Question

In: Statistics and Probability

3. In a Harris poll of 2036 adults, 40% said that they prefer to get their...

3. In a Harris poll of 2036 adults, 40% said that they prefer to get their news online.

a. Construct a 95% confidence interval estimate of the percentage of all adults who say that they prefer to get their news online.

b. Can we safely say that fewer than 50% of adults prefer to get their news online?

Expert Solution

Solution :

Given that,

n = 2036

Point estimate = sample proportion = = 0.40

1 - = 1- 0.40 =0.60

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96 ( Using z table )

Margin of error = E = Z/2 * $\sqrt$ ((( * (1 - )) / n)

= 1.96 ( $\sqrt$ ((0.4*0.6) /2036 )

E = 0.0213

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.40- 0.0213< p <0.40+ 0.0213

0.3787< p < 0.4213

The 95% confidence interval for the population proportion p is : 0.3787, 0.4213

orchestra answered 2 years ago

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