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Chapter 10: Problem 15

A soft-drink machine at a steak house is regulated so that the amount of drink dispensed is approximately normally distributed with a mean of 200 milliliters and a standard deviation of 15 milliliters. The machine is checked periodically by taking a sample of 9 drinks and computing the average content. If \(x\) falls in the interval \(191

Short Answer

Expert verified
The probability of committing a Type I error is approximately 0.072, and the probability of committing a Type II error is approximately 0.115.

Step by step solution

01

Calculate Probability of Type I Error

A type I error occurs when the null hypothesis (H0) is true, but it is rejected. The null hypothesis in this case is that the machine dispenses 200 ml (p = 200 ml). We're interested in the probability that the sample mean \(x\) does not fall in the interval \(191 < x < 209\) even though p = 200 ml. We can compute this by calculating 1 - the probability that \(x\) falls within that interval. The standard deviation of the sample mean \(\sigma_x\) is the population standard deviation \(\sigma\) divided by the square root of the sample size n (\(\sigma_x = \frac{\sigma}{\sqrt{n}} = \frac{15}{\sqrt{9}} = 5\)). We can now calculate Z-scores for 191 and 209 and find the probability that the sample mean falls within that range. \( Z_{191} = \frac{191 - 200}{5} = -1.8 \) and \( Z_{209} = \frac{209 - 200}{5} = 1.8 \). The probability that \( Z \) is between -1.8 and 1.8 is approximately 0.928. Therefore, the probability of a Type I error is 1 - 0.928 = 0.072.
02

Calculate Probability of Type II Error

A type II error occurs when the null hypothesis (H0) is false, but it is not rejected. The alternative in this case is that the machine dispenses 215 ml (p 鈮� 200 ml), but we do not reject the null hypothesis. This means that we find the sample mean \(x\) in the range \(191 < x < 209\) when it actually should be around 215. Therefore, the \(\mu = 215\). The standard deviation of the sample mean remains the same as step 1 (\(\sigma_x = 5\)). We now calculate the Z-scores for 191 and 209 with the new mean (215) and find the probability that the sample mean falls within that range. \( Z_{191} = \frac{191 - 215}{5} = -4.8 \) and \( Z_{209} = \frac{209 - 215}{5} = -1.2 \). The probability that \( Z \) is between -4.8 and -1.2 is approximately 0.115. Therefore, the probability of a Type II error is 0.115.

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Key Concepts

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

Type I Error
A Type I error occurs when you reject a true null hypothesis. In simpler terms, you think there's an effect or a difference when, in fact, there isn't one. This kind of error is often called a "false positive." Imagine you're testing a new drug and you conclude it works when it actually doesn鈥檛, that's a Type I error. In statistical testing, the probability of making a Type I error is represented by the symbol \( \alpha \). This is often set at a 5% significance level, meaning there's a 5% chance of incorrectly rejecting the null hypothesis. However, this threshold can vary based on the field of study or specific conditions of the testing. In the context of our exercise with the soft-drink machine, the null hypothesis is that the machine dispenses exactly 200 milliliters. A Type I error would happen if we conclude it does not, even when it actually does.
Type II Error
A Type II error occurs when the null hypothesis is false, but you fail to reject it. In other words, you think there's no effect or difference when in reality, there is one. This is often called a "false negative." Consider a medical test where you mistakenly assume a person doesn't have a disease when they actually do 鈥� that's a Type II error. The probability of making a Type II error is denoted by \( \beta \). Unlike the probability of Type I error, controlling the Type II error probability can be more complex because it depends on various factors, including the sample size and the effect size. In the soft-drink machine example, a Type II error would occur if we fail to detect that the machine is dispensing 215 milliliters instead of 200 milliliters, resulting in a decision that the machine is operating correctly when it is not.
Normal Distribution
The normal distribution, sometimes known as the "bell curve," is crucial in statistics for analyzing data. It refers to the way data clusters around a mean. In a perfectly normal distribution, the data is symmetric, with most of the observations clustering around the central peak, meaning the probability of values is higher near the mean. This distribution is characterized by its mean (average value) and its standard deviation (how spread out the numbers are). For the soft-drink machine scenario, we assume a normal distribution with a mean of 200 milliliters. The standard deviation tells us how much the drink amounts vary from one another, providing an understanding of the expected variability.
Hypothesis Testing
Hypothesis testing is a statistical method used to decide whether there is enough evidence to reject a hypothesis. It鈥檚 used to determine the likelihood that a given process or hypothesis is valid based on sample data.There are typically two hypotheses in testing:
  • Null Hypothesis \( H_0 \): Suggests no effect or no difference. For the soft-drink machine sample, it posits that the average dispensed amount is 200 milliliters.
  • Alternative Hypothesis \( H_a \): Indicates some effect or difference exists. Here, it states that the average dispensed might be different from 200 milliliters.
Testing involves calculating probabilities or "p-values" to see whether observed results likely happen by random chance within the context of the null hypothesis. If a result is extremely unlikely under the null hypothesis, we may reject \( H_0 \), providing evidence for \( H_a \). This is how we tackle the analysis in our soda machine problem.

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Most popular questions from this chapter

The hydrocarbon emissions are known to have decreased dramatically during the 1980 s. A study was conducted to compare the hydrocarbon emissions at idling speed, in parts per million (ppm), for automobiles of 1980 and \(1990 .\) Twenty cars of each year model were randomly selected and their hydrocarbon emission leycls were recorded. The data are as follows 1980 models: \(\begin{array}{llllllllll}141 & 359 & 247 & 940 & 882 & 494 & 306 & 210 & 105 & 880 \\ 200 & 223 & 188 & 940 & 241 & 190 & 300 & 435 & 241 & 380\end{array}\) 1990 models: \(\begin{array}{rrrrrrrrrr}140 & 160 & 20 & 20 & 223 & 60 & 20 & 95 & 360 & 70 \\\ 220 & 400 & 217 & 58 & 235 & 380 & 200 & 175 & 85 & 65\end{array}\) Test the hypothesis that \(\sigma_{1}=\sigma_{2}\) against the alternative that \(\sigma_{1} \neq \sigma_{2}\). Assume both populations are normal. Use a P-value.

A fuel oil company claims that one-fifth of the homes in a certain city are heated by oil. Do we have reason to believe that fewer than \(1 / 5\) are heated by oil if, in a random sample of 1000 homes in this city, it is found that 136 are heated by oil? Use a P-value in your conclusion.

In a research report by Richard H. Weindruch of the UCLA Medical School, it is claimed that mice with an average life span of 32 months will live to be about 40 months old when \(40 \%\) of the calories in their food are replaced by vitamins and protein. Is there any reason to believe that \(\mu<40\) if 64 mice that are placed on this diet have an average life of 38 months with a standard deviation of 5.8 months? Use a P-value in your conclusion.

A large manufacturing hirm is being charged with discrimination in its hiring practices. (a) What hypothesis is being tested if a jury commits a type I error by dueling the firm guilty? (b) What hypothesis is being tested if a jury commits a type Il error by finding the firm guilty?

In Exercise 1.19 on page \(28,\) test the goodness of fit between the observed class frequencies and the corresponding expected frequencies of a normal distribution with \(p=1.8\) and \(a=0.4,\) using a 0.01 level of significance.
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