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Friends, Updating here the recollected questions from June 2018 Exams. Wish you all the very best for your exam.
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In a survey of 150 people in a city, it was found that there were 50 smokers. Calculate the following based on the above data.
1. The estimate of population proportion
a. 0.3333
b. 0.5
c. 0.6666
d. 1.0
2. Estimated standard error of population
a. 0.0015
b. 0.2211
c. 0.0385
d. 0.4725
3. Binominal standard deviation of population
a. 1.76
b. 2.26
c. 5.77
d. 7.87
4. 95% confidence interval level of population proportion
a. 0.4326, 0.2340
b. 0.5468, 0.3178
c. 0.4088, 0.2578
d. 0.5568, 0.2778
5. 99% confidence interval level of population proportion
a. 0.4326, 0.2340
b. 0.5468, 0.3178
c. 0.4088, 0.2578
d. 0.5568, 0.2778
Ans - 1-a, 2-c, 3-c, 4-c, 5-a
Solution :
1. P = 50/150 = 0.3333
2. Estimated standard error = sqrt [P (1-P) / n]
P = 0.3333
1-P = 1 - 0.3333 = 0.6666
n = 150
So, sqrt [P (1-P) / n] = sqrt [0.3333 x 0.6666 / 150]
= sqrt (0.2222 / 150)
= sqrt (0.00148)
σx = 0.0385
3. Binominal standard deviation = √n*p*q
= √150*.333*.666
= √33.27
= 5.77
4. Level of Confidence and their Multiplier Number (z*) (Commonly used)
99% - 2.58
95% - 1.96
90% - 1.645
95% confidence interval = P +/- (1.96 (σx))
= 0.3333 + (1.96 x 0.0385)
= 0.3333 + 0.0755
= 0.4088
and
= 0.3333 - (1.96 x 0.0385)
= 0.3333 - 0.0755
= 0.2578
5. 99% confidence interval = P +/- (2.58 (σx))
= 0.3333 + (2.58 x 0.0385)
= 0.3333 + 0.0993
= 0.4326
and
= 0.3333 - (2.58 x 0.0385)
= 0.3333 - 0.0993
= 0.2340
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