/*! 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 79 A hot-tub manufacturer advertise... [FREE SOLUTION] | 91Ó°ÊÓ

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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 32 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 \(17.5\) min and \(2.2 \mathrm{~min}\), respectively. Does this data cast doubt on the company's claim? Compute the \(P\)-value and use it to reach a conclusion at level .05 (assume that the heating-time distribution is approximately normal).

Short Answer

Expert verified
The data casts doubt on the company's claim as the mean heating time is greater than 15 minutes.

Step by step solution

01

Define the Hypotheses

We begin by defining our null and alternative hypotheses. The null hypothesis, \( H_0 \), states that the true mean heating time is at most 15 minutes, \( \mu \leq 15 \). The alternative hypothesis, \( H_a \), proposes that the true mean heating time is greater than 15 minutes, \( \mu > 15 \).
02

Calculate the Test Statistic

Since the sample size is 32, we use the t-distribution for this hypothesis test. The test statistic is calculated using the formula: \[ t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} \] where \( \bar{x} = 17.5 \), \( \mu_0 = 15 \), \( s = 2.2 \), and \( n = 32 \). Substituting in our values, we compute the test statistic as follows: \[ t = \frac{17.5 - 15}{2.2/\sqrt{32}} \approx 6.27 \]
03

Determine the Degrees of Freedom

The degrees of freedom for this test are equal to \( n - 1 \), which is \( 32 - 1 = 31 \).
04

Find the Critical Value and P-Value

Using the t-distribution table or a calculator, we determine the critical value for a one-tailed test at \( \alpha = 0.05 \) with 31 degrees of freedom. The critical t-value is approximately 1.696. The P-value corresponds to the probability of observing a t-value of 6.27 or larger. Given this large t-value, the P-value will be very small, indicating that our result is extremely unlikely under the null hypothesis.
05

Make a Decision

Since the calculated test statistic (6.27) is much greater than the critical value (1.696), and the P-value is less than 0.05, we reject the null hypothesis. This means there is strong evidence against the company's claim that the mean heating time is at most 15 minutes.
06

Conclusion

With a P-value less than 0.05, we conclude that the data suggests the true mean heating time is greater than 15 minutes, casting doubt on the manufacturer's claim.

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Key Concepts

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

P-value Calculation
In hypothesis testing, the P-value helps us determine the strength of the evidence against the null hypothesis. Essentially, it tells us the probability of obtaining the observed data—or more extreme data—assuming the null hypothesis is true. If this probability is very low, it suggests that the observed data is unlikely under the null hypothesis.
To calculate the P-value, we first need the test statistic. In our exercise, we calculated a t-statistic of 6.27. This value indicates how many standard deviations our sample mean is from the hypothesized mean. We then use this t-statistic to find the P-value from the t-distribution table. A smaller P-value indicates stronger evidence against the null hypothesis.
In practice, we compare the P-value to a significance level, typically 0.05. If the P-value is less than this threshold, we reject the null hypothesis. In this exercise, the P-value is significantly less than 0.05, leading us to doubt the manufacturer's claim.
t-distribution
The t-distribution is crucial in situations where the sample size is small, and the population standard deviation is unknown. It's similar to a normal distribution but has heavier tails. This means it accounts for more variability in small samples. As the sample size increases, the t-distribution approximates the normal distribution more closely.
For our exercise, the sample size is 32, which allows us to use the t-distribution to calculate the test statistic. The t-distribution is defined by degrees of freedom, calculated as the sample size minus one. In our case, we have 31 degrees of freedom.
We used this distribution to find our critical t-value, which we compared to our test statistic to determine if our results were statistically significant. By understanding the behavior of the t-distribution, we can make better decisions about whether our sample data provides strong enough evidence to affect claims about the population.
One-tailed Test
In hypothesis testing, a one-tailed test is used when we are interested in deviations in only one direction. This means we expect the value to be either greater or less than a certain point, not both. Our exercise's alternative hypothesis, which stated the mean heating time to be greater than 15 minutes, is a classic example.
Using a one-tailed test affects the critical value that we use to make our decision about the hypothesis. Specifically, it focuses our region of rejection on one side of the distribution. For our specified significance level \( \alpha = 0.05 \), the entire 5% of the rejection region is on one tail— the right tail, in our example.
Choosing a one-tailed test can increase the test's power if our sample data shows a clear, expected deviation in one direction, as opposed to a two-tailed test which would split the significance level between both ends of the distribution.
Sample Statistics
Sample statistics provide us with estimates of the population parameters when it's impractical to study the entire population. Key sample statistics include the sample mean and sample standard deviation, which were used in our exercise.
The sample mean is the average of the observations in the sample. It serves as an estimator of the true population mean. In our hot tub exercise, the sample mean of 17.5 minutes suggests that it's longer than the manufacturer's claim of 15 minutes.
The sample standard deviation reflects the variability in the data around the sample mean. A smaller standard deviation indicates that the sample data points are closer to the sample mean, while a larger one indicates more variability. Our sample standard deviation is 2.2 minutes, which highlights the spread or variability of the heating times around the sample mean.
Collectively, these statistics fed into our test statistic calculation and ultimately helped us make a more informed decision about the manufacturer's claim.

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Most popular questions from this chapter

The calibration of a scale is to be checked by weighing a 10 -kg test specimen 25 times. Suppose that the results of different weighings are independent of one another and that the weight on each trial is normally distributed with \(\sigma=.200 \mathrm{~kg}\). Let \(\mu\) denote the true average weight reading on the scale. a. What hypotheses should be tested? b. Suppose the scale is to be recalibrated if either \(\bar{x} \geq\) \(10.1032\) or \(\bar{x} \leq 9.8968\). What is the probability that recalibration is carried out when it is actually unnecessary? c. What is the probability that recalibration is judged unnecessary when in fact \(\mu=10.1\) ? When \(\mu=9.8\) ? d. Let \(z=(\bar{x}-10) /(\sigma / \sqrt{n})\). For what value \(c\) is the rejection region of part (b) equivalent to the "two-tailed" region either \(z \geq c\) or \(z \leq-c\) ? e. If the sample size were only 10 rather than 25 , how should the procedure of part (d) be altered so that \(\alpha=.05\) ? f. Using the test of part (e), what would you conclude from the following sample data: $$ \begin{array}{rrrrr} 9.981 & 10.006 & 9.857 & 10.107 & 9.888 \\ 9.728 & 10.439 & 10.214 & 10.190 & 9.793 \end{array} $$ g. Reexpress the test procedure of part (b) in terms of the standardized test statistic \(Z=(\bar{X}-10) /(\sigma / \sqrt{n})\).

A new design for the braking system on a certain type of car has been proposed. For the current system, the true average braking distance at \(40 \mathrm{mph}\) under specified conditions is known to be \(120 \mathrm{ft}\). It is proposed that the new design be implemented only if sample data strongly indicates a reduction in true average braking distance for the new design. a. Define the parameter of interest and state the relevant hypotheses. b. Suppose braking distance for the new system is normally distributed with \(\sigma=10\). Let \(\bar{X}\) denote the sample average braking distance for a random sample of 36 observations. Which of the following three rejection regions is appropriate: \(R_{1}=\\{\bar{x}: \bar{x} \geq 124.80\\}, R_{2}=\\{\bar{x}: \bar{x} \leq 115.20\\}\), \(R_{3}=\\{\bar{x}\) : either \(\bar{x} \geq 125.13\) or \(\bar{x} \leq 114.87\\}\) ? c. What is the significance level for the appropriate region of part (b)? How would you change the region to obtain a test with \(\alpha=.001\) ? d. What is the probability that the new design is not implemented when its true average braking distance is actually \(115 \mathrm{ft}\) and the appropriate region from part (b) is used? e. Let \(Z=(\bar{X}-120) /(\sigma / \sqrt{n})\). What is the significance level for the rejection region \(\\{z: z \leq-2.33\\}\) ? For the region \(\\{z: z \leq-2.88\\} ?\)

For the following pairs of assertions, indicate which do not comply with our rules for setting up hypotheses and why (the subscripts 1 and 2 differentiate between quantities for two different populations or samples): a. \(H_{0}: \mu=100, H_{\mathrm{a}}: \mu>100\) b. \(H_{0}: \sigma=20, H_{\mathrm{a}}: \sigma \leq 20\) c. \(H_{0}: p \neq .25, H_{\mathrm{a}}: p=.25\) d. \(H_{0}: \mu_{1}-\mu_{2}=25, H_{\mathrm{a}}: \mu_{1}-\mu_{2}>100\) e. \(H_{0}: S_{1}^{2}=S_{2}^{2}, H_{\mathrm{a}}: S_{1}^{2} \neq S_{2}^{2}\) f. \(H_{0}: \mu=120, H_{\mathrm{a}}: \mu=150\) g. \(H_{0}: \sigma_{1} / \sigma_{2}=1, H_{\mathrm{a}}: \sigma_{1} / \sigma_{2} \neq 1\) h. \(H_{0}: p_{1}-p_{2}=-.1, H_{\mathrm{a}}: p_{1}-p_{2}<-.1\)

Pairs of \(P\)-values and significance levels, \(\alpha\), are given. For each pair, state whether the observed \(P\)-value would lead to rejection of \(H_{0}\) at the given significance level. a. \(P\)-value \(=.084, \alpha=.05\) b. \(P\)-value \(=.003, \alpha=.001\) c. \(P\)-value \(=.498, \alpha=.05\) d. \(P\)-value \(-.084, \alpha-.10\) e. \(P\)-value \(=.039, \alpha=.01\) f. \(P\)-value \(=.218, \alpha=.10\)

Two different companies have applied to provide cable television service in a certain region. Let \(p\) denote the proportion of all potential subscribers who favor the first company over the second. Consider testing \(H_{0}: p=.5\) versus \(H_{\mathrm{a}}: p \neq .5\) based on a random sample of 25 individuals. Let \(X\) denote the number in the sample who favor the first company and \(x\) represent the observed value of \(X\). a. Which of the following rejection regions is most appropriate and why? $$ \begin{aligned} &R_{1}=\\{x: x \leq 7 \text { or } x \geq 18\\}, R_{2}=\\{x: x \leq 8\\}, \\ &R_{3}=\\{x: x \geq 17\\} \end{aligned} $$ b. In the context of this problem situation, describe what type I and type II errors are. c. What is the probability distribution of the test statistic \(X\) when \(H_{0}\) is true? Use it to compute the probability of a type I error. d. Compute the probability of a type II error for the selected region when \(p=.3\), again when \(p=.4\), and also for both \(p=.6\) and \(p=.7\). e. Using the selected region, what would you conclude if 6 of the 25 queried favored company 1 ?

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