Question

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 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 \(17.5\) min and \(2.2\) min, respectively. Does this information cast doubt on the company's claim? Carry out a test of hypotheses using significance level \(.05 .\)

Short answer

The test suggests that the company's claim that their tubs' heating equipment can achieve 100-degree Fahrenheit in at most 15 minutes is incorrect, as the test statistic is more than the critical value at the significance level of .05. Therefore, the sample evidence casts doubt on the company's claim.

Step-by-step solution

Formulating the Hypotheses

The null hypothesis \(H_0\) is that the average heating time is 15 minutes, or equivalently, that the company claim is correct. Mathematically, if \( \mu \) is the average heating time, \(H_0: \mu \leq 15\). The alternative hypothesis \(H_a\) is that the average heating time is higher than 15 minutes, or equivalently, that the company's claim is incorrect. Formally, \(H_a: \mu > 15\).

Calculating the Test Statistic

The test statistic \(Z\) is given by \((\bar{x} - \mu_0) / (\sigma / \sqrt{n}) = (17.5 - 15) / (2.2 / \sqrt{25}) = 2.5 / (2.2 / 5) = 5.68\), where \( \bar{x} = 17.5 \) is the sample average time, \( \mu_0 = 15 \) is the value of \(\mu\) assumed in the null hypothesis, \( \sigma = 2.2 \) is the sample standard deviation, and \( n = 25 \) is the sample size.

Checking the Significance Level

For a significance level of \( .05 \), the critical value \(Z_{critical}\) is \(1.645\). Since the calculated test statistic \(Z = 5.68 \) is more than the Z-critical value, we reject the null hypothesis and support the alternative hypothesis.