# 10.67 The Pew Research Center conducted a poll in January 2014 ofonline adults who use social networking sites. According to this poll,89% of the 18-29 year olds and 82% of the 30-49 year olds who areonline use social networking sites (www.pewinternet.org). Supposethat this survey included 562 online adults in the 18-29 age group and624 in the 30-49 age group.andbe the proportion of all online adults in thea. LetP2P1age groups 18-29 and 30-49, respectively, who use socialnetworking sites. Construct a 95% confidence interval forP1 P2b. Using a 1% significance level, can you conclude that pi isdifferent from p2? Use both the critical-value and the p-valueapproaches

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Step 1

Solution:

a.Obtain the critical value at 95% confidence level:

From the given information, the proportion of olds between 18 and 29 years is 0.89 and proportion of olds between 30 and 49 years is 0.82.

The sample size of persons between 18 and 29 is 562 and sample size of persons between 30 and 49 years is 624.

The level of significance=(1–0.95)/2=0.025.

The required confidence level from left tail normal table=0.95+0.025=0.9750.

Procedure for finding the z-value is listed below:

1. From the table of standard normal distribution, locate the probability value of 0.9750.
2. Move left until the first column is reached. Note the value as 1.9
3. Move upward until the top row is reached. Note the value as 0.06.
4. The intersection of the row and column values gives the area to the two tail of z.

The critical value Z=1.96.

Step 2

The 95% confidence interval for p1 – p2 is calculated below:

Step 3

b.

The hypotheses are stated below:

Null hypothesis:

Ho=p1 = p2

Alternative Hypothesis:

Ha=p1 ≠ p2

The statistic value is obtained below:

...

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