In: Math
How do I do independent t test on the data set below and how do I know if its pooled t test or unrolled t test?
Health question for reference: to what extend does the age of MI patients vary by gender
Standard deviation: male = 13.944 female= 13.939
mean: male 65.353 female=73.628
Male female
65 88
77 81
78 82
76 66
40 81
83 73
58 64
43 53
39 69
66 67
61 89
49 85
85 81
54 85
82 84
68 83
78 76
56 77
72 84
50 43
75 87
61 70
48 64
82 59
62 91
39 60
45 80
65 72
68 73
73 85
64 80
80 79
74 48
80 32
92 86
51
41
90
83
61
64
82
48
63
81
52
65
74
62
71
73
43
80
72
57
76
53
44
71
64
86
60
63
74
56
Solution:
Here, we have to use two sample t test for the difference between two population means by assuming equal population variances.
We use pooled t test because standard deviations for the given two samples are approximately same and very close to each other.
Null hypothesis: H0: There is no significant difference in the average age of male and female.
Alternative hypothesis: Ha: There is a significant difference in the average age of male and female.
H0: µ1 = µ2 versus Ha: µ1 ≠ µ2
This is a two tailed test.
We assume level of significance = α = 0.05
Test statistic formula for pooled variance t test is given as below:
t = (X1bar – X2bar) / sqrt[Sp2*((1/n1)+(1/n2))]
Where Sp2 is pooled variance
Sp2 = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)
We are given
X1bar = 65.353
X2bar = 73.628
S1 = 13.944
S2 = 13.939
n1 = 65
n2 = 35
df = n1 + n2 – 2 = 65 + 35 – 2= 98
α = 0.05
Critical value = - 1.9845 and 1.9845
(by using t-table)
Sp2 = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)
Sp2 = [(65 – 1)* 13.944^2 + (35 – 1)* 13.939^2]/(65 + 35 – 2)
Sp2 = 194.3868
t = (X1bar – X2bar) / sqrt[Sp2*((1/n1)+(1/n2))]
t = (65.353 – 73.628) / sqrt[194.3868*((1/65)+(1/35))]
t = -8.2750 / 2.9231
t = -2.8309
P-value = 0.0056
(by using t-table/excel)
P-value < α = 0.05
So, we reject the null hypothesis
There is sufficient evidence to conclude that there is a significant difference in the average age of male and female.