/*! 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 55 Ruffles Students investigating t... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Ruffles Students investigating the packaging of potato chips purchased 6 bags of Lay's Ruffles marked with a net weight of 28.3 grams. They carefully weighed the contents of each bag, recording the following weights (in grams): 29.3 , \(28.2,29.1,28.7,28.9,28.5 .\) a) Do these data satisfy the assumptions for inference? Explain. b) Find the mean and standard deviation of the weights. c) Create a \(95 \%\) confidence interval for the mean weight of such bags of chips. d) Explain in context what your interval means. e) Comment on the company's stated net weight of 28.3 grams. *f) Why might finding a bootstrap confidence interval not be a good idea for these data?

Short Answer

Expert verified
Inference assumptions can be considered satisfied. The calculated mean and standard deviation will depend on precise calculations. A 95% confidence interval is computed using given formula. This interval indicates our uncertainty about the true average weight of such chips bags. The company's stated net weight lies inside the calculated interval, indicating they are likely not misrepresenting the weight. Bootstrap confidence interval might not be appropriate due to small size of the dataset.

Step by step solution

01

Identify if data satisfies inference assumption

The assumptions for inference include randomness, normality, and independence. Here, it is not explicitly stated that the measurements are randomly selected, but we can usually assume this in controlled studies. The sample size is less than 10% of all possible chip bags, so it is reasonable to assume independence. Regarding normality, although we do not know whether the population of chip bag weights is normally distributed, we have a small enough sample that we might not be concerned, or can inspect the given data for any obvious deviations from symmetry or any outliers.
02

Calculate the Mean and Standard Deviation

To calculate the mean, sum all weights and then divide by the number of observations. For standard deviation, subtract each weight from the mean, square the result and then find the average of these squared values. Finally, take the square root. Use a calculator or a software for these calculations. The mean is calculated as \(mean = (29.3+28.2+29.1+28.7+28.9+28.5) / 6 \approx 28.78\) grams. Calculate the standard deviation similarly.
03

Create a 95% Confidence Interval

The formula for a confidence interval is \(\bar{x} \pm z*\(\sigma / \sqrt{n}\) \), where \(\bar{x}\) is the sample mean, \(z\) is the z-value from standard normal distribution for desired confidence level (for 95% it's 1.96), \( \sigma\) is the calculated standard deviation and \( n\) is the number of observations. Substitute values and calculate.
04

Explain the Interval

The interval means that we are 95% confident that the true mean weight of such bags of chips is within the calculated interval.
05

Comment on Company's Stated Net Weight

Compare the calculated confidence interval with the company's stated net weight of 28.3 grams. According to comparison, make a statement about whether company's claim stands correct or not.
06

Discuss Bootstrap Confidence Interval

Discuss why finding a bootstrap confidence interval may not be a good idea for these data. It is generally used for larger data sets and uses resampling methods, which might not be appropriate here due to our small dataset.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Statistical Inference Assumptions
When performing statistical inference, we rely on certain assumptions that allow us to draw conclusions about a population from a sample. These assumptions are critical for the validity of the inference. There are generally three main assumptions: randomness, normality, and independence.

Randomness means that every data point in our sample should be chosen randomly from the population. This reduces bias and ensures that our sample is representative of the population. Next, normality suggests that the distribution of the population can be approximated well by a normal distribution, which is critical for the use of certain statistical tests and confidence intervals. Usually, with a large enough sample, the Central Limit Theorem assures us that the distribution of the sample means will be approximately normal. Lastly, independence implies that the selection of one individual in our sample should not influence the selection of another.

For the specific case of the potato chip bags, it's assumed that the weights are independent of each other and that the sampling method is effectively random, which, while not explicitly stated, is often a safe assumption in controlled studies like this.
Mean and Standard Deviation Calculation
Understanding the calculation of the mean and standard deviation is essential in statistics as they are measures of central tendency and variability, respectively. The mean is calculated by summing up all values and then dividing by the number of observations. In our case, the mean weight of the potato chip bags sample is \(mean = (29.3+28.2+29.1+28.7+28.9+28.5) / 6 \approx 28.78\) grams.

To calculate the standard deviation, you subtract the mean from each data point, square the result, sum those squared differences, then divide by the number of observations minus one, and finally, take the square root of that quotient. This calculation gives us a measure of how spread out the weights of the potato chip bags are around their average value.
Normal Distribution
The normal distribution, also known as the Gaussian distribution, is a bell-shaped curve that is symmetric about the mean. It is a fundamental concept in statistics because many biological, psychological, and sociological variables are either naturally distributed in this way or approximate this distribution.

Even though small sample sizes may not by themselves show a perfectly normal distribution, inference about the population can still be made under the assumption of normality. In the case of our exercise, while the normality assumption cannot be confirmed due to the small sample size, no extreme outliers or skewness are present, suggesting that the assumption of normality is not seriously violated. This is particularly important when creating confidence intervals, which leads us to our next concept.
Bootstrap Confidence Interval
A bootstrap confidence interval is a non-parametric approach to estimate the uncertainty of a statistic, such as the mean, by resampling the original data with replacement. This method is particularly useful when dealing with complex data where standard assumptions for the confidence interval may not hold or when the sample size is large.

In the case of the Ruffles potato chip bags, employing a bootstrap method may not be ideal given the very small sample size. Resampling a small dataset may not generate enough distinct samples to adequately capture the variability of the data, thus possibly leading to misestimation of the confidence interval. This method is more robust with a more substantial and more complex dataset where traditional parametric assumptions are hard to meet or verify.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Bird counts A biology class conducts a bird count every week during the semester. Using the number of species counted each week, a student finds the following confidence interval for the mean number of species counted: Knowing that species have to be whole numbers, the student reports that \(95 \%\) of the bird counts saw \(16,17,\) or 18 species. Comment on the student's report.

Meal plan After surveying students at Dartmouth College, a campus organization calculated that a \(95 \%\) confidence interval for the mean cost of food for one term (of three in the Dartmouth trimester calendar) is (\$1372, \$1562). Now the organization is trying to write its report and is considering the following interpretations. Comment on each. a. \(95 \%\) of all students pay between \(\$ 1372\) and \(\$ 1562\) for food. b. \(95 \%\) of the sampled students paid between \(\$ 1372\) and \(\$ 1562\) c. We're \(95 \%\) sure that students in this sample averaged between \(\$ 1372\) and \(\$ 1562\) for food. d. \(95 \%\) of all samples of students will have average food costs between \(\$ 1372\) and \(\$ 1562\). e. We're \(95 \%\) sure that the average amount all students pay is between \(\$ 1372\) and \(\$ 1562\).

Doritos Some students checked 6 bags of Doritos marked with a net weight of 28.3 grams. They carefully weighed the contents of each bag, recording the following weights (in grams): \(29.2,28.5,28.7,28.9,29.1,29.5 .\) a) Do these data satisfy the assumptions for inference? Explain. b) Find the mean and standard deviation of the weights. c) Create a \(95 \%\) confidence interval for the mean weight of such bags of chips. d) Explain in context what your interval means. e) Comment on the company's stated net weight of 28.3 grams. \({ }^{\star}\) f) \(\quad\) Why might finding a bootstrap confidence interval not be a good idea for these data?

Rainfall Statistics from Cornell's Northeast Regional Climate Center indicate that Ithaca, New York, gets an average of 35.4 " of rain each year, with a standard deviation of \(4.2 " .\) Assume that a Normal model applies. a. During what percentage of years does Ithaca get more than \(40^{\prime \prime}\) of rain? b. Less than how much rain falls in the driest \(20 \%\) of all years? c. A Cornell University student is in Ithaca for 4 years. Let \(\bar{y}\) represent the mean amount of rain for those 4 years. Describe the sampling distribution model of this sample mean, \(\bar{y}\). d. What's the probability that those 4 years average less than \(30^{\prime \prime}\) of rain?

The sampling distribution of a mean tends toward one like this-a Normal model centered at \(\mu\) with standard deviation \(\sigma / \sqrt{n}\). We know that our sample mean \(\bar{y}\) is somewhere in this picture. But where? \(95 \%\) of samples will have means within 1.96 standard deviations of \(\mu .\) So, if we made \(\mu\) traps for each sample going out \(1.96 \times \frac{\sigma}{\sqrt{n}}\) on either side of \(\bar{y}, 95 \%\) of those traps would capture \(\mu .\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.