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Online News In a Harris poll of 2036 adults, 40% said that they prefer to get their news online. Construct a 95% confidence interval estimate of the percentageof all adults who say that they prefer to get their news online. Can we safely say that fewer than 50% of adults prefer to get their news online?

Short Answer

Expert verified

The 95% confidence interval estimate of the percentage of all adults who say that they prefer to get their news online is 37.9%< p< 42.1.

Yes, you can safely say that fewer than 50% of adults prefer to get their news online.

Step by step solution

01

Given information

The number of adults is n=2036.

The percentage of adults who prefer to get their news online is 40%.

The level of confidence is 95%.

02

Compute the confidence interval

From the given information, the following points can be drawn:

The proportion of adults who prefer to get their news online isp^=0.40.

The level of confidence is 95%, which implies that the level of significance is 0.05.

From the Z table, the two-tailed critical value obtained at 0.05 level of significance is 1.96.

The margin of error is computed as follows:

role="math" localid="1648195038482" E=zα2p^(1-p^)n=1.96×0.40×1-0.402036=0.0213

Therefore, the margin of error is 0.213.

The 95% confidence interval is computed as follows:

role="math" localid="1648195006699" (p^-E,p^+E)=0.40-0.0213,0.40+0.0213=0.3787,0.42130.379,0.421

Therefore, the confidence interval estimate of the percentage of all adults who say that they prefer to get their news online is37.9% <p <42.1%.

03

Provide the conclusion

Since the actual percentage for the population of adults is within the confidence limits of 37.9% and 42.1%,it can be said that fewer than 50% of adults prefer to get their news online.

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