Question

1) The proportion of people at shopping mall who wear red shirts is p = 0.18. What is the mean of the sampling distribution given a sample size of 50? What is the standard deviation? (Central Limit Throrem).

2) What is the probability given a random sample of 50 people at shopping mall that more than 20% of them are wearing red shirts?

3) CONF. INT Suppose you take a random sample of 50 people at shopping mall and find that 9 were wearing red shirts. Construct a 90% confidence interval for the true proportion.

4) CONF. INT] Suppose you take a random sample of 50 people at shopping mall and find that 4 were wearing red shirts. Construct a 95% confidence interval for the true proportion.

5) [CONF. INT] If the true proportion is p = 0.18, which of our confidence intervals contains it? Which don’t?

Answer 1

1)

mena = 0.18

standard deviation = sqrt(0.18 *(1-0.18)/50)
= 0.0543

2)\
Here, μ = 0.18, σ = 0.0543 and x = 0.2. We need to compute P(X >= 0.2). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (0.2 - 0.18)/0.0543 = 0.37

Therefore,
P(X >= 0.2) = P(z <= (0.2 - 0.18)/0.0543)
= P(z >= 0.37)
= 1 - 0.6443 = 0.3557

3)

sample proportion, = 0.18
sample size, n = 50
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.18 * (1 - 0.18)/50) = 0.05433

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, Zc = Z(α/2) = 1.64

Margin of Error, ME = zc * SE
ME = 1.64 * 0.05433
ME = 0.0891

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.18 - 1.64 * 0.05433 , 0.18 + 1.64 * 0.05433)
CI = (0.091 , 0.269)

4)

sample proportion, = 0.08
sample size, n = 50
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.08 * (1 - 0.08)/50) = 0.03837

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

Margin of Error, ME = zc * SE
ME = 1.96 * 0.03837
ME = 0.0752

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.08 - 1.96 * 0.03837 , 0.08 + 1.96 * 0.03837)
CI = (0.005 , 0.155)

5)

90% interval contain sand 95% not