Question

Find the hypothesis testing on a two population mean. Need a T-test, interval and P-value focusing just on Gender not Rank or Enlisted

Description: A sample of 299 testing the Ethics scores for a sample of members of the U.S. Coast Guard by Gender (1=Male, 2=Female) and Rank (1=Officer, 2=Enlisted). Data simulated to match cell means and standard deviation. Variables/Columns Gender 8 Rank 16 Ethics Score 18-24

What is the T-test, P-value

Gender	Rank	Ethics Score
1	1	34.99
1	1	25.08
1	1	40.5
1	1	50.96
1	1	50.17
1	1	55.59
1	1	15.9
1	1	35.66
1	1	49.12
1	1	27.02
1	1	31.04
1	1	20.9
1	1	19.31
1	1	28.12
1	1	30.19
1	1	16.57
1	1	32.27
1	1	33.93
1	1	39.4
1	1	34.33
1	1	34.72
1	1	34.28
1	1	51.63
1	1	37.16
1	1	36.14
1	1	32.83
1	1	58.01
1	1	46.8
1	1	62.1
1	1	31.39
1	1	54.86
1	1	21.69
1	1	43.49
1	1	47.17
1	1	57.47
1	1	37.17
1	1	32.72
1	1	44.87
1	1	34.17
1	1	45.71
1	1	23.4
1	1	29.44
1	1	22.61
1	1	34.35
1	1	37.7
1	1	38.31
1	1	34.76
1	1	60.27
1	1	20.37
1	1	30.57
1	1	11.91
1	1	52.7
1	1	25.06
1	1	31.41
1	1	45.71
1	1	42.76
1	1	46.89
1	1	44.07
1	1	24.13
1	1	26.72
1	1	45.06
1	1	41.3
1	1	28.51
1	1	35.59
1	1	39.36
1	1	43.68
1	1	39.43
1	1	28.8
1	1	57.13
1	1	42.97
1	1	38.76
1	1	46.44
1	2	41.49
1	2	30.32
1	2	9.76
1	2	16.29
1	2	22.78
1	2	19.21
1	2	19.83
1	2	21.22
1	2	27.41
1	2	46.69
1	2	25.71
1	2	51.09
1	2	16.11
1	2	32.9
1	2	40.03
1	2	9.01
1	2	47.07
1	2	18.81
1	2	28.48
1	2	32.25
1	2	20.25
1	2	37.06
1	2	47.62
1	2	39.58
1	2	24.29
1	2	31.79
1	2	14.03
1	2	51.64
1	2	39.43
1	2	46.54
1	2	23.72
1	2	19.5
1	2	27.35
1	2	25.31
1	2	21.26
1	2	30.49
1	2	25.82
1	2	25.14
1	2	27.38
1	2	27.74
1	2	25.34
1	2	18.52
1	2	18.51
1	2	35.55
1	2	4.17
1	2	32.86
1	2	52.03
1	2	29.7
1	2	29.27
1	2	26.3
1	2	30.35
1	2	54.5
1	2	38.7
1	2	27.87
1	2	28.96
1	2	44.55
1	2	29.46
1	2	13.49
1	2	33.35
1	2	25.79
1	2	27.06
1	2	27.83
1	2	56.99
1	2	12.88
1	2	40.45
1	2	30.99
1	2	34.32
1	2	34.57
1	2	23.69
1	2	34.88
1	2	41.82
1	2	22.45
1	2	17.25
1	2	29.13
1	2	25.67
1	2	46
1	2	20.65
1	2	20.63
1	2	32.3
1	2	20.7
1	2	35.26
1	2	35.17
1	2	24.07
1	2	14.98
1	2	22.27
1	2	29.35
1	2	22.69
1	2	34.08
1	2	21.61
1	2	37.95
1	2	50.55
1	2	46.64
1	2	33.08
1	2	23.7
1	2	25.69
1	2	42.52
1	2	36.72
1	2	40.76
1	2	28.96
1	2	34.32
1	2	4.95
1	2	30.17
1	2	13.47
1	2	32.71
1	2	25.85
1	2	30.63
1	2	24.89
1	2	38.75
1	2	36.59
1	2	18.54
1	2	29.61
1	2	34.33
1	2	29.53
1	2	42.06
1	2	38.01
1	2	63.03
1	2	10.09
1	2	28.79
1	2	50.52
1	2	23.15
1	2	42.44
1	2	38.36
1	2	14.12
1	2	31.46
1	2	36.61
1	2	14.47
1	2	40.16
1	2	50.32
1	2	22.14
1	2	11.52
1	2	31.8
1	2	40.58
1	2	38.17
1	2	32.47
1	2	40.28
1	2	19.41
1	2	19.98
1	2	46.61
1	2	27.03
1	2	31.51
1	2	47.92
1	2	37.88
1	2	2.42
1	2	17.08
1	2	24.97
1	2	36.62
1	2	42.98
1	2	25.6
1	2	49.35
1	2	27.88
1	2	37.98
1	2	40.09
1	2	31.17
1	2	23.82
1	2	33.84
1	2	8.52
1	2	18.78
1	2	44.13
1	2	32.04
1	2	30.85
1	2	18.06
1	2	45.27
1	2	40.41
1	2	42.65
1	2	39.44
1	2	23.53
1	2	28.58
1	2	27.76
1	2	29.86
1	2	29.56
1	2	37.8
1	2	38.5
1	2	27.33
1	2	43.38
1	2	48.8
1	2	33.56
1	2	26.08
1	2	7.9
1	2	29.7
1	2	38.5
2	1	52.77
2	1	50.34
2	1	47.2
2	1	38.45
2	1	47.77
2	1	60.11
2	1	28.45
2	1	37.19
2	1	51.65
2	1	53.07
2	1	44.74
2	1	28.43
2	1	43.63
2	1	35.05
2	1	27.65
2	2	47.23
2	2	27.8
2	2	30.46
2	2	12.69
2	2	43.78
2	2	39.35
2	2	28.33
2	2	65.22
2	2	42.34
2	2	13.75
2	2	55.31
2	2	20.01
2	2	24.36
2	2	61.5
2	2	8.18
2	2	5.65
2	2	25.55
2	2	59.76
2	2	13.05
2	2	59.57
2	2	58.27
2	2	30.81
2	2	29.82
2	2	37.57
2	2	30.7
2	2	21.52
2	2	13.12
2	2	57.62
2	2	20
2	2	41.03
2	2	39.97
2	2	36.46

Answer 1

Solution:

Here, we have to use two sample t test for the difference between two population means assuming equal population variances. The null and alternative hypotheses for this test are given as below:

Null hypothesis: H₀: There is no any significant difference in the average ethics score of the male and female.

Alternative hypothesis: H_a: There is a significant difference in the average ethics score of the male and female.

H₀: µ₁ = µ₂ versus H_a: µ₁ ≠ µ₂

We assume level of significance = α = 0.05

Test statistic formula for pooled variance t test is given as below:

t = (X1bar – X2bar) / sqrt[S_p²*((1/n1)+(1/n2))]

Where S_p² is pooled variance

S_p² = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)

From given data, we are given

X1bar = 32.51429

X2bar = 37.17617

S1 = 11.60384

S2 = 15.72334

n1 = 252

n2 = 47

df = n1 + n2 – 2 = 252 + 47 – 2 = 297

α = 0.05

Critical values = - 1.9680 and 1.9680

(by using t-table or excel)

S_p² = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)

S_p² = [(252 – 1)* 11.60384^2 + (47 – 1)* 15.72334^2]/(252 + 47 – 2)

S_p² = 152.0849

t = (X1bar – X2bar) / sqrt[S_p²*((1/n1)+(1/n2))]

t = (32.51429 – 37.17617) / sqrt[152.0849*((1/252)+(1/47))]

t = -4.6619/ 1.9594

t = -2.3792

P-value = 0.0180

(by using t-table or 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 ethics score of the male and female.

Confidence interval for difference between two population means is given as below:

Confidence interval = (X1bar – X2bar) ± t*sqrt[S_p²*((1/n1)+(1/n2))]

Where S_p² is pooled variance

S_p² = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)

S_p² = 152.0849

(X1bar – X2bar) = -4.6619

We assume confidence level = 95%

df = 297

Critical t value = 1.9680

(by using t-table)

Confidence interval = -4.6619 ± 1.9680*1.9594

Confidence interval = -4.6619 ± 3.8561

Lower limit = -4.6619 - 3.8561 = -8.5180

Upper limit = -4.6619 + 3.8561 = -0.8058

Confidence interval = (-8.5180, -0.8058)

This confidence interval does not contain 0, so we reject the null hypothesis.

There is sufficient evidence to conclude that there is a significant difference in the average ethics score of the male and female.