Question

A study is designed to test Ho: P-0.50 against H: p>0.50, taking a random sample of size n-100, using a significance level of 0.05. Show that the rejection region consists of values of p> 0.582 a. Sketch a single picture that shows (i) the sampling distribution of p when Ho is true and (ii) the sampling distribution of p when p-0.60. Label each sampling distribution with its mean and standard error and highlight the rejection region. b. c. Find P(Type Il error) when p-0.60. d. Find the power of the study when p-0.60.

Answer 1

Solution

Back-up Theory

To test hypotheses: Null H₀ : p = p₀   Vs H_A : p > p₀, Test Statistic is

Z = (pcap - p₀)/?{ p₀(1 - p₀)/n} where pcap = sample proportion and n = sample size.

Under H₀, distribution of Z can be approximated by Standard Normal Distribution, provided

np₀ and np₀(1 - p₀) are both greater than 10.

Rejection Region is: Reject H₀, if Zcal > Zcrit, where Zcrit = upper ?% of N(0, 1) for level of significance of ?%.

Using Excel Functions of N(0, 1), Critical Value = 1.645.

Now, to work out the answer,

Part (a)

As described above, rejection region is: {(pcap – 0.5)/?(0.5 x 0.5/100)} > 1.645 or

(pcap – 0.5) > 1.645 x 0.05 or

Pcap > 0.58225 or

So, rejection region consists of all values of pcap > 0.582. DONE

Part (b)

Distribution of X = n(pcap) is B(100, p), where p = population proportion, which is given to be 0.5 under H₀ and 0.6 under H₀

The two sampling distributions are given below: [probabilities are obtained from Excel Function: BINOMDIST(Number_s:Trials:Probability_s:Cumulative]

x	pcap	Probability Under		Cum. Probability Under
		p = 0.5	p = 0.6	p = 0.5	p = 0.6
10	0.1	1.366E-17	1.604E-25	1.5316E-17	1.7297E-25
20	0.2	4.228E-10	2.864E-16	5.5795E-10	3.4204E-16
30	0.3	2.317E-05	9.0506E-10	3.9251E-05	1.2515E-09
40	0.4	0.0108439	2.4425E-05	0.02844397	4.2466E-05
50	0.5	0.0795892	0.01033751	0.53979462	0.0270992
58	0.58	0.0222923	0.07420719	0.95568696	0.37746732
60	0.6	0.0108439	0.08121914	0.9823999	0.53792466
70	0.7	2.317E-05	0.0100075	0.99998392	0.98522468
80	0.8	4.228E-10	1.0531E-05	1	0.99999412
90	0.9	1.366E-17	1.9612E-11	1	1
100	1	7.889E-31	6.5332E-23	1	1
Mean		0.5	0.6
SE		0.05	0.049
Rejection Region		In Bold		1 – (In Bold)	1 – (In Bold)

DONE

Part (c)

Probability of Type II Error = P(Accepting H₀ when H₁ is true, i.e., p = 0.6 given)

= P(pcap < 0.58225)

= 0.3775 [vide above Table last column against p = 0.58] ANSWER

Part (d)

Power = 1 - Probability of Type II Error = 0.6225 ANSWER