Question

In a sample of 200 Bend households, you nd that 110 have cable television. In a sample of 250 Portland households, you nd that 160 have cable television. You wish to know whether these data indicate that the percentage of Bend households (population 1) with cable is less than the percentage of Portland households (population 2) with cable. Use  = 0:05 in the hypothesis test.

1. Set up the null and alternative hypotheses and perform test statistics using normal model.

2. Are the conditions met for using the normal distribution? Explain.

3. Redo the hypothesis test using the R function prop.test, which does not use the normal model. Compare and explain the p-value of the test done using this method with the one using the normal model.

I've already ran the code and the results are below!

Results from Rstudio:

data: c(110, 160) out of c(200, 250)
X-squared = 3.3844, df = 1, p-value = 0.06582
alternative hypothesis: two.sided
95 percent confidence interval:
-0.185571915 0.005571915
sample estimates:
prop 1 prop 2
0.55 0.64

Answer 1

This is left tailed test you use two tailed in your code.

if we campare results of p.value for part a and c ,p.value in part c is slightly high but conclusion doesnt change we reject Ho in both parts