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When performing hypothesis testing, the t-statistic is a crucial value that helps determine whether to reject the null hypothesis. It measures the difference between the sample mean and the hypothesized population mean, adjusted for the sample size and variability. In the flextime problem, the sample mean number of hours worked was 6.7, while the hypothesized mean was 7. To compute the t-statistic, use the formula:\[t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} \]where \(\bar{x}\) is the sample mean, \(\mu_0\) is the hypothesized mean, \(s\) is the sample standard deviation, and \(n\) is the sample size. This calculation tells us how many standard errors the sample mean is from the hypothesized mean. In our exercise, this resulted in a t-statistic of approximately -1.744.The sign of the t-statistic indicates the direction of the difference. A negative t-statistic, like our -1.744, shows the sample mean is less than the hypothesized mean. This value is then compared against a critical value to make decisions about the hypotheses.

The critical value in hypothesis testing acts as the threshold for deciding whether to reject the null hypothesis. It's determined by the significance level \(\alpha\) and the degrees of freedom of the data. For a one-tailed test, like in this exercise, the critical value represents the boundary for unusually low or high test-statistic values.To find the critical value, we use a t-distribution table or statistical calculator. In the flextime problem, with \(\alpha = 0.05\) and \(79\) degrees of freedom (since degree of freedom \(n-1\)), the critical value was approximately -1.664.

• If the calculated t-statistic is more extreme than the critical value, we reject the null hypothesis.
• If it falls within the critical region (less than the critical value in this one-tailed test), there's enough evidence to support the alternative hypothesis.

In our case, the t-statistic of -1.744 was beyond the critical value of -1.664, leading us to reject the null hypothesis.

The null hypothesis, denoted as \(H_0\), is a statement that indicates no effect or no difference in the context of the research question. In hypothesis testing, it is usually the hypothesis that the test seeks to disprove.For the flextime scenario, the null hypothesis stated that the average number of hours worked on Mondays by assemblers is at least 7 hours: \(H_0: \mu \geq 7\). This represents the current assumption or the status quo.

• We begin by assuming the null hypothesis is true.
• We then gather sample data and calculate the test statistic to compare against a critical value.
• If the evidence (i.e., the test statistic) shows that the sample mean is significantly different (lower, in this case) from the hypothesized mean, we have grounds to reject \(H_0\).

The alternative hypothesis, indicated as \(H_a\) or \(H_1\), suggests a new effect or difference contrary to the null hypothesis. It is what the researcher aims to support with evidence from the sample.In our example of flextime hours, the alternative hypothesis proposed that the mean number of hours worked on Mondays is less than 7: \(H_a: \mu < 7\). This hypothesis implies a deviation from the routine hours.

Key Points about the Alternative Hypothesis

- It's a statement reflecting the presence of an effect or difference.- The test is designed to gather sufficient evidence in favor of the alternative hypothesis.With a significance level of 0.05, the test effectively said that the probability of observing the test statistic or more extreme, assuming the null is true, is less than 5%. As the calculated t-statistic (-1.744) was more extreme than the critical value (-1.664), there was sufficient evidence to support this hypothesis. In practical terms, this means assemblers likely work less than 7 hours on average per Monday under the current schedule.