |
134 |
1 |
138 |
1 |
145 |
2 |
134 |
2 |
123 |
1 |
136 |
1 |
133 |
1 |
167 |
1 |
158 |
2 |
149 |
2 |
143 |
2 |
145 |
1 |
147 |
1 |
142 |
a)
2 Sample t test
b)
Ho : µ1 - µ2 = 0
Ha : µ1-µ2 ╪ 0
c)
Level of Significance , α =
0.05
Sample #1 ----> Male
mean of sample 1, x̅1= 143.83
standard deviation of sample 1, s1 =
11.08
size of sample 1, n1= 35
Sample #2 ----> Female
mean of sample 2, x̅2= 136.58
standard deviation of sample 2, s2 =
9.44
size of sample 2, n2= 19
difference in sample means = x̅1-x̅2 =
143.8286 - 136.6 =
7.25
pooled std dev , Sp= √([(n1 - 1)s1² + (n2 -
1)s2²]/(n1+n2-2)) = 10.5426
std error , SE = Sp*√(1/n1+1/n2) =
3.0042
t-statistic = ((x̅1-x̅2)-µd)/SE = (
7.2496 - 0 ) /
3.00 = 2.413
d)
Degree of freedom, DF= n1+n2-2 =
52
p-value = 0.019376
(excel function: =T.DIST.2T(t stat,df) )
e)
Conclusion: p-value <α , Reject null
hypothesis
f)
There is enough evidence that male mean and female mean are
different.
Thanks in advance!
revert back for doubt
Please upvote