Question

Assume that you have a sample of n1=8​, with the sample mean X1=45​, and a sample standard deviation of S1=5, and you have an independent sample of n2=17 from another population with a sample mean of X2=35 and the sample standard deviation S2=8. Complete parts​ (a) through​ (d) below.

a. What is the value of the​ pooled-variance tSTAT test statistic for testing H0: μ1=μ2​?

tSTAT =

​(Round to two decimal places as​ needed.)

b. In finding the critical​ value, how many degrees of freedom are​ there?

​(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__

​(Round to two decimal places as​ needed.)

d. What is your statistical​ decision?

A. Do not reject H0 because the computed tSTAT test statistic is less than the​ upper-tail critical value.

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

C. 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

Solution:

Part a

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

H₀: µ₁ ≤ µ₂ versus H₁: µ₁ > µ₂

This is an upper tailed test.

We assume

µ₁ = population mean for first sample

µ₂ = population mean for second sample

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 have

X1bar = 45

X2bar = 35

S1 = 5

S2 = 8

n1 = 8

n2 = 17

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

S_p² = [(8 – 1)*5^2 + (17 – 1)*8^2]/(8 + 17 – 2)

S_p² = 52.1304

(X1bar – X2bar) = 45 - 35 = 10

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

t = 10 / sqrt[52.1304*((1/8)+(1/17))]

t = 10 / 3.0956

t = 3.2304

t_STAT = 3.23

Part b

We are given

n1 = 8

n2 = 17

df = n1 + n2 – 2 = 8 + 17 - 2 = 23

Degrees of freedom = 23

Part c

We are given

Test is an upper tailed test.

Level of significance = α = 0.10

df = 23

Critical value = 1.3195

(by using t-table/excel)

Critical value = 1.32

Part d

Test statistic value is greater than Critical value = 1.3195

So, we reject the null hypothesis

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