Question

A major newspaper conducted a random survey of 1,800 of its 75,000 subscribers. The subscribers were asked if the paper should increase its coverage of national news. Out of the 1,800 in the sample, 594 responded that they want more national news.
Construct a 95% confidence interval (z*-score = 1.96) for the proportion of subscribers who would like more national news coverage.
Complete the statements.

The estimated population proportion, , is ___%.

When the margin of error is calculated using the ___

Formula E = z* , to the nearest tenth of a percent, the result is %.
With 95% confidence, it can be said that the proportion of subscribers who would like more coverage of national news is between ___% and ___%

Answer 1

Answer:

33%

2.2%

30.8%

35.2%

Step-by-step explanation:

Answer 2

Answer:

Step-by-step explanation:

Given that a  major newspaper conducted a random survey of 1,800 of its 75,000 subscribers.

The sample proportion  of subscribers who would like more national news coverage.=$\frac{594}{1800} =0.33$

Hence

a) The estimated population proportion, , is _33__%.

b)

When the margin of error is calculated using the ___formula

E = $Z*\sqrt{\frac{p(1-p)}{(n} }$

=$1,96*\frac{0.33*0.67}{800} =0.0326$

c) With 95% confidence, it can be said that the proportion of subscribers who would like more coverage of national news is between _0.33-0.0326_=29.74_% and __0.33+0.0326=36.26_%