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# Statistics: Confidence intervals - don't understand problems?

The public opinion is split 65% against and 35% for increasing taxes to help balance the federal budget. Suppose also that 600 people from the population are selected randomly and interviewed.

a. Suppose that in the sample of 600, 120 people are in favour of increasing tax. Based on this information construct a 95% confidence interval for the true value of the proportion of people who are in favour of increasing taxes and remember you know its actual value is p = 35%.

b. Was your 95% confidence interval successful in capturing the value p = 35%, which is the true value of the proportion of people who are in favour of increasing taxes? How would you explain this?

Can someone teach me how to do these?

### 2 Answers

- Mike GLv 76 months agoFavorite Answer
a)

[p = actual value of the population proportion = 0.35]

p̂ = proportion of the sample in favour = 120/600 = 0.2

(1-p̂) = 0.8

n = sample size = 600

σ(sample) = √[p̂(1-p̂)/600)] = √[0.2*0.8/600] = 0.0163

From the z-table 95% level gives z = 1.96

Margin of error = 1.96*0.0163 = 0.032

0.2-0.032 = 0.168

0.2+0.032 = 0.232

0.168 ≤ p ≤ 0.232

b) No the populatio proportion was not captured

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- Anonymous6 months ago
I just finished this unit in stats. The premise of a 95% confidence interval is that we are 95% confident that the true proportion lies between two values. In this case, one must find what the proportion 120/600 in the given sample can tell us about the true proportion of the population who is in favor of raising taxes.

First, calculate the decimal that 120/600 represents. Then you have to find margin of error, given by the formula:

E=z(a/2)*((p̂q̂)/n)^1/2

where

z(a/2) is the critical value that separates an area of a/2 in the right tail of the standard normal distribution. a=1-confidence level (.95)

n is the sample size, in this case 600

p̂ is the sample proportion (expressed as a decimal), in this case 120/600

q̂ is simply 1-p̂, in this case 480/600

In this case, it is simply known from the critical value [z(a/2)] is 1.96. This can be proven with a graphing calculator or some other technology.

So your confidence interval is shown by p̂-E<p<p̂+E. In words, the true population proportion should lie between 1 margin of error in either direction of your sample proportion. Calculate the confidence interval. If it does capture the true value p=.35, then the sample is a good representation of the population. If not, then it isn't.

Remember to convert your percentages to decimals, and good luck!

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