Question

According to the N.Y. Times Almanac the mean family size in the U.S. is 3.18. A sample of a college math class resulted in the following family sizes:

4; 5; 5; 4; 4; 4; 6; 4; 3; 2; 5; 5; 6; 3; 3; 2; 7; 4; 5; 2; 2; 2; 3; 2.

At

α = 0.05

level, is the class's mean family size greater than the national average?

Note: If you are using a Student's t-distribution for the problem, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)

What is the p-value for the test?

Answer 1

In this problem, we are given the mean family size in the US as 3.18 and a sample of 24 family sizes obtained in a math class. Here, we make the following assumption to do the sum. Let us assume that the 24 family sizes

X1, X2,...,X24 is a random sample drawn from Normal distribution with mean   and variance , where both the parameters are unknown.

To prove this above assumption, we draw a histogram using minitab which is illustrated below-

From, this histogram we see this is a symmetric distribution and hence we can assume the underlying distribution follows normal distribution.

Hypothesis tested here is,

where, = 3.18

The appropriate test statistic used is Student's t statistic given by

where, = sample mean

  

s= sample standard deviation

  

We shall reject H0 vs H1 at level iff for the given sample where .

The calculated values are -

= 3.833

s = 1.4645

Therefore,

= 2.1736

From the t- table the value of tn-1 at level 0.05 is 1.714

Here, we see T>t so we reject the null hypothesis i.e, the class's mean family size is greater than the national average.

Now, we have to calculate the p-value of the test.

we know, for a right tailed test we reject null hypothesis id T>c where c is so chosen that PH0 [T>c] = i.e, the test is a level test. p-value of the test is defined as

p-value=PH0[T>t] where, t is the observed value of T.

Here, we see the obserd value t is given by 1.714 so by using the t-table we get the p-value as 0.02.