Question

Assume that you have a sample of n1=9​, with the sample mean X1=46​, and a sample standard deviation of S1=4, and you have an independent sample of n2=12 from another population with a sample mean of X2=31 and the sample standard deviation S2=5. Complete parts​ (a) through​ (d) below. a. What is the value of the​ pooled-variance tSTAT test statistic for testing H0: μ1=μ2​? tSTAT=nothing ​(Round to two decimal places as​ needed.) b. In finding the critical​ value, how many degrees of freedom are​ there? nothing degrees of freedom ​(Simplify your​ answer.) c. Using a significance level of α=0.1​, what is the critical value for a​ one-tail test of the hypothesis H0: μ1≤ μ2 against the alternative H1: μ1>μ2? The critical value is nothing. ​(Round to two decimal places as​ needed.) d. What is your statistical​ decision? A. Reject H0 because the computed tSTAT test statistic is greater than the​ upper-tail critical value. B. Reject H0 because the computed tSTAT test statistic is less than the​ upper-tail critical value. C. Do not reject H0 because the computed tSTAT test statistic is less than the​ upper-tail critical value. D. Do not reject H0 because the computed tSTAT test statistic is greater than the​ upper-tail critical value.

Answer 1

a. t`= 7.63

b.υ = [s₁²/n1 + s₂²/n2]²/ (s₁²/n1 )²/ n1-1 + (s₂²/n2)²/n2-1

= 18.885 ≈19 The degrees of freedom is always rounded in this calculation.

c. The critical region for one tailed test for alpha= 0.1 and 19 degrees of freedom is 1.328= 1.33

The critical region is ║t║` ≥ t 0.05,19 = 1.729= 1.73

A. Reject H0 because the computed tSTAT test statistic is greater than the​ upper-tail critical value

Explanation:

Answer:

Explanation:

We state our null and alternative hypothesis as

H0: μ1=μ2 against Ha: μ1 ≠ μ2

The level of significance is α=0.1

Since the population have unequal variances , the test statistic if H0 is true ,is

t`= x1`- x2`/ √s₁²/n1 + s₂²/n2

which has approximately a t- distribution with υ degrees of freedom where

υ = [s₁²/n1 + s₂²/n2]²/ (s₁²/n1 )²/ n1-1 + (s₂²/n2)²/n2-1

Computations:

t`= x1`- x2`/ √s₁²/n1 + s₂²/n2

t`= 46-31/ √4²/9 + 5²/12

t`= 15/√16/9 +25/12

t`= 15/ √64+75/36

t`=15/1.9649

t`=7.6339

t`= 7.63

υ = [s₁²/n1 + s₂²/n2]²/ (s₁²/n1 )²/ n1-1 + (s₂²/n2)²/n2-1

= [139/36]²/ (16/9)²/8 +(25/12)²/11

= 14.91/ 0.395 + 0.395

= 18.885

≈19 The degrees of freedom is always rounded in this calculation.

The critical region is ║t`║≥ t 0.05,19 = 1.729

Since the calculated value of t` falls in the critical region we reject our null hypothesis of equal means.