/*! 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 14 For Exercises 5 through \(20,\) ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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. Vitamin C in Fruits and Vegetables The amounts of vitamin C (in milligrams) for \(100 \mathrm{g}(3.57\) ounces) of various randomly selected fruits and vegetables are listed. Is there sufficient evidence to conclude that the standard deviation differs from \(12 \mathrm{mg}\) ? Use \(\alpha=0.10 .\) $$ \begin{array}{rrrrrr}{7.9} & {16.3} & {12.8} & {13.0} & {32.2} & {28.1} \\\ {46.4} & {53.0} & {15.4} & {18.2} & {25.0} & {5.2}\end{array} $$

Short Answer

Expert verified
Calculate and compare the Chi-Square test statistic to critical values to conclude.

Step by step solution

01

State the Hypotheses

First, we need to set up our null hypothesis (\(H_0\)) and the alternative hypothesis (\(H_1\)). Since we are testing if the standard deviation differs from 12 mg, we have: \(H_0: \sigma = 12\) and \(H_1: \sigma eq 12\).
02

Determine the Test Statistic

We use the Chi-Square test for variance. The test statistic is calculated using the formula \(\chi^2 = \frac{(n-1)s^2}{\sigma_0^2}\), where \(s^2\) is the sample variance, \(\sigma_0\) is the hypothesized standard deviation (12 in this case), and \(n\) is the sample size.
03

Calculate the Sample Standard Deviation

Compute the sample standard deviation (\(s\)). First, find the mean of the data set, then compute the variance and take the square root to find \(s\). The data set is: 7.9, 16.3, 12.8, 13.0, 32.2, 28.1, 46.4, 53.0, 15.4, 18.2, 25.0, 5.2.
04

Compute the Sample Variance

Using the sample standard deviation, compute the sample variance \(s^2\).
05

Compute the Test Statistic

Plug the values into the Chi-Square statistic formula: \[\chi^2 = \frac{(12-1)\times s^2}{12^2}\]. This gives us the critical test statistic value.
06

Determine the Critical Chi-Square Values

Look up the critical \(\chi^2\) values from the Chi-Square distribution table for \(\alpha = 0.10\) and with \(n-1 = 11\) degrees of freedom for a two-tailed test.
07

Compare Test Statistic to Critical Values

Compare the computed test statistic to the critical values. If the test statistic falls outside the critical values, we reject the null hypothesis.
08

Conclusion

Based on the comparison, decide whether there is sufficient evidence to reject the null hypothesis or not.

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 Significance
Statistical significance shows us whether an observed effect in a data set is likely due to a specific cause or simply a result of randomness. Imagine performing an experiment, and you see a noticeable result. Statistical significance helps decide if that result is truly meaningful.
In the context of hypothesis testing, we quantify statistical significance using a p-value. A small p-value, typically less than the chosen significance level (often denoted as \(\alpha\)), indicates strong evidence against the null hypothesis.
For example, in a hypothesis test to determine if the standard deviation of vitamin C content in fruits differs from 12 mg, a significance level of \(\alpha = 0.10\) is used. If the testing results show a significant difference, it means the observed values are likely not just due to chance.
Hence, statistical significance helps make informed decisions based on data, ensuring our actions are scientifically grounded.
Chi-Square Test
The Chi-Square test is a statistical method used to test the variance of a sample against a known or hypothesized variance. It is especially handy when assessing how well a sample matches a hypothesized standard deviation. In simpler terms, it checks if the spread of your data is normal.
To perform the Chi-Square test, you need a clear set of data and a hypothesis about the variance you're expecting. In the exercise, we're looking at vitamin C levels in 12 different fruits and vegetables. We want to ascertain if the variance of these levels differs from an assumed 12 mg.
Key steps involve:
  • Calculating the sample variance from your data.
  • Using the Chi-Square test formula: \(\chi^2 = \frac{(n-1)s^2}{\sigma_0^2}\).
  • Analyzing the result to determine if it falls within critical Chi-Square values from a distribution table.
Understanding the Chi-Square test can prevent incorrect assumptions about data and helps validate scientific research.
Sample Standard Deviation
Sample standard deviation (\(s\)) is a measure of how much the data points in a sample deviate from the sample mean. It is essential for understanding the spread of your data.
To find the sample standard deviation:
  • First, calculate the mean (average) of your data set.
  • Then, find the variance by subtracting the mean from each data point, squaring the result, and averaging those squares.
  • Lastly, take the square root of that average to get the standard deviation.
In the case of the vitamin C data, knowing the sample standard deviation helps to perform the Chi-Square test accurately.
It serves as a practical approach to anticipate how spread out or grouped your data values are around the mean. In hypothesis testing, the sample standard deviation is a cornerstone, ensuring accuracy in subsequent test calculations.
Null and Alternative Hypotheses
In hypothesis testing, we start by setting up two contrasting statements: the null hypothesis (\(H_0\)) and the alternative hypothesis (\(H_1\)). These hypotheses guide the testing procedure.
- Null Hypothesis (\(H_0\)): Assumes no effect or difference; it's the hypothesis we look to reject. For example, claiming the standard deviation of vitamin C is 12 mg.
- Alternative Hypothesis (\(H_1\)): Posits there is an effect or a difference. In our case, suggesting the standard deviation is not 12 mg.
These hypotheses provide a framework for statistical testing. By setting a predetermined significance level (\(\alpha\)), we can determine objectively whether to reject or fail to reject the null hypothesis based on collected data.
This process ensures that conclusions drawn from the test are robust and credible, allowing researchers to validate their findings scientifically.

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

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. Takeout Food A magazine article reported that \(11 \%\) of adults buy takeout food every day. A fast-food restaurant owner surveyed 200 customers and found that 32 said that they purchased takeout food every day. At \(\alpha=0.02,\) is there evidence to believe the article's claim? Would the claim be rejected at \(\alpha=0.05 ?\)

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. Soft Drink Consumption A researcher claims that the yearly consumption of soft drinks per person is 52 gallons. In a sample of 50 randomly selected people, the mean of the yearly consumption was 56.3 gallons. The standard deviation of the population is 3.5 gallons. Find the \(P\) -value for the test. On the basis of the \(P\) -value, is the researcher's claim valid?

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. Weight Loss of Newborns An obstetrician read that a newborn baby loses on average 7 ounces in the first two days of his or her life. He feels that in the hospital where he works, the average weight loss of a newborn baby is less than 7 ounces. A random sample of 32 newborn babies has a mean weight loss of 6.5 ounces. The population standard deviation is 1.8 ounces. Is there enough evidence at \(\alpha=0.01\) to support his claim?

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. Heights of Tall Buildings A researcher estimates that the average height of the buildings of 30 or more stories in a large city is at least 700 feet. A random sample of 10 buildings is selected, and the heights in feet are shown. At \(\alpha=0.025,\) is there enough evidence to reject the claim? $$ \begin{array}{ccccc}{485} & {511} & {841} & {725} & {615} \\ {520} & {535} & {635} & {616} & {582}\end{array} $$

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. Medical School Applications A medical college dean read that the average number of applications a potential medical school student sends is 7.8 . She thinks that the mean is higher. So she selects a random sample of 35 applicants and asks each how many medical schools they applied to. The mean of the sample is \(8.7 .\) The population standard deviation is \(2.6 .\) Test her claim at \(\alpha=0.01 .\)

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.