Question

For the given table of observed frequencies:

	1	2	3
A	29	8	14
B	9	10	9
C	47	8	10

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(a) Compute the row totals, the column totals, and the grand total.

(b) Construct the corresponding table of expected frequencies.

(c) Compute the value of the chi-square statistic.

(d) How many degrees of freedom are there?

(e) If appropriate, perform a test of independence, using the =α0.05 level of significance. If not appropriate, explain why.

2. HOW DO I FIND P VALUE WITH TI84?

The numbers of false fire alarms were counted each month at a number of sites. The results are given in the following table. Test the hypothesis that false alarms are equally likely to occur in any month. Use the =α0.05 level of significance and the P-value method with the TI-84 Plus calculator.

Month	Number of Alarms
January	44
February	28
March	47
April	27
May	26
June	43
July	37
August	32
September	44
October	26
November	32
December	46

3.

A statistics teacher claims that, on the average, 10%of her students get a grade of A, 20% get a B, 10% get a C, 35%get a D, and 25%get an F. The grades of a random sample of 100 students were recorded. The following table presents the results. Test the hypothesis that the grades follow the distribution claimed by the teacher. Use the 0.10 level of significance and the P-value method with the TI-84 Plus calculator.

Grade	A	B	C	D	F
Observed	8	8	15	46	23

Find the P Value with TI-84?

Answer 1

1)

a) The observed frequencies are

	1	2	3	Total
A	29	8	14	51
B	9	10	9	28
C	47	8	10	65
Total	85	26	33	144

b) The expected frequencies are

	1	2	3	Total
A	30.104	9.208	11.688	51.000
B	16.528	5.056	6.417	28.001
C	38.368	11.736	14.896	65.00
Total	85	26	33	144

c) The chi-square contribution values are

Oi	Ei	(Oi-Ei)^2 /Ei
29	30.104	0.0405
8	9.208	0.1585
14	11.688	0.4573
9	16.528	3.4288
10	5.056	4.8345
9	6.417	1.0397
47	38.368	1.942
8	11.736	1.1893
10	14.896	1.6092
	Total:	14.6998

Test Statistic, X^2: 14.7

d) Degrees of freedom: 2*2 = 4

P-Value: 0.0054

since P-value < alpha 0.05 so we reject H0

Thus we conclude that two attributes are not independent

2)
H0: false alarms are equally likely to occur in any month
H1: false alarms are not equally likely to occur in any month

Let the los be alpha = 0.05

From the given data

	Observed	Expected
Month	Freq (Oi)	Freq (Ei)
January	44	36
February	28	36
March	47	36
April	27	36
May	26	36
June	43	36
July	37	36
August	32	36
September	44	36
October	26	36
November	32	36
December	46	36
Total:	432	432

The expected frequencies is calculated as 432 / 12 = 36

TI-84: Stat -> Edit ->

Enter observed frequencies and expected frequencies in L1 and L2

Press: STAT, TESTS, Chi-squareGOFTest, enter df, Calculate

The output be

Since Chi-square test Statistic = 21.5556

P-value = 0.0281

df = 12 - 1 = 11

since P-value < alpha 0.05 so we reject H0

Thus we conclude that false alarms are not equally likely to occur in any month

3) H0: the grades follow the distribution claimed by the teacher

H1: the grades not follow the distribution claimed by the teacher

Let the los be alpha = 0.10

From the given data

	Observed		Expected
Grade	Freq (Oi)	Probability	Freq. (Ei)
A	8	0.10	10
B	8	0.20	20
C	15	0.10	10
D	46	0.35	35
E	23	0.25	25
Total:	100	1.00	100

TI-84: Stat -> Edit ->

Enter observed frequencies and expected frequencies in L1 and L2

Press: STAT, TESTS, Chi-squareGOFTest, enter df, Calculate

The output be

Chi-square Test Statistic = 13.717

P-value = 0.00825

Since P-value < alpha 0.10 so we reject H0

Thus we conclude that  the grades follow not the distribution claimed by the teacher