/*! 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 73 How much money do people spend o... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

How much money do people spend on graduation gifts? In \(2016,\) the National Retail Federation (www.nrf.com) surveyed 2511 consumers who reported that they bought one or more graduation gifts in 2016 . The sample was selected to be representative of adult Americans who purchased graduation gifts in 2016 . For this sample, the mean amount spent per gift was \(\$ 53.73 .\) Suppose that the sample standard deviation was \(\$ 20 .\) Construct and interpret a \(98 \%\) confidence interval for the mean amount of money spent per graduation gift in 2016 .

Short Answer

Expert verified
Based on the given data, we can be 98% confident that the mean amount spent on graduation gifts by adult Americans in 2016 was between \(52.80 and \)54.66.

Step by step solution

01

Find the critical value

Given the confidence level of 98%, the critical value (z*) can be found from the z-table or using calculator software. The z-value associated with the 99th percentile (since the area to the left is 0.99) is approximately 2.33.
02

Calculate the margin of error

The margin of error (E) is calculated using the formula: E = z* × (s / √n) Substituting the values: E = 2.33 × (20 / √2511) E ≈ 2.33 × (20 / 50.11) E ≈ $0.93
03

Determine the confidence interval

Now we can determine the 98% confidence interval by adding and subtracting the margin of error (E) from the sample mean (x̄): Lower limit of confidence interval = x̄ - E = \(53.73 - \)0.93 = $52.80 Upper limit of confidence interval = x̄ + E = \(53.73 + \)0.93 = $54.66 Hence, based on the given data, we can be 98% confident that the mean amount spent on graduation gifts by adult Americans in 2016 was between \(52.80 and \)54.66.

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Key Concepts

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

Statistical Inference
Understanding statistical inference is essential for interpreting data and making decisions based on samples. In essence, it involves using information from a sample to make generalizations about a larger population. This concept is especially important because it's rarely practical or possible to collect data from every individual in a population. Instead, we use sample data to estimate population parameters such as mean or proportion.

When we compute the average money spent on graduation gifts using a sample, we're doing statistical inference. We're inferring the population mean from the sample mean. Therefore, using statistical inference responsibly requires careful consideration of the sample's representativeness and the measures of uncertainty associated with estimates.
Sample Standard Deviation
Sample standard deviation provides a measure of the variability or dispersion of a sample data set. It gives us an idea of how spread out the individual observations are from the sample mean. A low standard deviation indicates that the data points tend to be close to the mean, while a high standard deviation suggests that the data points are spread out over a large range of values.

In the example of graduation gifts, a sample standard deviation of \(20 tells us that, on average, the amount spent differs from the mean (53.73) by about \)20. It's a crucial piece of information when constructing confidence intervals because it helps determine the precision of our estimate.
Z-Score
The z-score, or standard score, is a key concept in statistics that tells us how many standard deviations an element is from the mean. In the context of confidence intervals, we use the z-score to determine the critical value, typically noted as z*. This critical value corresponds to the desired confidence level.

A z-score of 2.33, as identified in the solution, indicates that the critical value lies 2.33 standard deviations above the mean on a standard normal distribution curve. This particular value is linked to the confidence level of 98%, serving as a multiplier when calculating the margin of error for our interval estimation.
Margin of Error
The margin of error represents the range of variability we expect due to sampling. In other words, it's the amount we add and subtract from the sample mean to create a confidence interval for the population mean. The margin of error is impacted by the sample size, the standard deviation, and the chosen confidence level.

By calculating a margin of error of $0.93, we're acknowledging that there's a small amount of uncertainty in estimating the true population mean. It reflects the level of certainty we have in our sample estimate and ensures that our confidence interval allows for this inherent uncertainty in the sampling process.
Confidence Level
The confidence level tells us the probability that the calculated confidence interval actually contains the population parameter we're estimating. Simply put, it's a measure of how 'confident' we can be in our interval estimate. For instance, a 98% confidence level means that if we were to take 100 different samples and compute a confidence interval for each of them, we would expect about 98 of those intervals to contain the true population mean.

In our exercise, the 98% confidence level reflects a high degree of certainty in our estimate. It assures us that there's only a 2% chance that our confidence interval does not contain the mean amount spent on graduation gifts by all adult Americans in 2016.

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Most popular questions from this chapter

An automobile manufacturer decides to carry out a fuel efficiency test to determine if it can advertise that one of its models achieves \(30 \mathrm{mpg}\) (miles per gallon). Six people each drive a car from Phoenix to Los Angeles. The resulting fuel efficiencies (in miles per gallon) are: \(\begin{array}{ll}30.3 & 29.6\end{array}\) \(\begin{array}{llll}27.2 & 29.3 & 31.2 & 28.4\end{array}\) Assuming that fuel efficiency is normally distributed, do these data provide evidence against the claim that actual mean fuel efficiency for this model is (at least) \(30 \mathrm{mpg}\) ?

The article "Americans' Big Debt Burden Growing, Not Evenly Distributed" (www.gallup.com, retrieved December 14, 2016) reported that for a representative sample of Americans born between 1965 and 1971 (known as Generation \(\mathrm{X}\) ), the sample mean number of credit cards owned was 4.5. Suppose that the sample standard deviation (which was not reported) was 1.0 and that the sample size was \(n=300\). a. Construct a \(95 \%\) confidence interval for the population mean number of credit cards owned by Generation \(\mathrm{X}\) Americans. b. The interval in Part (a) does not include \(0 .\) Does this imply that all Generation X Americans have at least one credit card? Explain.

The eating habits of \(n=12\) 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 \(\bar{x}=21.9\) minutes. Suppose that the standard deviation was \(s=7.7\) minutes. Is there convincing evidence that the mean supper time of a vampire bat whose meal consists of a frog is greater than 20 minutes? What assumptions must be reasonable for the one-sample \(t\) test to be appropriate?

The report "2016 Salary Survey Executive Summary" (National Association of Colleges and Employers, www.naceweb.org/uploadedfiles/files/2016/publications/executive- summary/2016-nace-salary-survey-fall-executive-summary. pdf, retrieved December 24,2016 ) states that the mean yearly salary offer for students graduating with mathematics and statistics degrees in 2016 was \(\$ 62,985 .\) Suppose that a random sample of 50 math and statistics graduates at a large university who received job offers resulted in a mean offer of \(\$ 63,500\) and a standard deviation of \(\$ 3300\). Do the sample data provide strong support for the claim that the mean salary offer for math and statistics graduates of this university is higher than the 2016 national average of \(\$ 62,985 ?\) Test the relevant hypotheses using \(\alpha=0.05\).

Suppose that a random sample of size 100 is to be drawn from a population with standard deviation 10 . a. What is the probability that the sample mean will be within 20 of the value of \(\mu\) ? b. For this example \((n=100, \sigma=10)\), complete each of the following statements by calculating the appropriate value: i. Approximately \(95 \%\) of the time, \(\bar{x}\) will be within of \(\mu\). ii. Approximately \(0.3 \%\) of the time, \(\bar{x}\) will be farther than from \(\mu\).

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