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The following data are the calories per half-cup serving for 16 popular chocolate ice cream brands reviewed by Consumer Reports (July 1999): \(\begin{array}{llllllll}270 & 150 & 170 & 140 & 160 & 160 & 160 & 290 \\ 190 & 190 & 160 & 170 & 150 & 110 & 180 & 170\end{array}\) Is it reasonable to use the \(t\) confidence interval to compute a confidence interval for \(\mu\), the true mean calories per half-cup serving of chocolate ice cream? Explain why on why not.

Short Answer

Expert verified
Given the Central Limit Theorem, it could be reasonable to use the t confidence interval to compute a confidence interval for the true mean calories per serving of chocolate ice cream, provided the sample is not strongly skewed or contains no significant outliers. However, it's noteworthy that our sample size is somewhat small (only 16 observations).

Step by step solution

01

Computing Sample Mean

First, add all the calorie values together and then divide by 16 (the total number of observations) to get the sample mean.\[\bar{x} = \frac{1}{n}(270+150+170+140+160+160+160+290+190+190+160+170+150+110+180+170)\]After calculating, the sample mean value is obtained.
02

Computing Sample Standard Deviation

Subtract each of the original observations from the sample mean, square the results, add these together and divide by (n-1), then take the square root of the result. This gives you the sample standard deviation, denoted by \(s\). It is calculated as \[s = \sqrt{\frac{1}{n-1}\sum_{i=1}^{n}(x_{i}-\bar{x})^2}\]Calculate the value by substituting \(x_{i}\) values, \(\bar{x}\) and \(n\).
03

Degrees of Freedom and Conclusion

The degrees of freedom is calculated as \(n-1\), which equals to 15 in this case. The shape of the t-distribution depends on degrees of freedom. As degrees of freedom increase, the t-distribution gets closer to the standard Normal distribution. Since the sample size is only 16, the distribution of the data might not represent a normal distribution perfectly. However, given the Central Limit Theorem, if the population is not strongly skewed or has no significant outliers, a t confidence interval could produce a reasonable approximation for confidence interval of the true mean population.

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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 dealing with small sample sizes and we don't know the population standard deviation, we turn to what's known as the t-distribution. It's a type of probability distribution that is symmetric and bell-shaped, similar to the normal distribution, but has heavier tails. This means there's a higher probability for extreme values, which accounts for the added uncertainty that comes with less data.

For the calories in chocolate ice cream, since we're working with 16 brands, which is considered a small sample size, the t-distribution is appropriate for constructing a confidence interval for the true mean. As the sample size (n) gets larger, the t-distribution looks more like the normal distribution. But for our purposes, its shape will be uniquely determined by the degrees of freedom (n-1), which in this case is 15.
sample mean
The sample mean, denoted as \(\bar{x}\), is a critical value in statistics representing the average of a sample. It's calculated by summing all observations in the sample and then dividing by the number of observations. For the chocolate ice cream calories, the mean gives us a central value around which all our individual ice cream calorie counts are distributed.

With the given data, by adding up the calorie amounts and dividing by 16, we obtain the average calories per half-cup serving. This value serves as our best estimate of the true mean calorie count, \(\mu\), from this particular sample of chocolate ice creams.
sample standard deviation
The sample standard deviation (s) gives us a measure of the variability or dispersion of the sample data points from the sample mean. To find s, we take the square root of the average of the squared deviations of each data point from the mean. In simpler terms, it tells us, on average, how far each ice cream brand's calorie count is from the mean calorie count.

The computation of s for the chocolate ice cream calories involves subtracting the sample mean from each observation, squaring these differences, summing them up, dividing by the degrees of freedom (n-1), and finally taking the square root of this quotient. It is a step that captures how much variety there is among the calorie counts of the different brands.
degrees of freedom
Degrees of freedom in statistics represent the number of values that are free to vary in the calculation of a statistic, such as the standard deviation. When calculating the sample standard deviation, we use n-1 degrees of freedom, which correctively adjusts for the bias that can occur when estimating population parameters from a sample.

In the context of the ice cream calories example, with a sample size of 16 brands, the degrees of freedom would be 15. This value is not only used in calculating the sample standard deviation but also is crucial when referencing the t-distribution table to determine the critical t-value for our confidence interval.
Central Limit Theorem
The Central Limit Theorem is a fundamental statistical principle stating that the distribution of sample means will approximate a normal distribution, regardless of the shape of the population distribution, as long as the sample size is sufficiently large. There's a bit of a caveat, though: the larger the sample size, the better the normal approximation.

For the ice cream calories example, even though we have a relatively small sample size of 16, the Central Limit Theorem would suggest that if we were to collect multiple samples of chocolate ice cream brands and calculate their means, the distribution of those sample means would tend to be normal. This is why we can use the t-distribution, which approximates the behavior of the sample means, to estimate the true mean calorie count of all chocolate ice creams, with the assurance that our estimation will be reasonable.

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Most popular questions from this chapter

An Associated Press article on potential violent behavior reported the results of a survey of 750 workers who were employed full time (San Luis Obispo Tribune, September 7,1999 ). Of those surveyed, 125 indicated that they were so angered by a coworker during the past year that they felt like hitting the coworker (but didn't). Assuming that it is reasonable to regard this sample of 750 as a random sample from the population of full-time workers, use this information to construct and interpret a \(90 \%\) confidence interval estimate of \(\pi\), the true proportion of fulltime workers so angered in the last year that they wanted to hit a colleague.

Why is an unbiased statistic generally preferred over a biased statistic for estimating a population characteristic? Does unbiasedness alone guarantee that the estimate will be close to the true value? Explain. Under what circumstances might you choose a biased statistic over an unbiased statistic if two statistics are available for estimating a population characteristic?

In a survey of 1000 randomly selected adults in the United States, participants were asked what their most favorite and what their least favorite subject was when they were in school (Associated Press, August 17,2005\() .\) In what might seem like a contradiction, math was chosen more often than any other subject in both categories! Math was chosen by 230 of the 1000 as the favorite subject, and it was also chosen by 370 of the 1000 as the least favorite subject. a. Construct a \(95 \%\) confidence interval for the proportion of U.S. adults for whom math was the favorite subject in school. b. Construct a \(95 \%\) confidence interval for the proportion of U.S. adults for whom math was the least favorite subject.

Because of safety considerations, in May 2003 the Federal Aviation Administration (FAA) changed its guidelines for how small commuter airlines must estimate passenger weights. Under the old rule, airlines used \(180 \mathrm{lb}\) as a typical passenger weight (including carry-on luggage) in warm months and \(185 \mathrm{lb}\) as a typical weight in cold months. The Alaska Journal of Commerce (May 25, 2003 ) reported that Frontier Airlines conducted a study to estimate average passenger plus carry-on weights. They found an average summer weight of \(183 \mathrm{lb}\) and a winter average of \(190 \mathrm{lb} .\) Suppose that each of these estimates was based on a random sample of 100 passengers and that the sample standard deviations were 20 lb for the summer weights and \(23 \mathrm{lb}\) for the winter weights. a. Construct and interpret a \(95 \%\) confidence interval for the mean summer weight (including carry-on luggage) of Frontier Airlines passengers. b. Construct and interpret a \(95 \%\) confidence interval for the mean winter weight (including carry-on luggage) of Frontier Airlines passengers. c. The new FAA recommendations are \(190 \mathrm{lb}\) for summer and \(195 \mathrm{lb}\) for winter. Comment on these recommendations in light of the confidence interval estimates from Parts (a) and (b).

The formula used to compute a large-sample confidence interval for \(\pi\) is $$ p \pm(z \text { critical value }) \sqrt{\frac{p(1-p)}{n}} $$ What is the appropriate \(z\) critical value for each of the following confidence levels? a. \(95 \%\) d. \(80 \%\) b. \(90 \%\) e. \(85 \%\) c. \(99 \%\)

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