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Chapter 9: Problem 57

(a) According to a report in the Roanoke Times \& World-News, approximately \(2 / 3\) of the 1600 adults polled by telephone said they think the space shuttle program is a good investment for the country. Find a \(95 \%\) confidence interval for the proportion of American adults who think the space shuttle program is a good investment for the country. (b) What can we assert with \(95 \%\) confidence about the possible size of our error if we estimate the proportion of American adults who think the space shuttle program is a good investment to be \(2 / 3 ?\)

Short Answer

Expert verified
The 95% confidence interval for the proportion of American adults who support the space shuttle program is given by \((p - 1.96*SE , p + 1.96*SE)\). The possible error of the estimate \(2/3\) is given by \(1.96*SE\).

Step by step solution

01

Determine the sample proportion

This is given by the fraction of adults who support the space shuttle program. This is given as \(2/3\). Let this be denoted by \(p\).
02

Determine the sample size

The exercise provides that 1600 adults were polled. Therefore, the sample size, denoted by \(n\), is 1600.
03

Calculate the standard error of the proportion

The standard error (SE) is given by the formula \(\sqrt{p(1-p)/n}\). Here \(p = 2/3\) (the sample proportion) and \(n = 1600\) (the sample size). Thus, SE = \(\sqrt{(2/3)*(1-(2/3))/1600}\).
04

Find the 95% confidence interval

Here, you need to subtract and add the value of \(1.96*SE\) from the sample proportion \(p\) to find the confidence interval. The number \(1.96\) is the critical value (z-value) for a 95% confidence interval. So the confidence interval is \((p - 1.96*SE , p + 1.96*SE)\).
05

Discuss the Meaning of the Confidence Interval

The confidence interval can be interpreted as follows: We can be 95% confident that the true proportion of American adults who support the space shuttle program is between the two values given by the confidence interval.
06

Discuss the Size of the Possible Error

The size of the possible error can be inferred from the width of the confidence interval. The larger the width, the greater the possible error. The possible error is given by the expression \(1.96*SE\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sample Proportion
Understanding the sample proportion is essential when conducting surveys or polls. It's a measure of frequency - specifically, it represents the percentage of the sample in which a certain characteristic is present.

In the context of the given exercise, the sample proportion is the fraction of adults who believe the space shuttle program is a good investment for the country. This was found to be \(2/3\), as roughly two-thirds of the 1600 adults polled expressed their support. It's crucial to identify this sample proportion before proceeding with any further statistical analysis because it serves as the basis for establishing confidence intervals and understanding public opinion.

It's also worth highlighting that while the sample proportion can give us a good estimate of the true proportion in the entire population, it's still an estimate and subject to uncertainty, which is why we go the extra step to calculate a confidence interval around it.
Standard Error
The concept of standard error (SE) is pivotal in statistics, as it indicates the precision of our sample proportion in estimating the true population proportion. In simpler terms, it reflects how much we would expect our sample estimate to vary if we were to take many samples.

The formula for the standard error of the sample proportion is \(\sqrt{p(1-p)/n}\), where \(p\) is the sample proportion and \(n\) is the sample size. For our exercise, using the given values \(p = 2/3\) and \(n = 1600\), the calculation of SE is a straightforward application of this formula.

The smaller the standard error, the more confident we can be that our sample proportion closely estimates the true population proportion. High precision is crucial for reliable statistics, which underlines the importance of a well-sized sample to minimize the standard error.
95% Confidence Level
The 95% confidence level plays a star role in statistical analysis, giving us the range within which we can expect the true population proportion to fall.

Specifically, after determining the SE, we calculate the confidence interval by taking the sample proportion and adding and subtracting the quantity \(1.96 \times SE\). The number 1.96 comes from the Z-distribution and corresponds to the 95% confidence level鈥攎eaning that if we were to repeat our survey many times, 95 out of 100 sample proportions would fall within this calculated range around the true population proportion.

The takeaway here is not that we are 95% certain the true proportion is within this exact interval, but rather, we have a method that produces intervals which include the true proportion 95% of the time. The 95% confidence level is chosen frequently because it strikes a balance between being sufficiently confident in our results and having an interval that is not too wide to be impractical for making decisions or inferences.

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Most popular questions from this chapter

If \(\mathrm{A}^{\prime}\) is a binomial random variable, show that (a) \(P=X / n\) is an unbiased estimator of \(p ;\) (b) \(P^{\prime}=\frac{X+\sqrt{n} / 2}{n+w}\) is a biased estimator of \(p\).

Consider the statistic \(S_{p}^{2}\), the pooled estimate of \(\sigma^{2}\). The estimator is discussed in Section 9.8 . It is used when one is willing to assume that \(\sigma_{1}^{2}=\sigma_{2}^{2}=a^{2}\). Show that the estimator is unbiased for \(a^{2}\) (i.e., show that \(E\left(S^{2}\right),=a^{2}\) ). You may make use of results from any theorem or example in Chapter 9.

An automotive company is considering two types of batteries for its automobile. Sample information on the life of the battery is being used. Twenty batteries of type \(A\) and twenty batteries of type \(B\) are being used. The summary statistics are \(X A=32.91\), \(x_{B}=30.47, s_{A}=1.57,\) and \(s_{B}-1.74 .\) Assume the data on each battery are normally distributed and assume \(\sigma_{A}=\sigma_{B}\) (a) Find a \(95 \%\) confidence interval on \(\mu_{A}-\mu_{B}\). (b) Draw some conclusion from (a) that provides some insight into whether \(A\) or \(B\) should be adopted.

Compute a \(98 \%\) confidence interval for the proportion of defective items in a process when it is found that a sample of size 100 yields 8 defectives.

A manufacturer of car batteries claims that his batteries will last, on average, 3 years with a variance of 1 year. If 5 of these batteries have lifetimes of 1.9 \(2.4,3.0,3.5,\) and 4.2 years, construct a \(95 \%\) confidence interval for \(\sigma^{2}\) and decide if the manufacturer's claim that \(\sigma^{2}=1\) is valid. Assume the population of battery lives to be approximately normally distributed.
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