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Chapter 9: Problem 79

A company claims that its 8 -ounce low-fat yogurt cups contain, on average, at most 150 calories per cup. A consumer agency wanted to check whether or not this claim is true. A random sample of 10 such cups produced the following data on calories. $$ \begin{array}{llllllllll} 147 & 159 & 153 & 146 & 144 & 161 & 163 & 153 & 143 & 158 \end{array} $$ Test using a \(2.5 \%\) significance level whether the company's claim is true. Assume that the numbers of calories for such cups of yogurt produced by this company have an approximate normal distribution.

Short Answer

Expert verified
Based on the t-test at a 2.5 % significance level, the data collected do not provide enough evidence to reject the company's claim that their low-fat yogurt cups contain, on average, at most 150 calories each.

Step by step solution

01

Formulate the Null and Alternative Hypotheses

In this exercise, we are examining whether the average calorie count exceeds the claimed 150 calories. The null hypothesis (\(H_0\)) is that the average calorie count is less than or equal to 150 calories (company's claim), while the alternative hypothesis (\(H_1\)) is that the average calorie count is, in fact, more than 150 calories. Formally, these are stated as follows: \(H_0: \mu \leq 150\) \(H_1: \mu > 150\)
02

Calculate the Sample Mean (\(\bar{x}\)) and Sample Standard Deviation (s)

Based on the given data, we can calculate the sample mean and the sample standard deviation as follows: \(\bar{x} = \frac{1}{n}\sum x_i = \frac{147 + 159 + 153 + 146 + 144 + 161 + 163 + 153 + 143 + 158}{10} = 152.7\) and \(s = \sqrt{\frac{1}{n-1}\sum(x_i - \bar{x})^2} = \sqrt{\frac{1}{10-1}\left[(147-152.7)^2+...+(158-152.7)^2\right]} = \approx 7.432\) where \(x_i\) represents each data point, and n is the number of data points(10 in this case).
03

Determine the Test Statistic (t)

The test statistic for this kind of hypothesis testing can be computed using the formula: \(t = \frac{\bar{x} - \mu_{0}}{s/\sqrt{n}}\) where \(\bar{x}\) is the sample mean, \(\mu_{0}\) is the claimed mean (150 in our case), s is the sample standard deviation, and n is the sample size. So, \(t = \frac{152.7 - 150}{7.432/\sqrt{10}} = 1.153\)
04

Determine Critical Value and Make Decision

At a 2.5 % level of significance and (n-1) degrees of freedom, the critical value (from t-distribution table) is approximately 1.833. Because the calculated t-value (1.153) is less than the critical value (1.833), we do not have evidence to reject the null hypothesis. Hence, the company's claim that the yogurt cups contain, on average, at most 150 calories per cup, is not contradicted by the data provided.

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Key Concepts

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

Null Hypothesis
In hypothesis testing, the null hypothesis is a statement that assumes no effect or no difference. It serves as the default or starting assumption. In this exercise, the null hypothesis (\(H_0\)) is that the average calorie count of the yogurt cups is less than or equal to 150 calories. This reflects the company's claim.
The null hypothesis often represents a status quo or a condition of no change. We test this hypothesis to determine whether there is enough statistical evidence to reject it.
  • The null hypothesis is expressed mathematically as: \(H_0: \mu \leq 150\).
  • The goal of testing is to see whether the sample data provides sufficient evidence to reject \(H_0\).
Rejecting \(H_0\) would imply that the true average is greater than 150 calories.
Alternative Hypothesis
The alternative hypothesis is what you may want to prove in a hypothesis test. It is the statement that contradicts the null hypothesis. In this situation, the alternative hypothesis (\(H_1\)) speculates that the average calorie count of the yogurt cups is more than 150 calories.
When performing the test, the alternative hypothesis is used to determine if there's enough evidence to support a claim against the null.
  • Mathematically, the alternative hypothesis is stated as: \(H_1: \mu > 150\).
  • Proving \(H_1\) suggests the company's claim is false.
This alternative hypothesis is called a one-tailed hypothesis because it tests the possibility of the mean being greater than a certain value, not both directions.
T-Test
A t-test is a statistical method used to determine if there is a significant difference between the means of two groups, which may be related through certain features. In this exercise, we conduct a one-sample t-test because we are comparing the sample mean to a known value鈥攖he company's claimed mean of 150 calories.
  • The test statistic is calculated as: \(t = \frac{\bar{x} - \mu_{0}}{s/\sqrt{n}}\), where \(\bar{x}\) is sample mean, \(\mu_{0}\) is the claimed mean, \(s\) is the sample standard deviation, and \(n\) is sample size.
  • For our data: \(t = \frac{152.7 - 150}{7.432/\sqrt{10}} = 1.153\).
This t-value is a measure of how many standard deviations the sample mean is from the claimed mean. The higher the absolute value of the t-statistic, the more evidence we have against the null hypothesis.
Significance Level
The significance level is a key concept in hypothesis testing, indicating the probability of rejecting the null hypothesis given that it is true. In other words, it's the risk of a Type I error, where evidence leads us to incorrectly rejecting the null hypothesis.
  • A common significance level is 5%, but in this exercise, a stricter level of 2.5% is used.
  • It is symbolized by \( \alpha \), and a 2.5% significance level means that there is a 2.5% risk of concluding a difference exists when there is none.
By selecting a 2.5% significance level, we set a high bar for rejecting the null hypothesis. It means we're ensuring that our conclusions are backed by strong evidence.
Normal Distribution
Normal distribution is a fundamental principle in statistics, often called the bell curve due to its shape. It is relevant because it allows us to make inferences about populations based on sample data.
  • The normal distribution is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean.
  • In this problem, it is assumed that the calorie counts are approximately normally distributed.
This assumption is important because it justifies the use of the t-test. The normal distribution ensures that the test statistic will follow the t-distribution, which is necessary for determining the critical value and making a decision on the hypothesis.

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Most popular questions from this chapter

Explain when a sample is large enough to use the normal distribution to make a test of hypothesis about the population proportion.

Consider \(H_{0}: \mu=80\) versus \(H_{1}: \mu \neq 80\) for a population that is normally distributed. a. A random sample of 25 observations taken from this population produced a sample mean of 77 and a standard deviation of 8. Using \(\alpha=.01\), would you reject the null hypothesis? b. Another random sample of 25 observations taken from the same population produced a sample mean of 86 and a standard deviation of 6 . Using \(\alpha=.01\), would you reject the null hypothesis? Comment on the results of parts a and \(\mathrm{b}\).

More and more people are abandoning national brand products and buying store brand products to save money. The president of a company that produces national brand coffee claims that \(40 \%\) of the people prefer to buy national brand coffee. A random sample of 700 people who buy coffee showed that 259 of them buy national brand coffee. Using \(\alpha=.01\), can you conclude that the percentage of people who buy national brand coffee is different from \(40 \%\) ? Use both approaches to make the test.

Consider the following null and alternative hypotheses: $$ H_{0}=p=.44 \text { versus } H_{1}: p<.44 $$ A random sample of 450 observations taken from this population produced a sample proportion of \(.39 .\) a. If this test is made at a \(2 \%\) significance level, would you reject the null hypothesis? Use the critical-value approach. b. What is the probability of making a Type I error in part a? c. Calculate the \(p\) -value for the test. Based on this \(p\) -value, would you reject the null hypothesis if \(\alpha=.01 ?\) What if \(\alpha=.025\) ?

According to a Bureau of Labor Statistics release of February \(20,2015,79 \%\) of American children under age 18 lived with at least one other sibling in 2014 . Suppose that in a recent sample of 2000 American children under age 18,1620 were living with at least one other sibling. a. Using the critical-value approach and \(\alpha=.05\), test if the current percentage of all American children under age 18 who live with at least one other sibling is different from \(79 \%\). b. How do you explain the Type I error in part a? What is the probability of making this error in part a? c. Calculate the \(p\) -value for the test of part a. What is your conclusion if \(\alpha=.05 ?\)
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