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Fat contents (in grams) for seven randomly selected hot dogs rated as very good by Consumer Reports (www .consumerreports.org) are shown. Is it reasonable to use these data and the \(t\) confidence interval of this section to construct a confidence interval for the mean fat content of hot dogs rated very good by Consumer Reports? Explain why or why not. \(\begin{array}{lllllll}14 & 15 & 11 & 10 & 6 & 15 & 16\end{array}\)

Short Answer

Expert verified
Using the t-distribution for the confidence interval is applicable if the normality assumption holds true. The mean and standard deviation values calculated will guide the construction of this interval. However, even if the assumption of normality is not strongly supported by our small data set, the t-distribution is still frequently used in such scenarios because it is robust against the violation of the normality assumption to some degree. Therefore, considering these factors, it may be reasonable to use the t confidence interval method here.

Step by step solution

01

List the Data

First, list down the fat content in grams of the seven hot dogs, which are: 14, 15, 11, 10, 6, 15, 16.
02

Find the Mean

Calculate the mean of the data set by adding up all the numbers and dividing by the count of numbers, which is 7. The result \(\mu = \frac{14+15+11+10+6+15+16}{7}\) should be calculated.
03

Find the Standard Deviation

Calculate the standard deviation, which is a measure of the amount of variation or dispersion of a set of values, using the sample standard deviation formula. Each data point should be subtracted from the mean, then squared, and these values should be added together. The result should be divided by (n-1), in this case 6, before taking the square root.
04

Check Normality Assumption

Assuming these 7 hot dogs are a simple random sample from the population of all hot dogs rated as very good, we need to check if the data is normally distributed. However, due to the small sample size we can't judge this directly from the data.
05

Final Analysis

Use the calculated mean, standard deviation and the normality assumption to decide if the t-confidence interval method is applicable. If the data is assumed to be normally distributed, it would be reasonable to use the t-distribution to calculate a confidence interval. For larger samples, the t-distribution approximates the normal distribution, but for small sample sizes like this, the assumption about normal distribution becomes more critical.

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Key Concepts

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

T-Distribution
When we're discussing confidence intervals and sample data, the t-distribution is a fundamental concept that often comes into play, especially with smaller sample sizes. Unlike the normal distribution, which is symmetrical and has fixed standard deviations, the t-distribution is slightly more spread out and has fatter tails. This makes it ideal for estimating population parameters when we have a limited amount of data.

The t-distribution's shape changes based on the degrees of freedom, which directly correlate to the sample size. With a larger sample size, the t-distribution looks more similar to the normal distribution. This variability makes the t-distribution an essential tool for constructing confidence intervals when working with small sample sets where the true standard deviation of the population is unknown, like in our hot dog fat content example.
Sample Mean
In any statistical analysis, understanding the sample mean is crucial. The sample mean, or the average of all the observations in your data, serves as a reliable estimate of the population mean. To calculate it, simply sum all the values and divide by the number of observations. For instance, in our hot dog example, we add up the fat content of each hot dog and then divide by seven, the total number of hot dogs surveyed.

The mean gives us a starting point for calculating other statistics, like the standard deviation, and is important when we're trying to make inferences about the population from which our sample was taken. It's the central value around which other calculations revolve.
Sample Standard Deviation
The sample standard deviation plays a pivotal role in assessing how varied the data is. It tells us how spread out the data points are from the mean. A higher standard deviation means data points are more scattered, while a lower value indicates they are closer to the mean.

To compute it, we subtract the sample mean from each data point, square the result, and sum these squared differences. We then divide by the number of observations minus one (to account for the fact that it's a sample rather than the entire population) and take the square root of the result. This calculation is crucial in confidence interval estimation because it impacts the margin of error - the range within which we expect the true population mean to lie.
Normality Assumption
A key assumption when using the sample mean and standard deviation to estimate a population parameter is the normality of the data. This means that we're assuming our data is roughly symmetrically distributed around the mean, much like the classic bell curve of a normal distribution.

For large samples, the Central Limit Theorem kicks in, suggesting that the distribution of the sample means tends to be normal regardless of the shape of the population distribution. However, with small samples such as our seven hot dogs, we can't rely on this theorem, and the normality assumption becomes more critical. If the data are not normally distributed, the confidence intervals calculated using the t-distribution may not be accurate, thereby limiting their usefulness in making inferences about the population.

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Most popular questions from this chapter

A study of fast-food intake is described in the paper "What People Buy From Fast-Food Restaurants" (Obesity [2009]:1369- 1374). Adult customers at three hamburger chains (McDonald's, Burger King, and Wendy's) in New York City were approached as they entered the restaurant at lunchtime and asked to provide their receipt when exiting. The receipts were then used to determine what was purchased and the number of calories consumed was determined. In all, 3,857 people participated in the study. The sample mean number of calories consumed was 857 and the sample standard deviation was 677 . a. The sample standard deviation is quite large. What does this tell you about number of calories consumed in a hamburgerchain lunchtime fast-food purchase in New York City? b. Given the values of the sample mean and standard deviation and the fact that the number of calories consumed can't be negative, explain why it is not reasonable to assume that the distribution of calories consumed is normal. c. Based on a recommended daily intake of 2,000 calories, the online Healthy Dining Finder (www.healthydiningfinder .com) recommends a target of 750 calories for lunch. Assuming that it is reasonable to regard the sample of 3,857 fast-food purchases as representative of all hamburger-chain lunchtime purchases in New York City, carry out a hypothesis test to determine if the sample provides convincing evidence that the mean number of calories in a New York City hamburger-chain lunchtime purchase is greater than the lunch recommendation of 750 calories. Use \(\alpha=0.01\). d. Would it be reasonable to generalize the conclusion of the test in Part (c) to the lunchtime fast-food purchases of all adult Americans? Explain why or why not. e. Explain why it is better to use the customer receipt to determine what was ordered rather than just asking a customer leaving the restaurant what he or she purchased.

The paper "The Curious Promiscuity of Queen Honey Bees (Apis mellifera): Evolutionary and Behavioral Mechanisms" (Annals of Zoology [2001]:255-265) describes a study of the mating behavior of queen honeybees. The following quote is from the paper: Queens flew for an average of \(24.2 \pm 9.21\) minutes on their mating flights, which is consistent with previous findings. On those flights, queens effectively mated with \(4.6 \pm 3.47\) males (mean \(\pm \mathrm{SD}\) ). The intervals reported in the quote from the paper were based on data from the mating flights of \(n=30\) queen honeybees. One of the two intervals reported was identified as a \(95 \%\) confidence interval for a population mean. Which interval is this? Justify your choice.

In a study of computer use, 1,000 randomly selected Canadian Internet users were asked how much time they spend online in a typical week (Ipsos Reid, August 9,2005 ). The sample mean was 12.7 hours. a. The sample standard deviation was not reported, but suppose that it was 5 hours. Carry out a hypothesis test with a significance level of 0.05 to decide if there is convincing evidence that the mean time spent online by Canadians in a typical week is greater than 12.5 hours. b. Now suppose that the sample standard deviation was 2 hours. Carry out a hypothesis test with a significance level of 0.05 to decide if there is convincing evidence that the mean time spent online by Canadians in a typical week is greater than 12.5 hours. c. Explain why the null hypothesis was rejected in the test of Part (b) but not in the test of Part (a).

Explain the difference between \(\bar{x}\) and \(\mu_{\bar{x}}\)

Give as much information as you can about the \(P\) -value of a \(t\) test in each of the following situations: a. Two-tailed test, \(n=16, t=1.6\) b. Upper-tailed test, \(n=14, t=3.2\) c. Lower-tailed test, \(n=20, t=-5.1\) d. Two-tailed test, \(n=16, t=6.3\)

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