Question

Suppose a government official in Phoenix, Arizona, wants to know what proportion of Phoenix residents are infected with the COVID-19 virus. Because of the limited availability of tests, she cannot test all the residents of Phoenix. A random sample of 180 Phoenix residents was taken, and these residents were tested for the virus. Of these 180 people, 36 were found to be infected with the virus. Find a 95% confidence interval for the true proportion of Phoenix residents who are infected. Also: Suppose the mayor of Phoenix had said that she suspected that fewer than 30 percent of Phoenix residents were infected at that time. Use a hypothesis test (with significance level 0.01) to determine whether the mayor's pronouncement is supported by these data. Verify that the sample size is large enough for the inferences in this analysis.

Answer 1

a) Sample proportion = 36 / 180 = 0.20

p̂ = X / n = 36/180 = 0.2
p̂ ± Z(α/2) √( (p * q) / n)
0.2 ± Z(0.05/2) √( (0.2 * 0.8) / 180)
Z(α/2) = Z(0.05/2) = 1.96
Lower Limit = 0.2 - Z(0.05) √( (0.2 * 0.8) / 180) = 0.142
upper Limit = 0.2 + Z(0.05) √( (0.2 * 0.8) / 180) = 0.258
95% Confidence interval is ( 0.142 , 0.258 )

b)

np0 = 180 * 0.30 = 54>= 10

n( 1 - p0) = 180 * 0.70 = 126

Since both conditions are satisfied, distribution is approximately normal.

H0: p >= 0.30 , Ha: p < 0.30

Test statistics

z = ( - p) / sqrt [ p ( 1 - p) / n ]

= ( 0.20 - 0.30) / sqrt ( 0.30 ( 1 - 0.30) / 180 )

= -2.93

p-value = P(Z < z)

= P(Z < -2.93)

= 0.0017

Since p-value < 0.01, Reject H0