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Chapter 10: Problem 82

A hot tub manufacturer advertises that with its heating equipment, a temperature of \(100^{\circ} \mathrm{F}\) can be achieved on average in 15 minutes or less. A random sample of 25 tubs is selected, and the time necessary to achieve a \(100^{\circ} \mathrm{F}\) temperature is determined for each tub. The sample mean time and sample standard deviation are \(17.5\) minutes and \(2.2\) minutes, respectively. Does this information cast doubt on the company's claim? Carry out a test of hypotheses using significance level \(.05\).

Short Answer

Expert verified
The solution process involves formulating the hypotheses based on the claim of the manufacturer, calculating the test statistic using the data given, determining the critical value at the given significance level, and making a decision regarding the null hypothesis. The final answer is then based on whether we reject or fail to reject the null hypothesis.

Step by step solution

01

Formulate the Hypotheses

Firstly, we need to formulate our null hypothesis (\(H_0\)) and alternative hypothesis (\(H_a\)). The null hypothesis is the statement that the manufacturer's claim is true. The alternative hypothesis is the statement we are testing for.\n\nSo, for this problem, the hypotheses are:\n\n- Null hypothesis (\(H_0\)): The mean time to heat the tubs to \(100^{\circ} F\) is 15 minutes. (i.e., \(\mu \le 15\))\n- Alternative hypothesis (\(H_a\)): The mean time to heat the tubs to \(100^{\circ} F\) is greater than 15 minutes. (i.e., \(\mu > 15\))
02

Calculate the Test Statistic

Since we're dealing with an unknown population standard deviation and a relatively large sample size (over 30), we will use a t-test. The formula for our test statistic in a t-test is: \n\n\[t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}}\] \n\nwhere \(\bar{x}\) is the sample mean, \(\mu_0\) is the claimed population mean, \(s\) is the sample standard deviation, and \(n\) is the sample size. Substituting the given values into the equation, we have\n\n\[t = \frac{17.5 - 15}{2.2/\sqrt{25}}\]
03

Determine the Critical Value and Make the Decision

To make our decision, we need to determine the critical value and compare our test statistic to it. The critical value corresponds to the given significance level, and it is obtained from the t-distribution table. For a one-tailed test (as we have here since we are testing if \(\mu > 15\)) and 24 degrees of freedom (sample size minus one), the critical value at a .05 significance level is approximately 1.711.\n\nIf our test statistic is greater than the critical value, we reject the null hypothesis. If not, we fail to reject the null hypothesis.
04

Draw Conclusions

Based on the calculation of the test statistic and the comparison to the critical value, we will reach a conclusion whether the company's claim is true or not. If we reject the null hypothesis then the information does cast doubt on the company's claim. If we do not reject it, then the data does not provide enough evidence to refute the claim.

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Key Concepts

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

t-test
A t-test is a statistical method used to determine whether there is a significant difference between the mean of a sample and a known value or another sample's mean. In this context, the t-test helps us assess the manufacturer's claim about the average time required to heat hot tubs. The t-test is particularly helpful when the population standard deviation is unknown, and you're working with a smaller sample size, typically under 30 units.

The formula for calculating the t-statistic is:
  • \[ t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} \]
Where:
  • \( \bar{x} \) is the sample mean,
  • \( \mu_0 \) is the population mean according to the null hypothesis,
  • \( s \) is the sample standard deviation,
  • \( n \) is the sample size.
Understanding how to calculate the t-statistic and interpret its result is crucial in hypothesis testing, allowing us to make evidence-based decisions.
null hypothesis
The null hypothesis, often represented as \( H_0 \), is a fundamental aspect of hypothesis testing. It serves as the starting point for the test, embodying the statement we wish to test for possible rejection based on statistical evidence. In our exercise, the null hypothesis states that the mean heating time is 15 minutes or less. This reflects the manufacturer's claim.

The purpose of the null hypothesis is to provide a baseline or default position that no effect or difference exists. It's important to note that failing to reject \( H_0 \) doesn't prove it true; rather, it means there's insufficient evidence to reject it based on the sample data.

In statistical testing, keeping the null hypothesis in mind helps in determining the direction and nature of the test we conduct.
alternative hypothesis
The alternative hypothesis, denoted as \( H_a \), is what we test against the null hypothesis. It suggests that there is a meaningful effect or difference that counters the null hypothesis. In this situation, the alternative hypothesis posits that the mean time to heat the tubs is greater than 15 minutes.

By proposing this, we are questioning the manufacturer's claim and examining whether the data provides sufficient evidence to support this alternative view.

The alternative hypothesis is complemented by the type of test being conducted鈥攊n our exercise, a one-tailed test is executed since we are specifically investigating whether the actual mean exceeds the claimed mean. Determining the alternative hypothesis accurately is critical as it influences the direction of the critical region of the test.
significance level
The significance level, often represented as \( \alpha \), is a pre-determined threshold that indicates the risk level we are willing to take for making a Type I error鈥攔ejecting the null hypothesis when it is actually true. In many cases, a significance level of 0.05, or 5%, is standard.

In hypothesis testing, the significance level helps determine the critical value, which in turn helps decide whether or not to reject the null hypothesis. A significance level of 0.05 means that there is a 5% risk of concluding that a difference exists when there is none.

Choosing the right significance level helps balance the risk of error with the robustness of the conclusions we draw, providing a practical framework within which statistical evidence can be evaluated. The selected significance level directly influences the critical value from the t-distribution table against which the test statistic is compared.

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Most popular questions from this chapter

Explain why the statement \(\bar{x}=50\) is not a legitimate hypothesis.

Water samples are taken from water used for cooling as it is being discharged from a power plant into a river. It has been determined that as long as the mean temperature of the discharged water is at most \(150^{\circ} \mathrm{F}\), there will be no negative effects on the river's ecosystem. To investigate whether the plant is in compliance with regulations that prohibit a mean discharge water temperature above \(150^{\circ} \mathrm{F}\), researchers will take 50 water samples at randomly selected times and record the temperature of each sample. The resulting data will be used to test the hypotheses \(H_{0}: \mu=150^{\circ} \mathrm{F}\) versus \(\left.H_{a}: \mu\right\rangle\) \(150^{\circ} \mathrm{F}\). In the context of this example, describe Type I and Type II errors. Which type of error would you consider more serious? Explain.

Assuming a random sample from a large population, for which of the following null hypotheses and sample sizes \(n\) is the large-sample \(z\) test appropriate: a. \(H_{0}: p=.2, n=25\) b. \(H_{0}: p=.6, n=210\) c. \(H_{0}: p=.9, n=100\) d. \(H_{0}: p=.05, n=75\)

A television manufacturer claims that (at least) \(90 \%\) of its TV sets will need no service during the first 3 years of operation. A consumer agency wishes to check this claim, so it obtains a random sample of \(n=100\) purchasers and asks each whether the set purchased needed repair during the first 3 years after purchase. Let \(\hat{p}\) be the sample proportion of responses indicating no repair (so that no repair is identified with a success). Let \(p\) denote the actual proportion of successes for all sets made by this manufacturer. The agency does not want to claim false advertising unless sample evidence strongly suggests that \(p<.9\). The appropriate hypotheses are then \(H_{0}: p=.9\) versus \(H_{a}: p<.9\). a. In the context of this problem, describe Type I and Type II errors, and discuss the possible consequences of each. b. Would you recommend a test procedure that uses \(\alpha=.10\) or one that uses \(\alpha=.01 ?\) Explain.

The paper "Playing Active Video Games Increases Energy Expenditure in Children" (Pediatrics [2009]: 534-539) describes an interesting investigation of the possible cardiovascular benefits of active video games. Mean heart rate for healthy boys age 10 to 13 after walking on a treadmill at \(2.6 \mathrm{~km} /\) hour for 6 minutes is 98 beats per minute (bpm). For each of 14 boys, heart rate was measured after 15 minutes of playing Wii Bowling. The resulting sample mean and standard deviation were \(101 \mathrm{bpm}\) and \(15 \mathrm{bpm}\), respectively. For purposes of this exercise, assume that it is reasonable to regard the sample of boys as representative of boys age 10 to 13 and that the distribution of heart rates after 15 minutes of Wii Bowling is approximately normal. a. Does the sample provide convincing evidence that the mean heart rate after 15 minutes of Wii Bowling is different from the known mean heart rate after 6 minutes walking on the treadmill? Carry out a hypothesis test using \(\alpha=.01\). b. The known resting mean heart rate for boys in this age group is \(66 \mathrm{bpm}\). Is there convincing evidence that the mean heart rate after Wii Bowling for 15 minutes is higher than the known mean resting heart rate for boys of this age? Use \(\alpha=.01\).
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