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Chapter 8: Problem 81

A hot-tub manufacturer advertises that with its heating equipment, a temperature of \(100^{\circ} \mathrm{F}\) can be achieved in at most \(15 \mathrm{~min}\). A random sample of 42 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 \(16.5 \mathrm{~min}\) 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.

Short Answer

Expert verified
The data casts doubt on the company's claim; the average time is more than 15 minutes.

Step by step solution

01

State the Hypotheses

The null hypothesis \(H_0\) is that the average time \( \mu \) to achieve \(100^{\circ} \mathrm{F} \) is \( \mu = 15 \) minutes. The alternative hypothesis \(H_a\) is \( \mu > 15 \) minutes, indicating it takes longer than claimed.
02

Identify the Sample Statistics

We have the sample mean \( \bar{x} = 16.5 \) minutes, sample standard deviation \( s = 2.2 \) minutes, and sample size \( n = 42 \).
03

Calculate the Test Statistic

For the mean of a sample, the test statistic \( t \) is calculated as follows: \[ t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} = \frac{16.5 - 15}{2.2/\sqrt{42}} \]Calculate this value to determine the observed test statistic.
04

Compute the Test Statistic

Compute the numerator and denominator separately:- Numerator: \( 16.5 - 15 = 1.5 \)- Denominator: \( \frac{2.2}{\sqrt{42}} \approx 0.339 \)Thus, \( t \approx \frac{1.5}{0.339} \approx 4.42 \).
05

Calculate the P-value

Using a t-distribution with \( n-1 = 41 \) degrees of freedom, find the \( P \)-value for \( t=4.42 \). A computational tool or t-table shows the \( P \)-value is very small, likely less than 0.001.
06

Compare to Significance Level

Since the \( P \)-value is much smaller than 0.05, it provides strong evidence against the null hypothesis.
07

Conclusion

We reject the null hypothesis. The data suggests that the average time to reach \(100^{\circ} \mathrm{F}\) is significantly greater than the claimed 15 minutes.

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Key Concepts

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

P-value
Understanding the concept of a P-value is crucial in hypothesis testing. A P-value helps us determine the strength of the evidence against the null hypothesis. It essentially tells us the probability of observing a test statistic as extreme or more extreme than what we calculated, just by random chance, assuming the null hypothesis is true.

When you find a very small P-value, it means that the observed data would be extremely unlikely if the null hypothesis were true. In this case, the P-value was less than 0.001, which is much smaller than the typical significance level of 0.05.
  • A small P-value (< 0.05) suggests strong evidence against the null hypothesis, leading to its rejection.
  • A larger P-value (> 0.05) implies insufficient evidence to reject the null hypothesis.
Therefore, a very small P-value, like the one we calculated, convinces us to reject the manufacturer's claim that the heating equipment reaches the desired temperature in at most 15 minutes.
Sample Statistics
Sample statistics are values calculated from a sample, which is a subset of a population, used to estimate the characteristics of the entire population.

In the problem of the hot-tub manufacturer, the sample statistics we calculated were:
  • Sample mean ( \( \bar{x} \) ): 16.5 minutes
  • Sample standard deviation ( \( s \) ): 2.2 minutes
  • Sample size ( \( n \) ): 42 tubs
These statistics are crucial because:
  • The sample mean gives us a measure of the central tendency, indicating how much time on average it took for the tubs to reach 100掳F.
  • The sample standard deviation shows the variability in the times recorded, informing us how spread out the individual measurements were around the sample mean.
  • The sample size impacts the reliability of the sample mean as an estimator of the population mean 鈥� larger samples typically provide more reliable estimates than smaller ones.
T-distribution
The t-distribution is a type of probability distribution used in statistical analysis, particularly when dealing with small sample sizes or when the population standard deviation is unknown.

In this exercise, we used the t-distribution because our sample size was 42, and we needed an appropriate method to calculate the probability of observing results as extreme as ours given that the population's true mean is 15 minutes.
  • The t-distribution is similar to the normal distribution, but it has heavier tails. This property accounts for additional uncertainty due to estimating the population standard deviation from a small sample.
  • The degrees of freedom, denoted as \( n-1 \) , play a crucial role in determining the shape of the t-distribution. In this case, the degrees of freedom were 41.
Using the t-distribution allowed us to calculate the test statistic and subsequently find the P-value to determine the validity of the manufacturer's claim.
Alternative Hypothesis
The alternative hypothesis, denoted as \( H_a \) , represents a new perspective that contradicts the null hypothesis. It's essentially what you're testing against.

In this scenario, the null hypothesis claimed that the hot tubs take at most 15 minutes to reach the desired temperature: \( H_0: \mu = 15 \) minutes.
The alternative hypothesis, on the other hand, suggested that the tubs actually take longer than 15 minutes: \( H_a: \mu > 15 \) minutes.
  • The alternative hypothesis is what researchers typically want to support with evidence from their data.
  • If we reject the null hypothesis in favor of the alternative, it suggests that the data provides sufficient evidence to support the new perspective.
By rejecting the null hypothesis, we supported the alternative hypothesis and suggested that the manufacturer's claim about the time needed for heating may not hold true.

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Most popular questions from this chapter

To obtain information on the corrosion-resistance properties of a certain type of steel conduit, 45 specimens are buried in soil for a 2-year period. The maximum penetration (in mils) for each specimen is then measured, yielding a sample average penetration of \(\bar{x}=52.7\) and a sample standard deviation of \(s=4.8\). The conduits were manufactured with the specification that true average penetration be at most 50 mils. They will be used unless it can be demonstrated conclusively that the specification has not been met. What would you conclude?

Because of variability in the manufacturing process, the actual yielding point of a sample of mild steel subjected to increasing stress will usually differ from the theoretical yielding point. Let \(p\) denote the true proportion of samples that yield before their theoretical yielding point. If on the basis of a sample it can be concluded that more than \(20 \%\) of all specimens yield before the theoretical point, the production process will have to be modified. a. If 15 of 60 specimens yield before the theoretical point, what is the \(P\)-value when the appropriate test is used, and what would you advise the company to do? b. If the true percentage of "early yields" is actually \(50 \%\) (so that the theoretical point is the median of the yield distribution) and a level \(.01\) test is used, what is the probability that the company concludes a modification of the process is necessary?

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 the 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 ?

The article "Caffeine Knowledge, Attitudes, and Consumption in Adult Women" (J. of Nutrition Educ., 1992: 179-184) reports the following summary data on daily caffeine consumption for a sample of adult women: \(n=47\), \(\bar{x}=215 \mathrm{mg}, s=235 \mathrm{mg}\), and range \(=5-1176\). a. Does it appear plausible that the population distribution of daily caffeine consumption is normal? Is it necessary to assume a normal population distribution to test hypotheses about the value of the population mean consumption? Explain your reasoning. b. Suppose it had previously been believed that mean consumption was at most \(200 \mathrm{mg}\). Does the given data contradict this prior belief? Test the appropriate hypotheses at significance level . 10 and include a \(P\)-value in your analysis.

The article "Uncertainty Estimation in Railway Track LifeCycle Cost" (J. of Rail and Rapid Transit, 2009) presented the following data on time to repair (min) a rail break in the high rail on a curved track of a certain railway line. A normal probability plot of the data shows a reasonably linear pattern, so it is plausible that the population distribution of repair time is at least approximately normal. The sample mean and standard deviation are \(249.7\) and \(145.1\), respectively. a. Is there compelling evidence for concluding that true average repair time exceeds \(200 \mathrm{~min}\) ? Carry out a test of hypotheses using a significance level of \(.05\). b. Using \(\sigma=150\), what is the type II error probability of the test used in (a) when true average repair time is actually \(300 \mathrm{~min}\) ? That is, what is \(\beta(300)\) ?
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