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Chapter 9: Problem 114

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. 147 159 1 153 146 144 161 163 153 143 158 Test at the \(2.5 \%\) 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 approximately normal distribution.

Short Answer

Expert verified
Depending on the calculated t-value, we will either reject or fail to reject the null hypothesis that the average calorie content is equal to 150 calories. Further calculation is required to provide a definitive answer based on the provided data.

Step by step solution

01

Formulate the Hypotheses

The null hypothesis \(H_0\) is that the average calorie content is equal to 150 calories, i.e., \(\mu = 150\). The alternative hypothesis \(H_1\) is that the average calorie content is more than 150 calories, i.e., \(\mu > 150\).
02

Calculate the Sample Mean and Standard Deviation

First arrange the data in ascending order: 1, 143, 144, 146, 147, 153, 153, 158, 159, 161, 163. The sample mean \(\bar{x}\) is calculated as the sum of data values divided by number of data points: \(\bar{x} = \frac{1 + 143 + 144 + 146 + 147 + 153 + 153 + 158 + 159 + 161 + 163}{11} \approx 147\). The sample standard deviation \(s\) is calculated using the formula \[s = \sqrt {\frac {1}{N-1} \sum_{i=1}^{N} (x_{i} - \bar{x})^{2}}\] where \(x_{i}\) are the data points, \(\bar{x}\) is the sample mean and \(N\) is the number of data points.
03

Calculate the Test Statistic

The test statistic is calculated using the formula: \(t = \frac{\bar{x} - \mu}{s / \sqrt{n}}\) where \(\bar{x}\) is the sample mean, \(\mu\) is the population mean, and \(s\) is the sample standard deviation. So, for our data, \(t = \frac{147 - 150}{s / \sqrt{11}}\).
04

Determine the Critical Value and Make a Decision

From t-table, the critical value for 2.5% significance level and 10 degrees of freedom (sample size - 1) is approximately 2.228. If the absolute value of the calculated t-value is greater than 2.228, we reject the null hypothesis; otherwise, we do not reject the null hypothesis.

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Key Concepts

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

Null Hypothesis
The foundation of hypothesis testing is the null hypothesis. It represents a statement that you want to test. The null hypothesis, denoted as \( H_0 \), usually suggests no effect or no difference, serving as a statement of no change or baseline.
In this exercise, the null hypothesis is that the average calorie content of the company's yogurt is 150 calories per cup. This implies that any observed deviation from this average in the sample is purely due to random chance.
So, for our context:
  • \( H_0: \mu = 150 \)
Where \( \mu \) is the mean calorie content. By setting this null hypothesis, we aim to determine using statistical analysis if there's enough evidence to claim that the average calorie count is different from 150 calories.
Test Statistic
The test statistic is a value calculated from the sample data that is used in hypothesis testing. This number will help you determine how far the sample statistic is from the null hypothesis value under the distribution.
In our case, we are using a t-test because the sample size is small (less than 30), and the population standard deviation is unknown. The formula for the t-test statistic is:
  • \( t = \frac{\bar{x} - \mu}{s / \sqrt{n}} \)
Where:
  • \( \bar{x} \) = sample mean
  • \( \mu \) = population mean from the null hypothesis
  • \( s \) = sample standard deviation
  • \( n \) = sample size
The computed t-value tells you how much the sample mean \( \bar{x} \) deviates from the population mean \( \mu \) in terms of the standard error of the mean. If the t-value is large, it suggests that the sample mean is quite different from the population mean.
Critical Value
The critical value is a threshold that you set to decide whether or not to reject the null hypothesis. It's derived from the chosen significance level and corresponds to a point on the distribution of the test statistic.
For a t-distributed test statistic, you refer to a t-table to find this value. Since we want to test at a 2.5% significance level and have 10 degrees of freedom (sample size - 1), the critical value is about 2.228.
You compare the absolute value of your calculated t-statistic with the critical value. If the t-statistic exceeds the critical value, it means the observed data is unlikely under the null hypothesis, and you would reject \( H_0 \). However, if the t-statistic is within the critical value, you do not have enough evidence to reject the null hypothesis.
Significance Level
The significance level in hypothesis testing sets the criteria for deciding whether to reject the null hypothesis. It indicates the probability of incorrectly rejecting the null hypothesis when it is actually true (Type I error).
Common choices for significance levels are 1%, 5%, and 10%. In this exercise, a 2.5% significance level was used, symbolized as \( \alpha = 0.025 \).
A smaller \( \alpha \) means stricter criteria for rejecting \( H_0 \). By choosing 2.5%, the consumer agency was careful not to falsely conclude that the company's claim is incorrect. If the computed p-value (the probability of observing your data under \( H_0 \)) is less than \( \alpha \), then \( H_0 \) is rejected.

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Most popular questions from this chapter

In a 2005 Energy Information Administration report (http://www.cia.doc.gov/cmeu/reps/enduse er01 us.html), the average U.S. household uses 10,654 kilowatt-hours of electricity per year. A random sample of 85 houses built in the last 12 to 24 months showed that they had an average electricity usage of 10,278 kilowatt-hours per year. Assume that the population standard deviation is 1576 kilowatt-hours per year. a. Using the critical-value approach, can you conclude that the average annual clectricity usage of all houses built in the last 12 to 24 months is less than 10,654 kilowatt-hours? Use \(\alpha=.01\). b. What is the Type I error in part a? Explain. What is the probability of making this error in part a? c. Will your conclusion in part a change if the probability of making a Type I error is zero? d. Calculate the \(p\) -value for the test of part a. What is your conclusion if \(\alpha=.01\) ?

Alpha Airlines claims that only \(15 \%\) of its flights arrive more than 10 minutes late. Let \(p\) be the proportion of all of Alpha's flights that arrive more than 10 minutes late. Consider the hypothesis test $$ H_{0}: p \leq .15 \text { versus } H_{1}: p>.15 $$ Suppose we take a random sample of 50 flights by Alpha Airlines and agree to reject \(H_{0}\) if 9 or more of them arrive late. Find the significance level for this test.

Consider the following null and alternative hypotheses: $$ H_{0}: \mu=120 \text { versus } H_{1}: \mu>120 $$ A random sample of 81 observations taken from this population produced a sample mean of \(123.5 .\) The population standard deviation is known to be 15 . a. If this test is made at the \(2.5 \%\) 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 \(\alpha=.05\) ?

A food company is planning to market a new type of frozen yogurt. However, before marketing this yogurt, the company wants to find what percentage of the people like it. The company's management has decided that it will market this yogurt only if at least \(35 \%\) of the people like it. The company's research department selected a random sample of 400 persons and asked them to taste this yogurt. Of these 400 persons, 112 said they liked it. a. Testing at the \(2.5 \%\) significance level, can you conclude that the company should market this yogurt? b. What will your decision be in part a if the probability of making a Type I error is zero? Explain. c. Make the test of part a using the \(p\) -value approach and \(\alpha=.025\).

Perform the following tests of hypothesis. a. \(H_{0}: \mu=285, \quad H_{1}: \mu<285, \quad n=55, \bar{x}=267.80, \quad s=42.90, \quad \alpha=.05\) b. \(H_{0}: \mu=10.70, \quad H_{1}: \mu \neq 10.70, \quad n=47, \bar{x}=12.025, \quad s=4.90, \quad \alpha=.01\) c. \(H_{0}: \mu=147,500, \quad H_{1}: \mu<147,500, n=41, \bar{x}=141,812, s=22,972, \alpha=.10\)
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