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The maker of a shampoo knows that customers like this product to have a lot of foam. Ten sample bottles of the product are selected at random and the foam heights observed are as follows (in millimeters): \(210,215,194,195,211,201,\) \(198,204,208,\) and 196 (a) Is there evidence to support the assumption that foam height is normally distributed? (b) Find a \(95 \%\) CI on the mean foam height. (c) Find a \(95 \%\) prediction interval on the next bottle of shampoo that will be tested. (d) Find an interval that contains \(95 \%\) of the shampoo foam heights with \(99 \%\) confidence. (e) Explain the difference in the intervals computed in parts (b), (c), and (d).

Short Answer

Expert verified
The foam heights are likely normally distributed. The 95% CI on the mean is centered around 203.2mm. The prediction interval considers an individual observation, while the CI (b) focuses on the mean. Interval (d) is wider due to the additional confidence level.

Step by step solution

01

Organize Data

First, list out all the foam heights in millimeters: \(210, 215, 194, 195, 211, 201, 198, 204, 208, 196\).
02

Check for Normality

To check if the foam heights are normally distributed, you can visualize the data using a histogram or calculate skewness and kurtosis. Alternatively, use a statistical test like Shapiro-Wilk. If using the histogram, look for the bell shape typical of normal distribution.
03

Calculate Mean and Standard Deviation

Find the mean by adding all foam heights and dividing by the number of samples (10). Mean = \(\frac{210 + 215 + 194 + 195 + 211 + 201 + 198 + 204 + 208 + 196}{10} = 203.2\). Calculate the standard deviation (s) using \ \( s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} \), where \( \bar{x} = 203.2 \) and \( n = 10 \).
04

Construct 95% Confidence Interval for the Mean

Using the formula for a confidence interval: \( \bar{x} \pm t_{0.025,9} \cdot \left(\frac{s}{\sqrt{n}}\right) \). Use a t-distribution table to find \( t_{0.025,9} \) and plug in values to calculate the interval bounds.
05

Construct 95% Prediction Interval for Next Observation

The prediction interval is calculated using: \( \bar{x} \pm t_{0.025,9} \cdot s \sqrt{1 + \frac{1}{n}} \). This gives the range for the next foam height observation.
06

Construct 95% Interval with 99% Confidence for Population

Calculate the interval which contains 95% of the population foam heights with 99% confidence using the formula: \( \bar{x} \pm (z_{0.005} \cdot s) \) where \( z_{0.005} \) is from the z-distribution.
07

Explain the Differences Between Intervals (b), (c), and (d)

The confidence interval estimates the mean foam height with 95% certainty, the prediction interval predicts the range for the next foam height observation, and the third interval estimates the range that contains 95% of foam heights with 99% certainty.

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Key Concepts

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

Prediction Interval
A prediction interval offers a range for predicting where a single new observation might fall. It is broader than a confidence interval because it accounts not only for the uncertainty in estimating the mean but also the natural variation among individual observations.
Imagine measuring the foam height for a new bottle. There’s more variability to account for since each bottle can differ slightly. This additional uncertainty makes prediction intervals wider.
  • Formula: \( \bar{x} \pm t_{0.025,9} \cdot s \sqrt{1 + \frac{1}{n}} \)
  • Includes variance in the population as well as variance in the sample mean.
  • Used for predicting single future observations.
This interval is especially useful when you're dealing with predictions or future estimates, not just estimating a population parameter. It shows a range indicating where the next observation is likely to fall, given our sample data. This is crucial when forecasting with uncertainties, such as customer satisfaction with the foaminess of shampoo bottles.
Normal Distribution
The normal distribution is a type of continuous probability distribution for a real-valued random variable. It is sometimes informally called a bell curve because of its characteristic shape. In this context, the foam heights are assumed to follow this normal distribution.
If data follows a normal distribution, the mean, median, and mode of the data set will be the same. The normal distribution is symmetrical, with a peak at the mean, and it gradually falls off in both directions from the mean.
  • Characteristics include symmetry and a bell shape.
  • The mean determines the center of the distribution.
  • The standard deviation controls the spread of the distribution.
Checking whether data is normally distributed involves visualizing the data through histograms or using statistical tests like Shapiro-Wilk. The normal distribution assumption is essential for constructing intervals such as confidence and prediction intervals, which rely on properties of the bell curve. Understanding this helps us know why certain statistical methods apply and when they might be reliable.
Mean and Standard Deviation
The mean and standard deviation are key statistical measures that provide insights into the data set. The mean represents the average value, calculated by summing all observations and dividing by the count. For the foam heights, the mean is 203.2 millimeters.
The mean gives us a central point around which the data is dispersed.The standard deviation measures the amount of variation or dispersion in a set of values. A low standard deviation means that the values tend to be close to the mean, while a high standard deviation indicates that the values are spread out over a larger range.
  • Mean (\( \bar{x} \)): sum of values divided by number of values.
  • Standard Deviation (s): \( s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} \).
  • Higher standard deviation = more spread out data.
These two metrics are crucial for estimating intervals. The mean indicates where intervals are centered, and the standard deviation affects their width. Calculating these correctly is fundamental to making sound statistical inferences, such as computing prediction and confidence intervals.

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Most popular questions from this chapter

A normal population has a known mean of 50 and unknown variance. (a) A random sample of \(n=16\) is selected from this population, and the sample results are \(\bar{x}=52\) and \(s=8 .\) How unusual are these results? That is, what is the probability of observing a sample average as large as 52 (or larger) if the known, underlying mean is actually \(50 ?\) (b) A random sample of \(n=30\) is selected from this population, and the sample results are \(\bar{x}=52\) and \(s=8\). How unusual are these results? (c) A random sample of \(n=100\) is selected from this population, and the sample results are \(\bar{x}=52\) and \(s=8\). How unusual are these results? (d) Compare your answers to parts (a)-(c) and explain why they are the same or different.

The 2004 presidential election exit polls from the critical state of Ohio provided the following results. The exit polls had 2020 respondents, 768 of whom were college graduates. Of the college graduates, 412 voted for George Bush. (a) Calculate a \(95 \%\) confidence interval for the proportion of college graduates in Ohio who voted for George Bush. (b) Calculate a \(95 \%\) lower confidence bound for the proportion of college graduates in Ohio who voted for George Bush.

Consider the confidence interval for \(\mu\) with known standard deviation \(\sigma\) : $$ \bar{x}-z_{\alpha_{1}} \sigma / \sqrt{n} \leq \mu \leq \bar{x}+z_{\alpha_{2}} \sigma / \sqrt{n} $$ where \(\alpha_{1}+\alpha_{2}=\alpha .\) Let \(\alpha=0.05\) and find the interval for \(\alpha_{1}=\alpha_{2}=\alpha / 2=0.025 .\) Now find the interval for the case \(\alpha_{1}=0.01\) and \(\alpha_{2}=0.04 .\) Which interval is shorter? Is there any advantage to a "symmetric" confidence interval?

Determine the \(\chi^{2}\) percentile that is required to construct each of the following CIs: (a) Confidence level \(=95 \%,\) degrees of freedom \(=24,\) onesided (upper) (b) Confidence level \(=99 \%,\) degrees of freedom \(=9,\) one-sided (lower) (c) Confidence level \(=90 \%,\) degrees of freedom \(=19,\) two-sided.

During the 1999 and 2000 baseball seasons, there was much speculation that the unusually large number of home runs hit was due at least in part to a livelier ball. One way to test the "liveliness" of a baseball is to launch the ball at a vertical surface with a known velocity \(V_{L}\) and measure the ratio of the outgoing velocity \(V_{0}\) of the ball to \(V_{L} .\) The ratio \(R=V_{0} / V_{L}\) is called the coefficient of restitution. Following are measurements of the coefficient of restitution for 40 randomly selected baseballs. The balls were thrown from a pitching machine at an oak surface. \(\begin{array}{llllll}0.6248 & 0.6237 & 0.6118 & 0.6159 & 0.6298 & 0.6192 \\\ 0.6520 & 0.6368 & 0.6220 & 0.6151 & 0.6121 & 0.6548 \\ 0.6226 & 0.6280 & 0.6096 & 0.6300 & 0.6107 & 0.6392 \\ 0.6230 & 0.6131 & 0.6223 & 0.6297 & 0.6435 & 0.5978 \\ 0.6351 & 0.6275 & 0.6261 & 0.6262 & 0.6262 & 0.6314 \\\ 0.6128 & 0.6403 & 0.6521 & 0.6049 & 0.6170 & \\ 0.6134 & 0.6310 & 0.6065 & 0.6214 & 0.6141 & \end{array}\) (a) Is there evidence to support the assumption that the coefficient of restitution is normally distributed? (b) Find a \(99 \%\) CI on the mean coefficient of restitution. (c) Find a \(99 \%\) prediction interval on the coefficient of restitution for the next baseball that will be tested. (d) Find an interval that will contain \(99 \%\) of the values of the coefficient of restitution with \(95 \%\) confidence. (e) Explain the difference in the three intervals computed in parts (b), (c), and (d).

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