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Chapter 8: Problem 56

Is America's romance with movies on the wane? In a Gallup Poll \(^{\star}\) of \(n=800\) randomly chosen adults, \(45 \%\) indicated that movies were getting better whereas \(43 \%\) indicated that movies were getting worse. a. Find a \(98 \%\) confidence interval for \(p\), the overall proportion of adults who say that movies are getting better. b. Does the interval include the value \(p=.50 ?\) Do you think that a majority of adults say that movies are getting better?

Short Answer

Expert verified
No, the interval does not include 0.50; thus, less than a majority believe movies are getting better.

Step by step solution

01

Identify the Population Proportion

The problem states that 45% of the survey respondents indicate movies are getting better. This gives us our sample proportion \( \hat{p} = 0.45 \).
02

Calculate the Standard Error

The standard error (SE) for a proportion is calculated using the formula: \[ SE = \sqrt{ \frac{\hat{p}(1-\hat{p})}{n} } = \sqrt{ \frac{0.45 \times 0.55}{800} } = \sqrt{ \frac{0.2475}{800} } \approx 0.0173 \]
03

Find the Critical Value

For a 98% confidence interval, we need the critical value that corresponds to \( \alpha = 0.02 \). From the standard normal distribution, the \( Z \)-score for 98% confidence is approximately \( Z = 2.33 \).
04

Calculate the Confidence Interval

The formula for the confidence interval is:\[ \hat{p} \pm Z \times SE = 0.45 \pm 2.33 \times 0.0173 \]Calculate the margin of error:\[ 2.33 \times 0.0173 \approx 0.0403 \]So, the confidence interval is:\[ 0.45 - 0.0403 \leq p \leq 0.45 + 0.0403 \]Thus, the interval is approximately:\( [0.4097, 0.4903] \).
05

Analyze if the Interval Includes 0.50

Since the confidence interval is \( [0.4097, 0.4903] \) and does not include 0.50, it suggests that less than 50% of the population thinks movies are getting better.

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Key Concepts

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

Population Proportion
The concept of population proportion is central when dealing with surveys and polls. In our example, the population proportion, represented by \( p \), symbolizes the overall percentage of adults who think movies are improving. Since it is usually impractical to ask every single person, we rely on a sample proportion, \( \hat{p} \).

In the given exercise, 45% of the surveyed adults stated that movies are getting better, resulting in \( \hat{p} = 0.45 \). Understanding this number is crucial. It's the best estimate of the true population proportion \( p \) based on the sample data.

Remember:
  • \( \hat{p} \) is not the exact proportion but an estimate.
  • We need further statistical tools to say anything about \( p \) with confidence.
Standard Error
The standard error is an important concept when it comes to estimating how much our sample proportion \( \hat{p} \) might differ from the true population proportion \( p \).

In simple terms, it measures the average distance between the sample proportion and the population proportion. The smaller the standard error, the closer our sample should be to the actual population proportion.

This is calculated using the formula: \[ SE = \sqrt{ \frac{\hat{p}(1-\hat{p})}{n} } \] Substitute the known values to find: \[ SE = \sqrt{ \frac{0.45 \times 0.55}{800} } \approx 0.0173 \]

Key points:
  • The larger the sample size \( n \), the smaller the SE.
  • A smaller SE indicates more precise estimates of \( p \).
Critical Value
To form a confidence interval, we need to calculate the critical value. This plays a pivotal role in determining how wide our interval will be, which tells us how confident we can be in our results.

The critical value is dependent on our confidence level. For a 98% confidence interval, we use the standard normal distribution. Here, the critical value, \( Z \)-score, is approximately \( 2.33 \).

Why it matters:
  • A higher confidence level means a larger critical value and thus a wider interval.
  • The critical value ensures the interval captures the true population proportion \( p \) with the desired level of confidence.
Margin of Error
The margin of error (MOE) tells us how much we expect our sample proportion \( \hat{p} \) to differ from \( p \). It combines both the standard error and critical value.

The margin of error is calculated as:\[ MOE = Z \times SE \approx 2.33 \times 0.0173 \approx 0.0403 \] This means our estimate can be 0.0403 above or below \( \hat{p} \).

Therefore, the 98% confidence interval becomes:
\[ \hat{p} \pm MOE = 0.45 \pm 0.0403 \approx [0.4097, 0.4903] \]

Why MOE is significant:
  • It tells us the range we believe the true proportion lies within.
  • The smaller the MOE, the more precise our estimate of \( p \).

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Most popular questions from this chapter

In a study to compare the perceived effects of two pain relievers, 200 randomly selected adults were given the first pain reliever, and \(93 \%\) indicated appreciable pain relief. Of the 450 individuals given the other pain reliever, \(96 \%\) indicated experiencing appreciable relief. a. Give an estimate for the difference in the proportions of all adults who would indicate perceived pain relief after taking the two pain relievers. Provide a bound on the error of estimation. b. Based on your answer to part (a), is there evidence that proportions experiencing relief differ for those who take the two pain relievers? Why?

Telephone pollisters often interview between 1000 and 1500 individuals regarding their opinions on various issues. Does the performance of colleges' athletic teams have a positive impact on the public's perception of the prestige of the institutions? A new survey is to be undertaken to see if there is a difference between the opinions of men and women on this issue. a. If 1000 men and 1000 women are to be interviewed, how accurately could you estimate the difference in the proportions who think that the performance of their athletics teams has a positive impact on the perceived prestige of the institutions? Find a bound on the error of estimation. b. Suppose that you were designing the survey and wished to estimate the difference in a pair of proportions, correct to within. \(02,\) with probability .9. How many interviewees should be included in each sample?

The reading on a voltage meter connected to a test circuit is uniformly distributed over the interval \((\theta, \theta+1),\) where \(\theta\) is the true but unknown voltage of the circuit. Suppose that \(Y_{1}, Y_{2}, \ldots, Y_{n}\) denote a random sample of such readings. a. Show that \(\bar{Y}\) is a biased estimator of \(\theta\) and compute the bias. b. Find a function of \(\bar{Y}\) that is an unbiased estimator of \(\theta\). c. Find \(\operatorname{MSE}(\bar{Y})\) when \(\bar{Y}\) is used as an estimator of \(\theta\)

The following statistics are the result of an experiment conducted by P. I. Ward to investigate a theory concerning the molting behavior of the male Gammarus pulex, a small crustacean. \(^{\star}\) If a male needs to molt while paired with a female, he must release her, and so loses her. The theory is that the male \(G\). pulex is able to postpone molting, thereby reducing the possibility of losing his mate. Ward randomly assigned 100 pairs of males and females to two groups of 50 each. Pairs in the first group were maintained together (normal); those in the second group were separated (split). The length of time to molt was recorded for both males and females, and the means, standard deviations, and sample sizes are shown in the accompanying table. (The number of crustaceans in each of the four samples is less than 50 because some in each group did not survive until molting time.) a. Find a \(99 \%\) confidence interval for the difference in mean molt time for "normal" males versus those "split" from their mates. b. Interpret the interval.

Solid copper produced by sintering (heating without melting) a powder under specified environmental conditions is then measured for porosity (the volume fraction due to voids) in laboratory. A sample of \(n_{1}=4\) independent porosity measurements have mean \(\bar{y}_{1}=.22\) and variance \(s_{1}^{2}=.0010 .\) A second laboratory repeats the same process on solid copper formed from an identical powder and gets \(n_{2}=5\) independent porosity measurements with \(\bar{y}_{2}=.17\) and \(s_{2}^{2}=.0020\) Estimate the true difference between the population means \(\left(\mu_{1}-\mu_{2}\right)\) for these two laboratories, with confidence coefficient. 95
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