Hello I am working on chaper 17 13 book is Statisticsforbe

Hello - I am working on chaper 17 # 13, book is Statisticsforbehovorial sciences 9th ed. On step 5 of the problem/solution can you explain step 5 of the solution the Chi-Square statistic is calculated. I am not understing how you got 21-15.

X^{2 = (fo-fe)2}

^{(21-15)2 /15 + (19-25)2/25+(9-15)2/15+(31-25)2/25.}

Thank you.

Solution

In chi-square distribution there are many tests.

1. Independence.

2. Goodness of fit. etc

Suppose we are test independence of two variables.

Null hypothesis is,

H0: two variables are independent.

Alternative hypothesis is,

H1: two variables are not independent,

The test statistic is,

X² = (O-E)² / E

where O is observed frequency.

E is expected frequency.

E can be calculated by using,

E = (Row total) * (column total) / sample size

It follows Chi-square distribution with (R-1) * (C-1) degrees of freedom.

where R is number of rows.

C is number of columns.

Here we are given that X² = (21-15)² / 15 = 2.4

Next is we have to find critical value or P-value using degrees of freedom.

If  X² > critical value

We have to set level of significance as 0.05 or 0.01 or 0.1.

Reject H0 at 5% level of significance.