Question

3) Find the test statistics x2 for the indicated goodness-of-fit test. (Do not write out the entire test) 3) Using the data below and a 0.05 significance level, test the claim that the responses occur with percentages of 15%, 20%, 25%, 25%, and 15% respectively. Response A B C D E Frequency 12 15 16 18 19 Fill out the table below and use it to find the test statistics x2. (0-E) E 0 E O-E (0-E) x2 = given contingency table. the row variable and the column variable

Answer 1

Observed frequency is given in the tabular format and the percentages are the expected frequencies that needs to be converted

now we first need to find out the expected percentages

total number of observed sample = 12+15+16+18+19=80

So expected frequency would be expected percentage multiplied by 80 and divided by 100

O	E	O-E	(O-E)^2	(O-E)^2/E
12	12	0	0	0
15	16	-1	1	0.0625
16	20	-4	16	0.8
18	20	-2	4	0.2
19	12	7	49	4.0833

So the chi square statistic is equal to

=0+0.0625+0.8+0.2+4.08333

=5.14583

The table value is given by chi square table which LOS 0.05 and degrees of freedom n-1=4

So

$\chi ^2=9.4877$

Hence as the table value is greater than observed value, at 5% LOS we can say that responses occur with the specified percentages.

Do comment if you have any doubt.

Thank you !!