Question

1,Suppose that 300 persons are selected at random from a large population and that each per- son in the sample is classified according to whether his blood type is O,A,B,AB and also according to whether blood type is Rh positive or Rh negative. The observed numbers are given below. Test whether blood type and Rh status are independent.

	O	A	B	AB
Rh +	82	89	54	19
Rh -	13	27	7	9

2,Let X1,...,Xn be a random sample from f(x;θ) = 1/θ e^-x/θ for x > 0.

(a) Find the method of moments estimate of θ using the data in (b).

(b) Find the maximum likelihood estimate of θ using the data in (b).

(c) What is the bias of the estimate in (c) and (d)?

3,Assume that X is following a normal distribution with mean μ and variance σ2. We want to test

H0 :μ=10vs.H1 :μ=15 based on a sample of size n. Consider the following decision rule:

If X ̄ > 13 claim that μ = 15, if X ̄ ≤ 13 ( X : mean ) claim that μ = 10. Consider n = 30 and σ2 = 5.

(a) What is the critical region of the test? (b) What is α, type I error of the test?

(c) What is the power of the test? Interpret the answer.

Answer 1

1:

Hypotheses are:

H0: Blood type and Rh status are independent.

Ha: Blood type and Rh status are not independent.

Following table shows the row total and column total:

	O	A	B	AB	Total
Rh +	82	89	54	19	244
Rh -	13	27	7	9	56
Total	95	116	61	28	300

Expected frequencies will be calculated as follows:

Following table shows the expected frequencies:

	O	A	B	AB	Total
Rh +	77.267	94.347	49.613	22.773	244
Rh -	17.733	21.653	11.387	5.227	56
Total	95	116	61	28	300

Following table shows the calculations for chi square test statistics:

O	E	(O-E)^2/E
82	77.267	0.289920522
89	94.347	0.303034638
54	49.613	0.387917864
19	22.773	0.625105564
13	17.733	1.263254328
27	21.653	1.3203902
7	11.387	1.690152718
9	5.227	2.723460685
Total		8.603236519

Following is the test statistics:

Degree of freedom: df =( number of rows -1)*(number of columns-1) = (2-1)*(4-1)=3

The p-value using excel function "=CHIDIST(8.603,3)" is: 0.0351

Since p-value is less than 0.05 so we reject the null hypothesis.