Question

39. Telephone Households In 1920 only 35% of U.S. households had telephones, but that rate is now much higher. A recent survey of 4276 randomly selected households showed that 94% of them had telephones (based on data from the U.S. Census Bureau). Using those survey results, construct a 99% confidence interval estimate of the proportion of households with telephones. Given that the survey involves only 4276 households out of 115 million households, do we really have enough evidence to say that the percentage of households with telephones is now more than the 35% rate in 1920?

Answer 1

Given :

Sample size = n = 4276

Sample proportion = ${\hat{p}}$ = 0.94

Significance level = $\alpha$ = 1-0.99 = 0.01

At 0.01 significance level, the critical value of Z is

2012= 20.0112 = 2.58 99./. CI; CI-P + Zal2x PC1-) * &12[picines 0.94 = 0.94 + 2.58x0.9461-0.94) 4276 = 0.94 +0.0094 - (0.9306, 0.9494) :-:99.lo contidence interval q proportion is co.9306,0.9494) Gwen år ns4276 x = 115 È = x - 115 = 0.0269 4276 .99.j. CI! CIS Ñ I Z412 x PCI-P) n 0.0269 + 2.58 2.58% 0.0269 C1-0.0269) 4 276 -0.0269 +0.0064 =c0.0205, 0.0333)

99% confidence interval is (0.0205, 0.0333)

Since confidence interval does not contain the hypothesized population proportion 0.35, we reject the null hypothesis.

We have enough evidence to say that the percentage of households with telephones is now more than the 35% rate in 1920.