Question

Use the contingency table to the right to complete parts​ (a) through​ (c) below.

                A             B             Total

1              58           12           70

2              12           18           30

Total      70           30           100

Find the expected frequency for each cell.

	A	B	Total
1			70
2			30
Total	70	30	100

​(Type integers or​ decimals.)

b. Compare the observed and expected frequencies for each cell.

Choose the correct answer below.

1. The expected values are always greater than the observed values.

2. The totals for the observed and expected frequencies are the same.

3. The totals for the observed values are always greater than the totals for the expected values.

D. The observed values are always greater than the expected values.

c. Compute χ2STAT. Is it significant at α= 0.05​?

First let π1 be the proportion of all events of interest in​ A, and let π2 be the proportion of all events of interest in B.

Determine the hypotheses. Choose the correct answer below.

1. H0: π1≥π2

H1: π1<π2

2. H0: π1=π2

H1: π1≠π2

3. H0: π1≠π2

H1: π1=π2

4. H0: π1≤π2

H1: π1>π2

Calculate the test statistic.

χ2STAT=

​(Round to two decimal places as​ needed.)

What is the​ p-value?

​p-value=

​(Round to three decimal places as​ needed.)

Is this value significant at α=0.05​?

1. Yes, because the​ p-value is less than or equal to the alpha level.​ Therefore, the null hypothesis should be rejected.be rejected.
2. No, because the​ p-value is greater than the alpha level.​ Therefore, the null hypothesis should not be rejected.
3. No​, because the​ p-value is greater than the alpha level.​ Therefore, the null hypothesis should be rejected.be rejected.
4. Yes, because the​ p-value is less than or equal to the alpha level.​ Therefore, the null hypothesis should not be rejected.

Answer 1

given

Observed	O	A	B	Total
	1	57	13	70
	2	13	17	30
	Total	70	30	100

Expected	E=row total*column total/grand total	A	B	Total
	1	49.000	21.000	70
	2	21.000	9.000	30
	Total	70	30	100

hi square	=(O-E)^2/E	A	B	Total
	1	1.306	3.048	4.354
	2	3.048	7.111	10.159
	Total	4.354	10.159	14.512

Now the expected frequency is given in expected table

b) The total for the observed and expected frequencies are the same.

c) Now the chi square states from above =14.512

and critical value of chi statistics at 1 degree of freedom = 6.634

d) Finally p - value for above = 0.0001

So therefore the p value is < 0.01 level

Therefore we reject the null hypothesis; and

accept the test significance   =  0.01