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"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?

Short Answer

Expert verified
The student should sample 4 bottles of ketchup.

Step by step solution

01

Determine the known parameters

From the problem, we can determine the following parameters: The range of amount of ketchup in 20-oz bottles varied from \(19.9\) to \(20.3\) oz, so the standard deviation (\(σ\)) can be estimated by dividing the range by 4, which gives \(σ = 0.1\) oz. The desired margin of error (E) is \(0.1\) oz. The Z-Value for \(95\%\) confidence level is \(1.96\), which you can find in a standard normal distribution table.
02

Use the formula for calculating sample size

We will use the formula for the sample size n, which is \(n = (Z^2 * σ^2)/E^2)\). Note that, \(Z^2\) is the square of the Z-value from the confidence level, \(σ^2\) is the variance, and \(E^2\) is the square of the margin of error.
03

Substitute values into the formula and solve

Now, substitute all known variables into the sample size formula and solve. So, \(n = ((1.96)^2 * (0.1)^2)/(0.1)^2 = 3.8416\). Round this value up to the nearest whole number, as we cannot have a fraction of a bottle. So, the number of bottles needed would be \(4\).

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Key Concepts

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

Confidence Interval
When working with data, it's important to have a reliable estimate of the parameter you are studying. The confidence interval gives you a way to express this estimate in a range of values, providing a sense of how accurate or precise the estimate might be.

A confidence interval for a sample mean is an interval estimate, calculated using the sample data and designed to include the true population mean a certain percentage of the time if you were to repeat your sampling process numerous times. For instance, in our exercise, a confidence interval of 95% means that if we repeated the study many times, approximately 95 out of 100 confidence intervals calculated from these studies would contain the true mean amount of ketchup in the bottles.

Creating a confidence interval involves the estimated mean, the standard error, and checking against a table of the normal distribution (Z-table). This requires selection of the appropriate Z-value, which represents how many standard deviations away you are from the mean.
Standard Deviation
Understanding standard deviation is essential for interpreting data variability. It measures how spread out the numbers are in a data set. In other words, it tells you how much the individual data points vary from the average or mean.

In the Heinz ketchup scenario, the standard deviation is a crucial figure. The data shown varies between 19.9 and 20.3 ounces, so to estimate the standard deviation (\(\sigma\)), you can use the range/4 rule, which gives \(\sigma = 0.1\) oz. This estimation helps in understanding how much the individual measurements might deviate from the mean, and it's integral when calculating other metrics like the margin of error.

Remember that a low standard deviation means that most of the numbers are close to the average, while a high standard deviation means that the numbers are more spread out.
Margin of Error
The margin of error in statistical terms represents the range, above and below the sample statistic, that you believe your study will fall within when estimating a population parameter.

For Heinz’s ketchup case, the desired level of precision was to estimate the true mean amount of ketchup to within 0.1 oz. This 0.1 oz is your margin of error and is used to gauge the accuracy of your sample mean in reflecting the actual population mean. The smaller the margin of error, the more precise your estimate will be, but this typically requires a larger sample size.

Using the formula for sample size, you incorporate the margin of error along with the standard deviation and Z-value to determine how many data points (or bottles of ketchup, in this case) you need to accurately reflect the population's mean.
Z-Value
The Z-value, also known as a Z-score, is a crucial component when working with confidence intervals. It helps determine how many standard deviations your parameter is from the mean, based on the probability distribution.

For the Heinz ketchup exercise, the Z-value for a 95% confidence interval is 1.96, meaning we expect the sample mean to lie within 1.96 standard deviations of the true population mean 95% of the time. This Z-value is sourced from a standard normal distribution table, which provides a value corresponding to a chosen level of confidence.

Determining the correct Z-value is integral, as it's used in the formula to calculate the necessary sample size, reflecting how many observations are needed to achieve a specified level of precision within the estimated interval.

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Most popular questions from this chapter

One thousand randomly selected adult Americans participated in a survey conducted by the Associated Press (June, 2006). When asked "Do you think it is sometimes justified to lie or do you think lying is never justified?" \(52 \%\) responded that lying was never justified. When asked about lying to avoid hurting someone's feelings, 650 responded that this was often or sometimes OK. a. Construct a \(90 \%\) confidence interval for the proportion of adult Americans who think lying is never justified. b. Construct a \(90 \%\) confidence interval for the proportion of adult American who think that it is often or sometimes OK to lie to avoid hurting someone's feelings. c. Based on the confidence intervals from Parts (a) and (b), comment on the apparent inconsistency in the responses given by the individuals in this sample.

For each of the following choices, explain which would result in a wider large-sample confidence interval for \(\pi\) : a. \(90 \%\) confidence level or \(95 \%\) confidence level b. \(n=100\) or \(n=400\)

The Chronicle of Higher Education (January 13, 1993) reported that \(72.1 \%\) of those responding to a national survey of college freshmen were attending the college of their first choice. Suppose that \(n=500\) students responded to the survey (the actual sample size was much larger). a. Using the sample size \(n=500\), calculate a \(99 \%\) confidence interval for the proportion of college students who are attending their first choice of college. b. Compute and interpret a \(95 \%\) confidence interval for the proportion of students who are not attending their first choice of college. c. The actual sample size for this survey was much larger than 500 . Would a confidence interval based on the actual sample size have been narrower or wider than the one computed in Part (a)?

The following data on gross efficiency (ratio of work accomplished per minute to calorie expenditure per minute) for trained endurance cyclists were given in the article "Cycling Efficiency Is Related to the Percentage of Type I Muscle Fibers" (Medicine and Science in Sports and Exercise [1992]: \(782-88\) ): \(\begin{array}{llllllll}18.3 & 18.9 & 19.0 & 20.9 & 21.4 & 20.5 & 20.1 & 20.1\end{array}\) \(\begin{array}{llllllll}20.8 & 20.5 & 19.9 & 20.5 & 20.6 & 22.1 & 21.9 & 21.2\end{array}\) \(\begin{array}{lll}20.5 & 22.6 & 22.6\end{array}\) a. Assuming that the distribution of gross energy in the population of all endurance cyclists is normal, give a point estimate of \(\mu\), the population mean gross efficiency. b. Making no assumptions about the shape of the population distribution, estimate the proportion of all such cyclists whose gross efficiency is at most 20 .

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 \(-1.81\)

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