Question

A certain pen has been designed so that true average writing lifetime under controlled conditions (involving the use of a writing machine) is at least 10 hours. A random sample of 18 pens is selected, the writing lifetime of each is determined, and a normal probability plot of the resulting data supports the use of a one-sample \(t\) test. The relevant hypotheses are \(H_{0}: \mu=10\) versus \(H_{a}: \mu<10 .\) a. If \(t=-2.3\) and \(\alpha=.05\) is selected, what conclusion is appropriate? b. If \(t=-1.83\) and \(\alpha=.01\) is selected, what conclusion is appropriate? c. If \(t=0.47,\) what conclusion is appropriate?

Short answer

Based on the calculation, for scenario 'a' we reject the null hypothesis indicating mean pen lifetime is less than 10 hours. However, for scenarios 'b' and 'c', we fail to reject the null hypothesis due to insufficient evidence that the mean pen lifetime is under 10 hours.

Step-by-step solution

Understanding the Hypothesis Test

The objective of this test is to check the validity of two opposite hypotheses. Here \(H_{0}: \mu=10\) represents null hypothesis suggesting that the mean pen lifetime is equal to 10 hours. The alternative hypothesis \(H_{a}: \mu<10\) suggests that the mean pen lifetime is less than 10 hours. The t-values provided are the test statistics.

Finding Critical Value for a

The critical value for the one-tailed t test can be found in the t-table at (n - 1) degree of freedom. In this case, degree of freedom is (18 - 1) = 17. If not found directly in the table, we should take the closest lower degree of freedom. For the alpha level \(\alpha=0.05\), the critical value is approximately -1.740 and for \(\alpha=0.01\) the critical value is approximately -2.567.

Comparison and Conclusion for a

Compare the given t-value with the calculated critical value. Here, the critical value is -1.740, and the t-value is -2.3. Since -2.3 is to the left of -1.740 on a normal curve, the result lies into the critical region. So, we conclude by rejecting the null hypothesis \(H_{0}\). It appears that the mean pen lifetime is less than 10 hours.

Comparison and Conclusion for b

Here, the critical value for \(\alpha=0.01\) is -2.567, and the given t-value is -1.83. Since -1.83 is to the right of -2.567 on a normal curve, the observed value falls in the acceptance region. Therefore, we fail to reject the null hypothesis \(H_{0}\) for this case. It cannot be stated that the mean pen lifetime is less than 10 hours.

Comparison and Conclusion for c

The t-value given here is 0.47 which lies in the acceptance region for any reasonable value of \(\alpha\). Therefore, we again fail to reject the null hypothesis \(H_{0}\). There is insufficient evidence to conclude that the mean pen lifetime is less than 10 hours.

Key concepts

Hypothesis Testing

To understand the essence of a one-sample t-test, one must first grasp the concept of hypothesis testing. Hypothesis testing is a method used in statistics to decide between two competing hypotheses regarding a population parameter, based on sampled data.

In the context provided, the objective was to determine whether the average writing lifetime of a pen is at least 10 hours. To accomplish this, the exercise takes us through a process where we have a null hypothesis (\(H_0\)) that asserts the pen’s lifetime is 10 hours, and an alternative hypothesis (\(H_a\)) that it is less than 10 hours. The verification of these claims is done by using the test statistic, in this case, a t-value, and comparing it to a critical value that corresponds with a pre-chosen level of significance, denoted by \(\alpha\).

• If the test statistic falls into the critical region (usually a portion of the t-distribution tails), the null hypothesis is rejected.
• If the test statistic falls outside the critical region, the null hypothesis is not rejected.

Statistical Significance

Statistical significance is a term that helps us determine if the result of a hypothesis test is not due to random chance. In the presented exercise, this is analyzed by setting a significance level noted as \(\alpha\).

The significance level, often chosen as 0.05 or 0.01, represents a threshold beyond which we may conclude the observed data would be very unlikely if the null hypothesis were true.

• For \(\alpha = 0.05\), we are saying there is a 5% risk we are rejecting the null hypothesis when it is actually true (Type I error).
• For a more stringent \(\alpha = 0.01\), the risk is reduced to 1%.

Choosing a suitable \(\alpha\) level is critical as it affects whether we should reject or not reject the null hypothesis, as seen in the various scenarios of the exercise.

Null Hypothesis

The null hypothesis (\(H_0\)) plays a pivotal role in hypothesis testing. It is a statement about the population parameter that indicates no effect, no difference, or a state of skepticism. Essentially, it's the default assumption to test against.

In the exercise we've observed, the null hypothesis was that the mean writing lifetime of the pens is exactly 10 hours (\(H_0: \mu = 10\)). Unless the sample data provides sufficient evidence to the contrary, we do not discard this assumption. The null hypothesis thus provides a benchmark for comparing the evidence provided by the sample data. If the evidence is strong enough to prove that the parameter is different from the value stated in \(H_0\), only then we consider rejecting \(H_0\).

Alternative Hypothesis

Complementing the null hypothesis is the alternative hypothesis (\(H_a\)), which represents what we aim to support by the test. It's a statement that contradicts the null hypothesis, and in this test scenario, it posits that the average writing lifetime of the pens is less than 10 hours (\(H_a: \mu < 10\)).

The alternative hypothesis becomes the focus if we find enough evidence to reject the null hypothesis. In the given exercise, when the test statistic was -2.3 with \(\alpha = 0.05\), we had sufficient evidence to support \(H_a\) and concluded that the writing lifetime is indeed less than 10 hours. Alternatively, for other given \(t\)-values with different \(\alpha\) levels, we did not find enough evidence to reject \(H_0\) and hence, could not substantiate \(H_a\).