Question

Two assemblages of lithic scrapters were excavated from different levels in a cave site

Level 1:

Sample mean: 14.83

Variance: 7.05

Sample Size: 63

Level 2:

Sample Mean: 14.17

Variance: 5.49

Sample Size: 47

From superficial examination, they seem to differ. However, your job is to test whether the difference between the samples from the two levels represent a significant difference between the two occupations. Lets use a 0.05 significance level for this analysis

Step 1: Frame this is a way amenable to testing, with both research and null hypotheses.  

Step 2: Calculate the pooled standard error of means

Step 3: Calculate the t value

Step 4: Calculate the degrees of freedom, and find the corresponding critical t-value from a t-table (keep in mind that we are using the two-tailed test).

Step 5: Compare the calculated and critical values. What is your conclusion?  

Answer 1

Solution:

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 hypotheses for this test are given as below:

Null hypothesis: H0: There is no significant difference between the two occupations.

Alternative hypothesis: Ha: There is a significant difference between the two occupations.

H0: µ1 = µ2 versus Ha: µ1 ≠ µ2

This is a two tailed test.

Test statistic formula for pooled variance t test is given as below:

t = (X1bar – X2bar) / sqrt[Sp2*((1/n1)+(1/n2))]

Where Sp2 is pooled variance

Sp2 = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)

We are given

X1bar = 14.83

X2bar = 14.17

S1^2 = 7.05

S2^2 = 5.49

n1 = 63

n2 = 47

df = n1 + n2 – 2 = 63 + 47 – 2 = 108

α = 0.05

Critical value = -1.9822 and 1.9822

(by using t-table)

Sp2 = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)

Sp2 = [(63 – 1)*7.05 + (47 – 1)*5.49]/(63 + 47 – 2)

Sp2 = 6.3856

Pooled standard error = sqrt[Sp2*((1/n1)+(1/n2))]

Pooled standard error = sqrt[6.3856*((1/63)+(1/47))]

Pooled standard error = 0.4871

t = (X1bar – X2bar) / sqrt[Sp2*((1/n1)+(1/n2))]

t = (14.83 – 14.17)/ 0.4871

t = 1.3551

Test statistic = 1.3551

Test statistic value is lies between critical values -1.9822 and 1.9822.

So, we do not reject the null hypothesis

There is not sufficient evidence to conclude that there is a significant difference between the two occupations.