A study is designed to test Ho: P-0.50 against H: p>0.50, taking a random sample
ID: 3371650 • Letter: A
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.Explanation / Answer
Solution
Back-up Theory
To test hypotheses: Null H0 : p = p0 Vs HA : 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
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)
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