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Chapter 9: Problem 43

A soft-drink manufacturer claims that its 12 -ounce cans do not contain, on average, more than 30 calories. A random sample of 64 cans of this soft drink, which were checked for calories, contained a mean of 32 calories with a standard deviation of 3 calories. Does the sample information support the alternative hypothesis that the manufacturer's claim is false? Use a significance level of \(5 \%\). Find the range for the \(p\) -value for this test. What will your conclusion be using this \(p\) -value and \(\alpha=.05\) ?

Short Answer

Expert verified
Based on the extremely small p-value, we can reject the null hypothesis and conclude that the manufacturer's claim is false. The mean number of calories per can is significantly greater than 30, at a 5% level of significance.

Step by step solution

01

State the null and alternative hypotheses

The null hypothesis, denoted by \(H_0\), is that the mean calorie content per can is not more than 30. In mathematical terms: \(H_0: \mu \leq 30\). The alternative hypothesis, denoted by \(H_a\), is that the mean calorie content per can is more than 30: \(H_a: \mu > 30\).
02

Calculate the test statistic

Calculate the z-value using the formula \(Z = \frac{\bar{x} - \mu_{0}}{\frac{\sigma}{\sqrt{n}}}\), where \(\bar{x} = 32\) is the sample mean, \(\mu_{0} = 30\) is the claimed population mean, \(\sigma = 3\) is standard deviation and \(n = 64\) is the sample size. Setting these values into the formula gives \(Z = \frac{32 - 30}{\frac{3}{\sqrt{64}}} = 5.33\).
03

Find the p-value

The p-value is the probability that a z-score is greater than the calculated z-value (5.33). For the standard normal distribution, a z-value of 5.33 is extremely high. Therefore, the p-value is practically 0.
04

Make your decision

Since the p-value is less than the significance level, \( \alpha = 0.05\), we reject the null hypothesis. This supports the alternative hypothesis that the mean calorie content per can is more than 30.

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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 serves as the starting point for your statistical analysis. It's essentially a statement that suggests there is no effect or difference, and it assumes any observed difference is due to sampling variability or chance.

In this exercise about calorie content in soft-drink cans, the null hypothesis (\( H_0 \)) asserts that the average number of calories per can is not more than 30. Mathematically, it is stated as \( H_0: \mu \leq 30 \).

This hypothesis reflects the manufacturer's claim and suggests there's no deviation above this calorie count. You will test against this hypothesis to determine if it's reasonable to reject it based on the sample data you have.
Alternative Hypothesis
The alternative hypothesis is the opposite of the null hypothesis. It suggests that there is a genuine effect or difference. In other words, it argues against the null hypothesis and asserts that there is some non-random cause to any observed differences.

For our soft-drink example, the alternative hypothesis (\( H_a \)) claims that the true average number of calories per can is more than 30. This is represented mathematically as \( H_a: \mu > 30 \).

The role of the alternative hypothesis is to challenge the null hypothesis, pushing us to test whether the evidence in our sample data is strong enough to reject the status quo.
P-Value
The p-value plays a crucial role in hypothesis testing. It quantifies the strength of the evidence against the null hypothesis. Specifically, it's the probability of observing a test statistic as extreme as, or more extreme than, the one obtained, assuming the null hypothesis is true.

In the context of this exercise, the p-value was calculated for a z-score of 5.33. This z-score is significantly high, indicating that the deviation from the hypothesized mean of 30 calories is very unlikely to have occurred by random chance alone.

When the p-value is exceedingly low, as it is here (practically 0), it signals strong evidence against the null hypothesis. Thus, you would be more inclined to reject the null hypothesis.
Significance Level
The significance level, denoted by the Greek letter alpha \( \alpha \), is a threshold set by the researcher to determine when to reject the null hypothesis. It's a pre-determined probability that defines how much risk of a Type I error (incorrectly rejecting a true null hypothesis) the researcher is willing to take.

In this scenario, a significance level of 0.05 (or 5%) was chosen. This means you would be comfortable with a 5% chance of wrongfully rejecting the null hypothesis if it were indeed true.

If your p-value is less than or equal to the significance level, as it is here, you reject the null hypothesis. A significance level helps you interpret the p-value and make a determination about the hypotheses.

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Most popular questions from this chapter

Write the null and alternative hypotheses for each of the following examples. Determine if each is a case of a two-tailed, a left-tailed, or a right-tailed test. a. To test if the mean amount of time spent per week watching sports on television by all adult men is different from \(9.5\) hours b. To test if the mean amount of money spent by all customers at a supermarket is less than \(\$ 105\) c. To test whether the mean starting salary of college graduates is higher than \(\$ 47,000\) per year d. To test if the mean waiting time at the drive-through window at a fast food restaurant during rush hour differs from 10 minutes e. To test if the mean time spent per week on house chores by all housewives is less than 30 hours

Alpha Airline 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 Airline and agree to reject \(H_{0}\) if 9 or more of them arrive late. Find the significance level for this test.

The manager of a service station claims that the mean amount spent on gas by its customers is \(\$ 15.90\) per visit. You want to test if the mean amount spent on gas at this station is different from \(\$ 15.90\) per visit. Briefly explain how you would conduct this test when \(\sigma\) is not known.

The past records of a supermarket show that its customers spend an average of \(\$ 95\) per visit at this store. Recently the management of the store initiated a promotional campaign according to which each customer receives points based on the total money spent at the store, and these points can be used to buy products at the store. The management expects that as a result of this campaign, the customers should be encouraged to spend more money at the store. To check whether this is true, the manager of the store took a sample of 14 customers who visited the store. The following data give the money (in dollars) spent by these customers at this supermarket during their visits. \(\begin{array}{rrrrrrr}109.15 & 136.01 & 107.02 & 116.15 & 101.53 & 109.29 & 110.79 \\ 94.83 & 100.91 & 97.94 & 104.30 & 83.54 & 67.59 & 120.44\end{array}\) Assume that the money spent by all customers at this supermarket has a normal distribution. Using a \(5 \%\) significance level, can you conclude that the mean amount of money spent by all customers at this supermarket after the campaign was started is more than \(\$ 95\) ? (Hint: First calculate the sample mean and the sample standard deviation for these data using the formulas learned in Sections \(3.1 .1\) and \(3.2 .2\) of Chapter 3 . Then make the test of hypothesis about \(\mu .\) )

What is the difference between the critical value of \(z\) and the observed value of \(z\) ?
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