/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 According to an AP-Ipsos poll (J... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Fat contents (in percentage) for 10 randomly selected hot dogs were given in the article "Sensory and Mechanical Assessment of the Quality of Frankfurters" (Journal of Texture Studies \([1990]: 395-409\) ). Use the following data to construct a \(90 \%\) confidence interval for the true mean fat percentage of hot dogs: \(\begin{array}{lllllllllllll}25.2 & 21.3 & 22.8 & 17.0 & 29.8 & 21.0 & 25.5 & 16.0 & 20.9 & 19.5\end{array}\)

Given a variable that has a \(t\) distribution with the specified degrees of freedom, what percentage of the time will its value fall in the indicated region? a. \(10 \mathrm{df}\), between \(-1.81\) and \(1.81\) b. \(10 \mathrm{df}\), between \(-2.23\) and \(2.23\) c. \(24 \mathrm{df}\), between \(-2.06\) and \(2.06\) d. \(24 \mathrm{df}\), between \(-2.80\) and \(2.80\) e. 24 df, outside the interval from \(-2.80\) to \(2.80\) f. \(24 \mathrm{df}\), to the right of \(2.80\) g. \(10 \mathrm{df}\), to the left of \(-1.81\)

"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 article "National Geographic, the Doomsday Machine," which appeared in the March 1976 issue of the Journal of Irreproducible Results (yes, there really is a journal by that name -it's a spoof of technical journals!) predicted dire consequences resulting from a nationwide buildup of National Geographic magazines. The author's predictions are based on the observation that the number of subscriptions for National Geographic is on the rise and that no one ever throws away a copy of National Geographic. A key to the analysis presented in the article is the weight of an issue of the magazine. Suppose that you were assigned the task of estimating the average weight of an issue of National Geographic. How many issues should you sample to estimate the average weight to within \(0.1 \mathrm{oz}\) with \(95 \%\) confidence? Assume that \(\sigma\) is known to be 1 oz.

The eating habits of 12 bats were examined in the article "Foraging Behavior of the Indian False Vampire Bat" (Biotropica \([1991]: 63-67) .\) These bats consume insects and frogs. For these 12 bats, the mean time to consume a frog was \(\bar{x}=21.9 \mathrm{~min}\). Suppose that the standard deviation was \(s=7.7 \mathrm{~min} .\) Construct and interpret a \(90 \%\) confidence interval for the mean suppertime of a vampire bat whose meal consists of a frog. What assumptions must be reasonable for the one-sample \(t\) interval to be appropriate?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.