Confidence Interval and Statistical Tests

Question 4
n = 277
x = 69
^p= x
n = 69
277 =0.2491
79.6% confidence interval for portion of adult female shopper who request a rain check in such
situation
= ^p ¿ ¿α/2 * √ ^p(1− ^p)
n )
Zα/2 value of 79.6% confidence level is determined as follows;
The confidence level would be subtracted from 1 and the result divided by two in order to find
the value of alpha level which will be representing the one tail area.
(1 – 0.796)/2 = 0.102
Subtract the result from one, the area is then looked at the middle of the z – table in order to
determine the value of the z score.
1 – 0.102 = 0.898
Checking it at z – table as shown
Z – score = 1.27) as highlighted on the z- table diagram below)

= ^p ¿ ¿α/2 * √ ^p(1− ^p)
n )
= 0.2491 (1.27∗
√ 0.2491(1−0.2491)
277 )− ¿
+ ¿¿ ¿
= 0.2491 0.33002− ¿+¿ ¿ ¿
79.6% CI = (0.2821, 0.2161)
b)
^p= 69
277 =0.2491
q = 1 - ^p = 1 – 0.2491 = 0.7509
Margin error = 0.1/2 = 0.05
For 95% confidence, the value of z will be equivalent to 1.96

Size of sample required will be equivalent to
n = pq*(z− score
E )2
= 0.2491 * 0.7509*( 1.96
0.05 )
2
n = 287.427
n ≈ 287 (rounded off to next whole number)
Question 7
P(number 2 or less) = 2 / 6 = 1 / 3
Mean = np = 500 * 1/3 = 166.6667
Standard deviation = √(np(1− p))
= √( 500∗1
3 ∗2/3)
= 10.5409
Using normal approximation.,
P( X ≤ x) = P( Z < x+0.5 - mean / SD)
So,
P(X≤ 150) = P( Z ≤ ¿150.5 - 166.6667 / 10.5409))
= P (Z ≤ -1.5337)
From the z-table shown on the diagram below, the probability will be equivalent to;