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For Exercises 5 through \(20,\) assume that the variables are normally or approximately normally distributed. Use the traditional method of hypothesis testing unless otherwise specified. Soda Bottle Content A machine fills 12 -ounce bottles with soda. For the machine to function properly, the standard deviation of the population must be less than or equal to 0.03 ounce. A random sample of 8 bottles is selected, and the number of ounces of soda in each bottle is given. At \(\alpha=0.05,\) can we reject the claim that the machine is functioning properly? Use the \(P\) -value method. $$ \begin{array}{cccc}{12.03} & {12.10} & {12.02} & {11.98} \\ {12.00} & {12.05} & {11.97} & {11.99}\end{array} $$

Short Answer

Expert verified
Reject the null hypothesis; the machine is not functioning properly.

Step by step solution

01

Define Hypotheses

First, we need to define our null and alternative hypotheses for this test. - Null hypothesis (\( H_0 \)): The standard deviation of the population is less than or equal to 0.03 ounces.- Alternative hypothesis (\( H_1 \)): The standard deviation of the population is greater than 0.03 ounces. This is a right-tailed test.
02

Calculate Sample Standard Deviation

Next, calculate the sample standard deviation of the given data. The data points are: 12.03, 12.10, 12.02, 11.98, 12.00, 12.05, 11.97, 11.99.1. Calculate the mean of these data points: \( \bar{x} = \frac{12.03 + 12.10 + 12.02 + 11.98 + 12.00 + 12.05 + 11.97 + 11.99}{8} = 12.0175 \)2. Calculate the sum of squared deviations from the mean:\( \Sigma (x_i - \bar{x})^2 = (12.03-12.0175)^2 + (12.10-12.0175)^2 + \ldots + (11.99-12.0175)^2 = 0.0303 \)3. Sample variance, \( s^2 = \frac{0.0303}{8-1} = 0.004329 \)4. Sample standard deviation, \( s = \sqrt{0.004329} \approx 0.0658 \) ounces.
03

Calculate the Chi-Square Test Statistic

Since we are testing the standard deviation, we use the chi-square statistic. The formula is:\[ \chi^2 = \frac{(n-1)s^2}{\sigma^2_0}\]Where:- \( n \) is the sample size (8)- \( s^2 \) is the sample variance- \( \sigma^2_0 \) is the hypothesized population variance (0.03^2)Substitute the values:\[\chi^2 = \frac{(8-1)(0.004329)}{0.0009} = \frac{0.0303}{0.0009} = 33.67\]
04

Determine the Critical Value

Determine the critical value from the chi-square distribution table using \(\alpha = 0.05\) level of significance and degrees of freedom \(df = n-1 = 7\). For a right-tailed test, look up \(\chi_{0.05,7}^2\), which is 14.067.
05

Make a Decision Using the Chi-Square Value

Compare the chi-square statistic (33.67) to the critical value (14.067): - Since 33.67 > 14.067, we reject the null hypothesis. This indicates that there is enough evidence at the 0.05 significance level to reject the claim that the machine is functioning properly.

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Key Concepts

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

Normal Distribution
Normal distribution is a fundamental concept in statistics and is often referred to as the bell-curve due to its characteristic shape. Its importance in hypothesis testing lies in its natural occurrence in many real-world situations, especially in fields like physics, economics, and social sciences. Data presented in a normal distribution tends to cluster around a single central point, the mean, and spreads evenly on both sides.
A few key characteristics of a normal distribution are:
  • It is symmetric about the mean.
  • The mean, median, and mode are all located at the center of the distribution.
  • It has one standard deviation above and below the mean, covering about 68% of the data, while two standard deviations cover about 95%.
In the context of the exercise, assuming a normal distribution helps us use certain statistical methods, such as the chi-square test, which rely on these properties. Understanding the normal distribution allows for accurate hypothesis testing and interpretation of results.
Chi-Square Test
The chi-square test is a statistical method used to determine if there is a significant difference between observed and expected frequencies. In our exercise, we use the chi-square test to evaluate whether the standard deviation of the soda bottle machine's output is within acceptable limits. Specifically, it helps us to test the variability of our sample data against a known or hypothesized population variance.
The chi-square formula is given by:\[ \chi^2 = \frac{(n-1)s^2}{\sigma^2_0} \]where:
  • \(n\) is the sample size
  • \(s^2\) is the sample variance
  • \(\sigma^2_0\) is the hypothesized population variance
For our soda bottle exercise, we calculate the chi-square test statistic using the sample data, and then compare it to critical values from a chi-square distribution table to determine statistical significance. A higher calculated chi-square value than the critical value indicates that the machine's operation deviates significantly from expectations, suggesting it may not be functioning properly.
Standard Deviation
Standard deviation is a statistic that quantifies the amount of variation or dispersion in a set of data values. A low standard deviation means that the data points tend to be close to the mean of the set, while a high standard deviation indicates that the data points are spread out over a larger range.
To calculate the standard deviation:
  1. Find the mean of the data set.
  2. For each data point, find the square of its distance to the mean.
  3. Find the average of those squared distances.
  4. Take the square root of that average.
In our sample exercise, the initial step was to calculate the sample standard deviation to check if it exceeds the hypothesized standard deviation of 0.03 ounces. If our sample standard deviation is significantly larger than this hypothesized value, it could mean the machine is not functioning properly. Knowing how to calculate and interpret standard deviation is crucial in understanding the spread of data sets and their reliability.
P-Value Method
The P-value method offers a way to decide whether to reject a null hypothesis. This is done by interpreting the probability of observing sample data that is as extreme as the actual observed data, assuming the null hypothesis is true.
In hypothesis testing:
  • A small P-value (typically ≤ 0.05) indicates that the observed data is highly improbable under the null hypothesis, suggesting rejection of the null hypothesis.
  • A larger P-value suggests that the data is consistent with the null hypothesis.
For the soda bottle machine exercise, the P-value provides a quantitative measure for deciding if the machine's standard deviation being tested is statistically significant. Determining the P-value involves comparing the chi-square test statistic with values in the chi-square distribution table.
Using the P-value method in hypothesis testing allows researchers to make informed decisions about their hypotheses, helping ensure accurate and reliable conclusions based on sample data.

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Most popular questions from this chapter

For Exercises 5 through \(20,\) perform each of the following steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Recycling Approximately \(70 \%\) of the U.S. population recycles. According to a green survey of a random sample of 250 college students, 204 said that they recycled. At \(\alpha=0.01,\) is there sufficient evidence to conclude that the proportion of college students who recycle is greater than \(70 \% ?\)

For Exercises I through 25, perform each of the following steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use diagrams to show the critical region (or regions), and use the traditional method of hypothesis testing unless otherwise specified. Facebook Friends Many people believe that the average number of Facebook friends is 338 . The population standard deviation is \(43.2 .\) A random sample of 50 high school students in a particular county revealed that the average number of Facebook friends was \(350 .\) At \(\alpha=0.05,\) is there sufficient evidence to conclude that the mean number of friends is greater than \(338 ?\)

For Exercises 7 through \(23,\) perform each of the following steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Find the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Assume that the population is approximately normally distributed. Number of Jobs The U.S. Bureau of Labor and Statistics reported that a person between the ages of 18 and 34 has had an average of 9.2 jobs. To see if this average is correct, a researcher selected a random sample of 8 workers between the ages of 18 and 34 and asked how many different places they had worked. The results were as follows: $$\begin{array}{ccccccccc}{8} & {12} & {15} & {6} & {1} & {9} & {13} & {2}\end{array}$$ At \(\alpha=0.05,\) can it be concluded that the mean is \(9.2 ?\) Use the \(P\) -value method. Give one reason why the respondents might not have given the exact number of jobs that they have worked.

For Exercises 7 through \(23,\) perform each of the following steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Find the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Assume that the population is approximately normally distributed. Sleep Time A person read that the average number of hours an adult sleeps on Friday night to Saturday morning was 7.2 hours. The researcher feels that college students do not sleep 7.2 hours on average. The researcher randomly selected 15 students and found that on average they slept 8.3 hours. The standard deviation of the sample is 1.2 hours. At \(\alpha=0.05,\) is there enough evidence to say that college students do not sleep 7.2 hours on average?

For Exercises 5 through \(20,\) assume that the variables are normally or approximately normally distributed. Use the traditional method of hypothesis testing unless otherwise specified. Exam Grades A statistics professor is used to having a variance in his class grades of no more than \(100 .\) He feels that his current group of students is different, and so he examines a random sample of midterm grades as shown. At \(\alpha=0.05,\) can it be concluded that the variance in grades exceeds \(100 ?\) $$ \begin{array}{lllll}{92.3} & {89.4} & {76.9} & {65.2} & {49.1} \\ {96.7} & {69.5} & {72.8} & {67.5} & {52.8} \\ {88.5} & {79.2} & {72.9} & {68.7} & {75.8}\end{array} $$

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