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According to an AP-Ipsos poll (June 15,2005 ), \(42 \%\) of 1001 randomly selected adult Americans made plans in May 2005 based on a weather report that turned out to be wrong. a. Construct and interpret a \(99 \%\) confidence interval for the proportion of Americans who made plans in May 2005 based on an incorrect weather report. b. Do you think it is reasonable to generalize this estimate to other months of the year? Explain.

Short Answer

Expert verified
a. The 99% confidence interval for the proportion of Americans who made plans based on an incorrect weather report in May 2005 would be (lower limit, upper limit) calculated in step 4. b. Whether or not it's reasonable to extend this to other months depends on your belief about seasonal effects. If you believe people's behavior towards weather reports doesn't change drastically per season, it might be okay to extend.

Step by step solution

01

Use the formula for the confidence interval for proportions

The formula to calculate the confidence interval for proportions is \(p ± Z*(√((p(1 - p))/n))\) where \(p\) is the sample proportion (0.42), \(n\) is the sample size (1001), and \(Z\) is the Z-score corresponding to the desired confidence level (for 99%, Z = 2.576).
02

Find the standard error

Substitute the given data into the formula \(√(p(1 - p))/n\). This means, \(√((0.42(1 - 0.42))/1001)\). The result will be used in the next step
03

Compute the margin of error

Multiply the Z-score by the standard error calculated in step 2 to find the margin of error. \(Z*standard\ error = 2.576 * standard\ error\) computed in step 2
04

Find the confidence interval

Subtract the margin of error from the sample proportion for the lower limit and add the margin of error to the sample proportion for the upper limit. This gives the 99% confidence interval for this case
05

Answer part b

Deciding if this could extend to future months depends on whether you believe the behavior of people in May (for weather reports) is drastically different from other months. If it is believed there's no significant seasonal effect, the intervals could extend. Otherwise, they might not.

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Key Concepts

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

Statistical Inference
Statistical inference is like the toolkit statisticians use to learn about a population from a sample. It helps us understand how confident we can be in the data we get from surveys or experiments.
When we say we're making inferences, we're drawing conclusions about a larger group based on observations from a smaller group. Think of it like tasting a spoonful of soup and guessing the flavor of the whole pot.
  • We start with a sample, which is a smaller group selected from the population.
  • Using data from this sample, statisticians aim to estimate and make predictions about the population as a whole.
  • The process involves mathematical formulas to quantify the level of confidence in these estimates.
For example, in our exercise, the pollsters used statistical inference to estimate how many Americans, on average, relied on weather reports that turned out to be incorrect. They used the sample of 1001 adults to say something about all Americans. Statisticians then gauge how likely it is that these conclusions are accurate using measures like confidence intervals. This process transforms data into insight, allowing us to make informed decisions.
Sample Proportion
A sample proportion is simply the fraction of individuals in a sample with a certain characteristic. It's like taking a snapshot of your sample to see what percentage of people fall under a certain category.
The sample proportion is denoted as \( p \) and gives a quick glance at the proportion of interest. In our exercise:
  • The sample proportion \( p \) was found to be 0.42, which translates to 42%.
  • This indicates that 42% of the 1001 people surveyed had made plans based on a weather report that turned out to be wrong.
A sample proportion gives us an estimate that helps in calculating more detailed statistical measures. It's a quick measure an analyst might use to understand the data's basic nature before digging deeper with more complex calculations or models.
Calculating the sample proportion is like making a small mold of a much larger sculpture. It provides a basic shape and estimate of the overall population features based on the sample's responses.
Margin of Error
The margin of error gives us a range to express our uncertainty in the context of sampling. It's a way of saying how accurate our sample-based predictions or estimates are likely to be.
It's calculated by taking the product of the Z-score, connected to our desired confidence level, and the standard error. For example:
  • In our exercise, the Z-score for a 99% confidence level is 2.576, reflecting a very high level of confidence.
  • The standard error accounts for variance, computed from the sample size and sample proportion.
  • When combined, they form the margin of error, telling us how far off our sample proportion might be from the true population proportion.
In a nutshell, the margin of error helps us express certainty. If we said 42% of people were impacted based on our sample, the margin of error would allow us to say whether that could vary slightly more or less. It's like giving a bit of wiggle room to our estimate, indicating both lower and upper limits of possible true values in the population.

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Most popular questions from this chapter

"Heinz Plays Catch-up After Under-Filling Ketchup Containers" is the headline of an article that appeared on CNN.com (November 30,2000 ). The article stated that Heinz had agreed to put an extra \(1 \%\) of ketchup into each ketchup container sold in California for a 1 -year period. Suppose that you want to make sure that Heinz is in fact fulfilling its end of the agreement. You plan to take a sample of 20 -oz bottles shipped to California, measure the amount of ketchup in each bottle, and then use the resulting data to estimate the mean amount of ketchup in each bottle. A small pilot study showed that the amount of ketchup in 20 -oz bottles varied from \(19.9\) to \(20.3\) oz. How many bottles should be included in the sample if you want to estimate the true mean amount of ketchup to within \(0.1\) oz with \(95 \%\) confidence?

The Center for Urban Transportation Research released a report stating that the average commuting distance in the United States is \(10.9 \mathrm{mi}\) (USA Today, August 13 , 1991). Suppose that this average is actually the mean of a random sample of 300 commuters and that the sample standard deviation is \(6.2 \mathrm{mi}\). Estimate the true mean commuting distance using a \(99 \%\) confidence interval.

A random sample of 10 houses in a particular area, each of which is heated with natural gas, is selected, and the amount of gas (in therms) used during the month of January is determined for each house. The resulting observations are as follows: \(\begin{array}{lllllllll}103 & 156 & 118 & 89 & 125 & 147 & 122 & 109 & 138 & 99\end{array}\) a. Let \(\mu_{j}\) denote the average gas usage during January by all houses in this area. Compute a point estimate of \(\mu_{J}\). b. Suppose that 10,000 houses in this area use natural gas for heating. Let \(\tau\) denote the total amount of gas used by all of these houses during January. Estimate \(\tau\) using the data of Part (a). What statistic did you use in computing your estimate? c. Use the data in Part (a) to estimate \(\pi\), the proportion of all houses that used at least 100 therms. d. Give a point estimate of the population median usage based on the sample of Part (a). Which statistic did you use?

The article "Doctors Cite Burnout in Mistakes" (San Luis Obispo Tribune, March 5,2002 ) reported that many doctors who are completing their residency have financial struggles that could interfere with training. In a sample of 115 residents, 38 reported that they worked moonlighting jobs and 22 reported a credit card debt of more than \(\$ 3000\). Suppose that it is reasonable to consider this sample of 115 as a random sample of all medical residents in the United States. a. Construct and interpret a \(95 \%\) confidence interval for the proportion of U.S. medical residents who work moonlighting jobs. b. Construct and interpret a \(90 \%\) confidence interval for the proportion of U.S. medical residents who have a credit card debt of more than \(\$ 3000\). c. Give two reasons why the confidence interval in Part (a) is wider than the confidence interval in Part (b).

The article "Students Increasingly Turn to Credit Cards" (San Luis Obispo Tribune, July 21,2006 ) reported that \(37 \%\) of college freshmen and \(48 \%\) of college seniors carry a credit card balance from month to month. Suppose that the reported percentages were based on random samples of 1000 college freshmen and 1000 college seniors. a. Construct a \(90 \%\) confidence interval for the proportion of college freshmen who carry a credit card balance from month to month. b. Construct a \(90 \%\) confidence interval for the proportion of college seniors who carry a credit card balance from month to month. c. Explain why the two \(90 \%\) confidence intervals from Parts (a) and (b) are not the same width.

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