Answer:
a)
So the rejection zone would be given by In our case since the estimated proportion is 0.08 we are not on the rejection zone so we FAIL to reject the null hypothesis.
b) prop.test(40, 500, p = 0.05,alternative = c("less"),conf.level = 0.95, correct = FALSE)
With the following output
1-sample proportions test without
continuity correction
data: 40 out of 500, null probability 0.05
X-squared = 9.4737, df = 1, p-value = 0.999
alternative hypothesis: true p is less than 0.05
95 percent confidence interval:
0.000000 0.102291
sample estimates:
p
0.08
Step-by-step explanation:
Data given and notation
n=500 represent the random sample taken
X=40 represent the customers who would qualify for membership
estimated proportion of adults that said that it is morally wrong to not report all income on tax returns
is the value that we want to test
represent the significance level
z would represent the statistic (variable of interest)
Alternative hypothesis:
When we conduct a proportion test we need to use the z statisitc, and the is given by:
(1)
The One-Sample Proportion Test is used to assess whether a population proportion is significantly different from a hypothesized value
.
Part a
The population proportion have the following distribution
And we know that the z score is given by:
And we are interested to find the value of
And if we solve for
So the rejection zone would be given by In our case since the estimated proportion is 0.08 we are not on the rejection zone so we FAIL to reject the null hypothesis.
Part b
prop.test(40, 500, p = 0.05,alternative = c("less"),conf.level = 0.95, correct = FALSE)
With the following output
1-sample proportions test without
continuity correction
data: 40 out of 500, null probability 0.05
X-squared = 9.4737, df = 1, p-value = 0.999
alternative hypothesis: true p is less than 0.05
95 percent confidence interval:
0.000000 0.102291
sample estimates:
p
0.08