A local cable company claims that the proportion of people who have Internet access is less than 63%. To test this claim, a random sample of 800 people is taken and its determined that 478 people have Internet access. The following is the setup for this hypothesis test: H0:p=0.63 Ha:p<0.63 Find the p-value for this hypothesis test for a proportion and round your answer to 3 decimal places.

Step-by-step explanation:

For the null hypothesis,

H0 : p = 0.63

For the alternative hypothesis,

Ha : p < 0.63

This is a left tailed test

Considering the population proportion, probability of success, p = 0.63

q = probability of failure = 1 - p

q = 1 - 0.63 = 0.37

Considering the sample,

Sample proportion, P = x/n

Where

x = number of success = 478

n = number of samples = 800

P = 478/800 = 0.6

We would determine the test statistic which is the z score

z = (P - p)/√pq/n

z = (0.6 - 0.63)/√(0.63 × 0.37)/800 = - 1.76

From the normal distribution table, the area below the test z score in the left tail 0.039

Thus

p = 0.039

idk

step-by-step explanation:

answer: $1794

step-by-step explanation: