Question

4. Where people turn for news is different for various age groups. (Data extracted from "Cellphone Users Who Access News on Their Phones," USA Today, March 1, 2010, p. 1 A.) A study was conducted on the use of cellphones for accessing news. The study reported that 47% of users under age 50 and 15% of users age 50 and over accessed news on their cellphones. Suppose that the survey consisted of 1,000 users under age 50, of whom 470 accessed news on their cellphones and 891 users age 50 and over, of whom 134 accessed news on their cellphones. a. Is there evidence of a significant difference in the proportion of users under age 50 and users 50 years and older that accessed the news on their cellphones? (use a -0.05). b. Determine the p-value in (a) and interpret its meaning. c. Construct and interpret a 95% confidence interval estimate for the difference between the population proportion of users under 50 years old and those 50 years or older who access the news on their cellphones.

Answer 1

(a)

n1 = 1000

p1 = 470/1000 = 0.47

n2 = 891

p2 = 134/891 = 0.1504

Pooled Proportion is given by:

P= nlpl + n2p2 470 + 134 nl + n21000 + 891 = 0.3194

q=1-p=0.6806

SE= 0.3194 x 0.6806 x (1/1000 + 1/891) = 0.0215

Test Statistic is given by:

Z = (0.47 - 0.1504)/0.0215

= 14.8651

$\alpha$ = 0.05

From Table, critical values of Z = $\pm$ 1.96

Since calculated value of Z = 14.8651 is greater than critical value of Z = 1.96, the difference is significant. Reject null hypothesis.

Conclusion:
The data support the claim that there is evidence of significant difference in the proportion of users under age 50 and users of 50 years and older that accessed the news on their cellphones.

(b)

Z score = 14.8651

By Technology, Cumulative Area Under Standard Normal Curve = 1 nearly.

So,

p - value = 1 - 1 nearly = 0.0000

Interpretation:

p - value = 0.0000 indicates strong evidence against the null hypothesis, so we reject the null hypothesis.

(c)

Confidence Interval:

(0.47 - 0.1505) $\pm$ (1.96 X 0.0215)

= 0.3195 $\pm$ 0.0421

= ( 0.2774 ,0.3616)

Confidence interval:

0.2774 < P1 - P2 < 0.3616

Interpretation:

The 95% Confidence Interval 0.2774 < P1 - P2 < 0.3616 is a range of values we are 95% certain will capture the population proportion of difference users under age 50 and users of 50 years and older that accessed the news on their cellphones.