Question

Using = 0.05, test if the mean is different from 8.2 if a sample of size 36 yielded a mean of 7.4 and a sample std dev )s) od 4.

Answer 1

Solution:

Given the hypothesis

H0: = 8.2 ....null hypothesis

H1:     8.2 ....alternative hypothesis

n = 36 ....sample size

d.f. = n -1 = 36 -1 = 35

= 7.4 ...sample mean

s = 4 ....Sample standard deviation

= 0.05   .... level of significance

/2 = 0.025

Since population SD is unknown,we use t test.

The test statistics t is given by ..

t =  

= (7.4 - 8.2)/(4/36)

= -1.20

| t | = 1.20

Now, sign in H1 indicates that the two tailed test.

So, the critical values are   i.e.  

The critical region : t <   or t >  

  =   _0.025,35  = 2.030   (using t table)

   | t | <   

We fail to reject the null hypothesis H₀ at 0.05 level.

Not sufficient evidence to support the claim that the mean is different from 8.2.