Question

Xbar in the first sample: 0.6

Standard Deviation in the first sample: 1.2

Size of the first sample (n1): 13

Xbar in the second sample: 5.2

Standard Deviation in the second sample: 3.0

Size of the second sample (n2): 17

Use the conservative t-test to test the null hypothesis of equality of means. Submit the p-value of your test of significance.

Answer 1

Answer:

Given,

To determine the p value

consider,

H0 : mu1 - mu2 = 0

Ha: mu1 - mu2 != 0

let us consider the test statistics:

Sp = sqrt(((n1-1)*s1^2+(n2-1)*s2^2)/(n1+n2-2))

substitute values

= sqrt(((13 - 1) * 1.2^2 + ( 17-1) * 3^2)/(13+17 -2))

Sp = 2.40

t = (x1 -x2)/Sp *sqrt(1/n1+1/n2)

substitute the given values in above expression

= ( 0.6 - 5.2)/(2.40 *sqrt(1/13 + 1/17))

t = - 5.202

So for t = - 5.202 the p value is 0.00001

Hence we reject the null hypothesis H0 at 0.05 level of significance