Question

4.) Consider the hypothetical hypothesis test of: VS HA: H1-H2 < 0 Hot-H2 = 0 If we know that sample 1 has a mean of 4.7 with a variance of 4; and sample 2 has a mean of 7.8 and a variance of 6.25 and both samples are of size 16; we are also allowed to assume that both populations sampled from are normal and that the true deviations of each population are equal. Use a 1% level for this test. Show all work! Step 1: Step 2: Step 3: Step 4: Step 5: P-value: Step 6 Step 7:

Answer 1

Solution:

Given:

Sample 1:

$\bar{x_{1}}=4.7$

s1²=4 therefore, s1 = 2

n1 = sample size = 16

Sample 2:

.22 = 7.8

s2²=6.25 therefore, s2 = 2.5

n2 = sample size = 16

Degrees of freedon = n1+n2-2 = 30

The following null and alternative hypotheses:

$H0:\mu_{1}-\mu_{2}=0$

$Ha:\mu_{1}-\mu_{2}<0$

This corresponds to a left-tailed test, for which a t-test for two population means, with two independent samples.

Since it is assumed that the population variances are equal so, the t-statistic:

t= X - X2 (n-1)sj+(12-1)3 (+ n+n2-2 12 4.7 – 7.8 (16-1)22+(16-1)2.52 (1+1) 16-16-2 3.873

P-value approach:

P(T<|t|) = P(T<3.873) = 0.0003 ...Using excel formula, =TDIST(3.873,30,1)

Here, the significance level is α=0.01

Decision: Since P-value < 0.01, reject H0.

Conclusion: There is enough evidence to claim that population mean μ1​ is less than μ2​, at the 0.01 significance level.

Done