Question

The average family size was reported as 3.18. A random sample of families in a particular school district resulted in the following family sizes:

5 4 5 4 4 3 6 4 3 3 5 6 3 3 2 7 4 5 2 2 2 3 5 2

Use a 0.05 significance level to test the claim that the mean family size differs from 3.18.

a)State the claim and opposite symbolically.

b) State the Null and alternate hypotheses symbolically.

c) Identify the significance level.

d) Find the P-value

e) State the decision

f) State the conclusion

Answer 1

Solution :

We are given a data of sample size n = 24

5,4,5,4,4,3,6,4,3,3,5,6,3,3,2,7,4,5,2,2,2,3,5,2

Using this, first we find sample mean() and sample standard deviation(s).

=   

= (5 + 4.......+ 2)/24

= 3.83

Now ,

s=   

Using given data, find Xi - for each term.Take square for each.Then we can easily find s.

s = 1.43

1) The null and alternative hypothesis are

H0 : = 3.18

Ha :    3.18

2) The test statistic t is

t =   = [3.83 - 3.18]/[1.43 /24] = 2.227

The value of the test​ statistic t = 2.227

3) Now , d.f. = n - 1 = 24 - 1 = 23

sign in Ha indicates that the test is TWO TAILED.

t = 2.227

So , using calculator ,

p value = 0.036

4) p value is less than the significance level 0.05.

Decision: Reject the null hypothesis H0

5) The critical value approach

Two tailed test

Critical value is  ,df

.df  = 0.05,23 = 2.069

The critical value +/- 2.069

Rejection region,

>   .df

= 2.227 >   .df  = 2.069

Decision: Reject the null hypothesis H0

Conclusion : There is sufficient evidence to support the claim that the mean family size differs from 3.18.