Question

Suppose in July of 2008, the true proportion of U.S. adults that thought unemployment would increase was 47%. In November of 2008, the same question was asked to a simple random sample of 1000 U.S. adults and 432 of them thought unemployment would increase. Can we conclude that the true proportion of U.S. adults that thought unemployment would increase in November is less than the proportion in July? Use a 5% significance to test.
Round to the fourth
H0: Select an answer x̄ p̂ μ p  Select an answer = < > ≠  
HA: Select an answer x̄ p̂ μ p  Select an answer = < > ≠  
What's the minimum population size required?
How many successes were there?
Test Statistic:
P-value:
Did something significant happen? Select an answer Significance Happened Nothing Significant Happened
Select the Decision Rule: Select an answer Reject the Null Accept the Null Fail to Reject the Null
Select an answer: is or is not  enough evidence to conclude Select an answer that the true proportion of U.S. adults that thought unemployment would increase in November is less than 0.47 that the true proportion of U.S. adults that thought unemployment would increase in November is more than 0.47 that the true proportion of U.S. adults that thought unemployment would increase in November is 0.47

Build a 90% confidence interval and decide if you can conclude the same. Use your calculator to do this and round to the fourth decimal place.
( , )
Can we conclude the same as our Hypothesis Test?
Select an answer no yes  because the true proportion of U.S. adults in November 2008 that thought unemployment would increase

Answer 1

Below are the null and alternative Hypothesis,
Null Hypothesis: p = 0.47
Alternative Hypothesis: p < 0.47

432 are the successes

1000 is the population size

Test statistic,
z = (pcap - p)/sqrt(p*(1-p)/n)
z = (0.432 - 0.47)/sqrt(0.47*(1-0.47)/1000)
z = -2.4077

P-value Approach
P-value = 0.0080

Significance Happened

Reject the Null

is or enough evidence to conclude Select an answer that the true proportion of U.S. adults of U.S. adults that thought unemployment would increase in November is less than 0.47

sample proportion, = 0.432
sample size, n = 1000
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.432 * (1 - 0.432)/1000) = 0.0157

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 * 0.0157
ME = 0.0257

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.432 - 1.64 * 0.0157 , 0.432 + 1.64 * 0.0157)
CI = (0.4063 , 0.4577)

yes because the true proportion of U.S. adults in November 2008 that thought unemployment would increase