Question

2. One study wants to test whether the number of televisions in the far western United States families distribution fits the distribution of the American families. At the 1% significance leve, does it appear that the distribution "number of televisions" of far western United States families is different from the distribution for the American population as a whole? Observed Number of TV's Expected 60 96 0 бо 119 330 340 003 60 48 15

Goodness of fit. What are the steps to solve this problem?

Answer 1

	Observed(Oi)	Expected(Ei)	Difference(Oi-Ei)	Difference Sq.(Oi-Ei)2	(Oi-Ei)2.Ei
	66	60	6.00	36.00	0.60
1	119	96	23.00	529.00	5.51
2	340	330	10.00	100.00	0.30
3	60	66	-6.00	36.00	0.55
4+	15	48	-33.00	1089.00	22.69
					29.646

Based on the information provided, the significance level is α=0.01, the number of degrees of freedom is df=5−1=4, so then the rejection region for this test is R={χ²:χ²>13.277}.

Test Statistics

The Chi-Squared statistic is computed as follows:

The test statistic χ² equals 29.646

Decision about the null hypothesis

Since it is observed that χ²=29.646>χ_c²​=13.277, it is then concluded that the null hypothesis is rejected.

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that some of the population proportions differ from those stated in the null hypothesis, at the α=0.01 significance level.