Question

In making plans for an executive traveler’s club, an airline would like to estimate the proportion of its current customers who would qualify for membership. A random sample of 500 customers yielded 40 who would qualify.
(a) Test the hypothesis H0 : p = 0.05 vs Ha : p < 0.05 by constructing reject region at α = 0.1 state what action the airline should take.
(b) Compute the p-value using R.

Answer 1

Answer:

a) $\hat p = 0.05 -1.28 \sqrt{\frac{0.05 (1-0.05)}{500}}=0.0375$

So the rejection zone would be given by $\hat p$ 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

$\hat p=\frac{40}{500}=0.08$ estimated proportion of adults that said that it is morally wrong to not report all income on tax returns

$p_o=0.05$ is the value that we want to test

$\alpha$ represent the significance level

z would represent the statistic (variable of interest)

$p_v{/tex} represent the p value (variable of interest) Concepts and formulas to use We need to conduct a hypothesis in order to test the claim that the true proportion is less than 0.05.: Null hypothesis:[tex]p\geq 0.05$  

Alternative hypothesis:$p < 0.05$  

When we conduct a proportion test we need to use the z statisitc, and the is given by:  

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)  

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$.

Part a

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

And we know that the z score is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$

And we are interested to find the value of $hat p[\tex] in order to reject the null hypothesis. So we need to find a z value that accumulates 0.1 of the area on the left and this value is [tex]z_{0.1}=-1.28$

$-1.28=\frac{\hat p -0.05}{\sqrt{\frac{0.05 (1-0.05)}{500}}}$

And if we solve for $hat p[\tex] we got:[tex]\hat p = 0.05 -1.28 \sqrt{\frac{0.05 (1-0.05)}{500}}=0.0375$

So the rejection zone would be given by $\hat p$ 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