/*! 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 93 A hot tub manufacturer advertise... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A hot tub manufacturer advertises that with its heating equipment, a temperature of \(100^{\circ} \mathrm{F}\) can be achieved in at most 15 min. 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 average time and sample standard deviation are \(17.5 \mathrm{~min}\) and \(2.2 \mathrm{~min}, \mathrm{re}-\) spectively. Does this information cast doubt on the company's claim? Carry out a test of hypotheses using significance level \(.05 .\)

Short Answer

Expert verified
Based on the hypothesis test results, we have sufficient evidence to cast doubt on the company's claim that their hot tubs can reach a temperature of \(100^{\circ} \mathrm{F}\) in at most 15 mins.

Step by step solution

01

Set up the Null and Alternative Hypothesis

The null hypothesis (\(H_0\)) is the manufacturer's claim, that is the tub reaches the specified temperature within 15 mins on average, or in other words the population mean (\(\mu\)) is less than or equal to 15. The alternative hypothesis (\(H_a\)), on the other, refers to what the test aims to detect, that is the tub takes longer than 15 mins to reach the specified temperature, or in other words the population mean is greater than 15. So, \(H_0 = \mu ≤ 15\) and \(H_a = \mu > 15\).
02

Find the Test Statistic

We calculate the test statistic, which is a standardized measurement comparing the observed sample mean (\(x̄\)) to its hypothesized population mean under \(H_0\). The formula to find this for a single-sample t-test is: \(t = \frac{x̄ - \mu} {s/\sqrt{n}}\), where x̄ is the sample mean (17.5 mins), \(\mu\) is the population mean under \(H_0\) (15 mins), s is the sample standard deviation(2.2 mins), and n is the sample size (25). So, \(t = \frac{17.5 - 15} {2.2/\sqrt{25}} = 5.68\)
03

Find the Critical Value

We can find the critical value from a t-distribution table or using statistical software. We need to find the value for a one-sided test (since \(H_a\) is greater than) with a significance level of 0.05 and degrees of freedom \(df = n - 1 = 25 - 1 = 24\). The critical value of t for a one-tailed test with \(df = 24\) and \(\alpha = 0.05\) is approximately 1.7109.
04

Compare Test Statistic and Critical Value

Now, compare the test statistic to the critical value. The test statistic (5.68) is greater than the critical value (1.7109), this means that we reject the null hypothesis.
05

Make the Final Interpretation

Since the test statistic lies in the rejection region, we have sufficient evidence to reject the null hypothesis at a 0.05 significance level. Hence, this information casts doubt on the company's claim that their hot tubs can achieve a temperature of \(100^{\circ} \mathrm{F}\) in at most 15 mins.

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.

Significance Level
In hypothesis testing, the significance level (\( \alpha \), pronounced "alpha") is a critical threshold that helps you decide whether to reject the null hypothesis. It represents the probability of committing a Type I error, which occurs when you incorrectly reject a true null hypothesis.
It acts as a cutoff point. If the probability of observing a test statistic as extreme as the one calculated is less than \( \alpha \), you reject the null hypothesis. Common significance levels include 0.05, 0.01, and 0.10, with 0.05 being one of the most widely used.
In the exercise, a significance level of 0.05 means that if the probability of observing the results under the null hypothesis is less than 5%, we reject it. This provides a clear boundary for decision-making in statistical hypothesis testing.
Null Hypothesis
The null hypothesis, denoted as \( H_0 \), is a statement we aim to test. It's typically a statement of no effect or no difference, essentially a default or "nothing new" position.
In the context of the exercise, the null hypothesis asserts that the mean time it takes to heat the tubs to 100°F is 15 minutes or less.
Formally: \( H_0: \mu \leq 15 \).
This hypothesis is set up to be the one we challenge, attempting to gather sufficient evidence to reject it in favor of the alternative hypothesis. It is crucial because it forms the basis of the test, allowing us to use statistical methods to decide if we should maintain or reject it.
t-Distribution
The t-distribution is a continuous probability distribution that is used when the sample size is small and/or when the population standard deviation is unknown.
It's similar to the normal distribution but has thicker tails, which means it reflects more variability. As the sample size grows, the t-distribution approaches the normal distribution.
In the exercise, the t-distribution helps calculate the critical value needed to determine the rejection region of the null hypothesis. The calculation involves degrees of freedom, which is the sample size minus one (\( df = n - 1 \)). Using a t-table or statistical software, one can find the critical value based on the significance level (0.05) and degrees of freedom (24 in this case).
The t-distribution plays a vital role in deciding whether the observed data is statistically significant under the null hypothesis.
Sample Mean
The sample mean is a measure of central tendency that provides an average of the data in your sample. It serves as an estimate of the population mean, especially when dealing with larger samples.
In the problem statement, the sample mean is 17.5 minutes, which represents the average time taken by the tubs to reach 100°F in the sample observed.
The sample mean is crucial for calculating the test statistic, as it enables the comparison between observed and hypothesized population means.
To compute the sample mean, you sum all observed values and divide by the number of observations in the sample. The formula is: \( \bar{x} = \frac{1}{n}\sum_{i=1}^{n}x_i \).
In hypothesis testing, understanding the relationship between the sample mean, the null hypothesis, and the significance level is essential for making informed statistical decisions.

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 which of the following \(P\) -values will the null hypothesis be rejected when performing a level .05 test: a. 001 d. . 047 b. 021 e. \(.148\) c. 078

A random sample of \(n=44\) individuals with a B.S. degree in accounting who started with a Big Eight accounting firm and subsequently changed jobs resulted in a sample mean time to change of \(35.02\) months and a sample standard deviation of \(18.94\) months ("The Debate over Post-Baccalaureate Education: One University's Experience," Issues in Accounting Education \([1992]: 18-36)\). Can it be concluded that the true average time to change exceeds 2 years? Test the appropriate hypotheses using a significance level of \(.01\).

A manufacturer of hand-held calculators receives large shipments of printed circuits from a supplier. It is too costly and time-consuming to inspect all incoming circuits, so when each shipment arrives, a sample is selected for inspection. Information from the sample is then used to test \(H_{0}: \pi=.05\) versus \(H_{a}: \pi>.05\), where \(\pi\) is the true proportion of defective circuits in the shipment. If the null hypothesis is not rejected, the shipment is accepted, and the circuits are used in the production of calculators. If the null hypothesis is rejected, the entire shipment is returned to the supplier because of inferior quality. (A shipment is defined to be of inferior quality if it contains more than \(5 \%\) defective circuits.) a. In this context, define Type I and Type II errors. b. From the calculator manufacturer's point of view, which type of error is considered more serious? c. From the printed circuit supplier's point of view, which type of error is considered more serious?

Pizza Hut, after test-marketing a new product called the Bigfoot Pizza, concluded that introduction of the Bigfoot nationwide would increase its sales by more than \(14 \%\) (USA Today, April 2, 1993). This conclusion was based on recording sales information for a random sample of Pizza Hut restaurants selected for the marketing trial. With \(\mu\) denoting the mean percentage increase in sales for all Pizza Hut restaurants, consider using the sample data to decide between \(H_{0}: \mu=14\) and \(H_{a}: \mu>14\). a. Is Pizza Hut's conclusion consistent with a decision to reject \(H_{0}\) or to fail to reject \(H_{0}\) ? b. If Pizza Hut is incorrect in its conclusion, is the company making a Type I or a Type II error?

Are young women delaying marriage and marrying at a later age? This question was addressed in a report issued by the Census Bureau (Associated Press, June 8 , 1991). The report stated that in 1970 (based on census results) the mean age of brides marrying for the first time was \(20.8\) years. In 1990 (based on a sample, because census results were not yet available), the mean was \(23.9\). Suppose that the 1990 sample mean had been based on a random sample of size 100 and that the sample standard deviation was \(6.4\). Is there sufficient evidence to support the claim that in 1990 women were marrying later in life than in 1970 ? Test the relevant hypotheses using \(\alpha=.01\). (Note: It is probably not reasonable to think that the distribution of age at first marriage is normal in shape.)

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.