Question

Hypothesis tests for mean comparison: one‐tail or two‐tail (1 test on 2 main variables) on the information provided below:

Sample 1: Mean: 29,264.1000. SD: 8446.84853. N=30

19,400
24,380
32,500
25,400
34,395
19,500
23,570
30,930
32,000
29,000
34,000
36,800
45,000
39,500
41,830
45,000
35,800
35,950
40,000
26,810
26,995
23,710
17,999
19,000
26,000
24,000
12,875
31,000
18,000
26,579

Sample 2: Mean: 57640.8333. SD: 29995.85446. N=30

$48,950
$34,900
$102,600
$54,500
$33,900
$26,415
$39,785
$18,645
$22,995
$24,595
$30,750
$26,790
$66,500
$40,250
$89,900
$52,950
$75,150
$32,700
$113,900
$64,900
$54,600
$51,400
$113,900
$69,900
$49,700
$31,950
$54,900
$124,300
$91,100
$85,000

We need to use the unequal-variances t-test for mean comparison using this data. Please help! Thank you

Answer 1

The two-tailed test is:

The hypothesis being tested is:

H₀: µ₁ = µ₂

H₁: µ₁ ≠ µ₂

Sample 1	Sample 2
29264.1	57640.8333	mean
8446.84853	59995.85446	std. dev.
30	30	n
	30	df
	-28,376.7333000	difference (Sample 1 - Sample 2)
	11,061.7234981	standard error of difference
	0	hypothesized difference
	-2.565	t
	.0155	p-value (two-tailed)

The p-value is 0.0155.

Since the p-value (0.0155) is less than the significance level (0.05), we can reject the null hypothesis.

Therefore, we can conclude that µ₁ ≠ µ₂.

The one-tailed test is:

The hypothesis being tested is:

H₀: µ₁ = µ₂

H₁: µ₁ < µ₂

Sample 1	Sample 2
29264.1	57640.8333	mean
8446.84853	59995.85446	std. dev.
30	30	n
	30	df
	-28,376.7333000	difference (Sample 1 - Sample 2)
	11,061.7234981	standard error of difference
	0	hypothesized difference
	-2.565	t
	.0078	p-value (one-tailed, lower)

The p-value is 0.0078.

Since the p-value (0.0078) is less than the significance level (0.05), we can reject the null hypothesis.

Therefore, we can conclude that µ₁ < µ₂.