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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

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 \(\bar{x}=21.9\) 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?

What percentage of the time will a variable that has a \(t\) distribution with the specified degrees of freedom fall in the indicated region? a. \(10 \mathrm{df}\), between -2.23 and 2.23 b. 24 df, between -2.80 and 2.80 c. \(24 \mathrm{df}\), to the right of 2.80

What percentage of the time will a variable that has a \(t\) distribution with the specified degrees of freedom fall in the indicated region? (Hint: See discussion on page 496 ) a. 10 df, between -1.81 and 1.81 b. 24 df, between -2.06 and 2.06 c. 24 df, outside the interval from -2.80 to 2.80 d. 10 df, to the left of -1.81

The article "Most Canadians Plan to Buy Treats, Many Will Buy Pumpkins, Decorations and/or Costumes" (Ipsos-Reid, October 24,2005\()\) summarized a survey of 1,000 randomly selected Canadian residents. Each individual in the sample was asked how much he or she anticipated spending on Halloween. The resulting sample mean and standard deviation were \(\$ 46.65\) and \(\$ 83.70\), respectively. a. Explain how it could be possible for the standard deviation of the anticipated Halloween expense to be larger than the mean anticipated expense. b. Is it reasonable to think that the distribution of anticipated Halloween expense is approximately normal? Explain why or why not. c. Is it appropriate to use the one-sample \(t\) confidence interval to estimate the mean anticipated Halloween expense for Canadian residents? Explain why or why not. d. If appropriate, construct and interpret a \(99 \%\) confidence interval for the mean anticipated Halloween expense for Canadian residents.

The formula used to calculate a confidence interval for the mean of a normal population is $$ \bar{x} \pm(t \text { critical value }) \frac{s}{\sqrt{n}} $$ What is the appropriate \(t\) critical value for each of the following confidence levels and sample sizes? a. \(95 \%\) confidence, \(n=17\) b. \(99 \%\) confidence, \(n=24\) c. \(90 \%\) confidence, \(n=13\)

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