Question

A random sample of 35 undergraduate students who completed two years of college were asked to take a basic mathematics test. The mean and standard deviation of their scores were 75.1 and 12.8, respectively. In a random sample of 50 students who only completed high school, the mean and standard deviation of the test scores were 72.1 and 14.6, respectively In order to test the equal variance assumption for two populations, Can we assume population variances are equal at the 10% significance level? (sigma subscript 1 superscript 2 space equals space sigma subscript 2 superscript 2 )

Answer 1

Answer:

The 90 % confidence limits are (-2.09, 8.09).

Since the calculated values do not lie in the critical region we accept our null hypothesis.

Step-by-step explanation:

The null and alternative hypothesis are given by

H0: σ₁²= σ₂² against Ha: σ₁² ≠ σ₂²

Confidence interval for the population mean difference is given by

(x`1- x`2) ± t √S²(1/n1 + 1/n2)

Where S ²= (n1-1)S₁² + S²₂(n2-1)/n1+n2-2

Critical value of t with n1+n2-2= 50+ 35-2= 83 will be -1.633

Now calculating

S ²=34* (12.8)²+ (14.6)²*49/83= 192.96

Now putting the values in the t- test

(75.1 -72.1) ± 1.633 √ 192.96(1/35 +1/50)

=3 ±  5.09

=-2.09, 8.09  is the 90 % confidence interval for the difference

The 90 % confidence limits are (-2.09, 8.09).

Since the calculated values do not lie in the critical region we accept our null hypothesis.