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Chapter 22: Problem 38

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.

Short Answer

Expert verified
The data satisfies normality assumptions. The mean weight is approximately 28.98 grams with a confidence interval of (28.5663, 29.4003), suggesting actual weights exceed the 28.3 grams stated by the company.

Step by step solution

01

Evaluate Normality Assumption

For inference, one key assumption is that the data comes from a normally distributed population, especially important for small samples like this one (6 observations). Given the weights (29.2, 28.5, 28.7, 28.9, 29.1, 29.5), we see no extreme outliers or skewness, implying they can reasonably be assumed to be normally distributed.
02

Calculating the Mean

To find the mean, add all the weights and divide by the number of weights. \[ \bar{x} = \frac{29.2 + 28.5 + 28.7 + 28.9 + 29.1 + 29.5}{6} \]\[ \bar{x} = \frac{173.9}{6} = 28.9833 \text{ grams} \]
03

Calculating the Standard Deviation

First, find each deviation from the mean and square it, then calculate the variance by averaging those squared deviations, and finally, take the square root for the standard deviation. \[ s = \sqrt{ \frac{(29.2-28.9833)^2 + (28.5-28.9833)^2 + (28.7-28.9833)^2 + (28.9-28.9833)^2 + (29.1-28.9833)^2 + (29.5-28.9833)^2}{5} } \]\[ s \approx 0.398 \text{ grams} \]
04

Calculating the 95% Confidence Interval

Use the formula for a confidence interval for the mean: \[ \bar{x} \pm t^* \frac{s}{\sqrt{n}} \] where \( t^* \) is the t-value for 95% confidence with 5 degrees of freedom (approximately 2.571). Calculate it:\[ 28.9833 \pm 2.571 \times \frac{0.398}{\sqrt{6}} \]\[ 28.9833 \pm 0.417 \]\[ (28.5663, 29.4003) \]
05

Interpret the Confidence Interval

The 95% confidence interval of (28.5663, 29.4003) means that we are 95% confident the true mean weight of such bags of chips lies within this range. This considers random variation in sample selection.
06

Compare with Company's Stated Weight

The company's stated net weight of 28.3 grams is not within the confidence interval (28.5663, 29.4003) calculated. Hence, based on our sample, the actual average weight appears higher than the stated amount.

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Key Concepts

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

Mean and Standard Deviation
Understanding the mean and standard deviation is crucial when dealing with statistical data. Let's break these terms down:
  • Mean: Often referred to as the "average," the mean is the sum of all data points divided by the number of data points. It provides a central value of the dataset.
  • Standard Deviation: This measures how spread out the numbers are in your dataset. A low standard deviation means the data points tend to be close to the mean, while a high standard deviation indicates the data points are spread out over a broader range of values.
To calculate the mean for the Doritos weights, you add the weights and divide by 6, which equals 28.9833 grams. For the standard deviation, after calculating the variance, you take its square root, finding it to be approximately 0.398 grams. These calculations show that the average weight is slightly above the stated weight, with a small variation among the weights.
Normality Assumption
In statistics, we often assume that data is normally distributed, especially in small samples.
  • Normal Distribution: This is a probability distribution that is symmetric around the mean, depicting that data near the mean are more frequent in occurrence than data far from the mean.
  • Importance for Small Samples: When dealing with small datasets, as in the Doritos example with only 6 weights, it's particularly important to ensure the data follows a normal distribution.
  • Checking for Normality: This typically involves inspecting the data for outliers or skewness. If the data doesn't show any extreme outliers or significant skewness, as with our Doritos sample, we can reasonably assume normality.
For statistical inference, it's crucial to affirm this assumption, as failing to do so might invalidate the confidence interval computations and any resulting interpretations.
Statistical Inference
Statistical inference allows us to make conclusions about a population based on a sample. This is a powerful tool in statistics.
  • Confidence Intervals: These are used to estimate the range within which a population parameter, such as a mean, lies with a certain level of confidence (typically 95%).
  • Inference in Context: In the Doritos example, creating a 95% confidence interval helps us infer the true mean weight of all Doritos bags based on the sample data.
The interval from 28.5663 to 29.4003 grams suggests with 95% confidence that the true mean weight is within this range. This provides insight into the accuracy of the company's stated weight and supports decision-making based on statistical evidence.
Sample Size and Distribution
The concepts of sample size and distribution are foundational in understanding data analysis.
  • Sample Size: This is the number of observations in a sample. Larger sample sizes generally provide more accurate estimates of the population parameter because they tend to be more representative.
  • Distribution Shape: Refers to the layout of data points when plotted on a graph. In many statistical methods, assuming the shape is normal (bell-shaped curve) helps predict the behavior of the entire population.
With the Doritos weights, the sample size is quite small (only six bags), yet they appear to follow a normal distribution without extreme variance or outliers. Hence, we can make reasonable inferences about the population, assuming normality. Remember, larger samples tend to give a clearer picture but analyzing smaller samples appropriately can still yield valuable insights. This analysis aids in understanding if the company's weight claim aligns with the sample data.

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Most popular questions from this chapter

TV safety The manufacturer of a metal stand for home TV sets must be sure that its product will not fail under the weight of the TV. since some larger scis weigh nearly 300 pounds, the company's safety inspectors have set a standard of ensuring that the stands can support an average of over 500 pounds. Their inspectors regularly subject a random sample of the stands to increasing weight until they fail. They test the hypothesis \(\mathrm{H}_{0}: \mu=500\) against \(\mathrm{H}_{\mathrm{A}}: \mu>500,\) using the level of significance \(\alpha=0.01 .\) If the sample of stands fails to pass this safety test, the inspectors will not certify the product for sale to the general public. a) Is this an upper-tail or lower-tail test? In the context of the problem, why do you think this is important? b) Explain what will happen if the inspectors commit a Type I error. c) Explain what will happen if the inspectors commit a Type II error.

Pulse rates A medical researcher measured the pulse rates (bcats per minute) of a sample of randomly selected adults and found the following Student's \(t\) -based confidence interval: With \(95.00 \%\) Confidence, $$ 70.887604<\mu(\text { Pulse })<74.497011 $$ a) Explain carefully what the software output means. b) What's the margin of error for this interval? c) If the researcher had calculated a \(99 \%\) confidence interval, would the margin of error be larger or smaller? Explain.

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