91Ó°ÊÓ

The null hypothesis, often symbolized as \( H_0 \), is a foundational concept in hypothesis testing. It represents a default position that there is no effect or no difference in the situation being tested. In this context, the null hypothesis is set based on the company's claim that the average time anyone familiar with computers takes to learn the software is not more than 2 hours. This claim translates to \( H_0: \mu \leq 2 \). The purpose of null hypothesis testing is to determine whether there is sufficient statistical evidence to reject this assumption.

It's essential to use an accurate null hypothesis because it sets the baseline for further statistical analysis. Generally, researchers carry out tests to see if they can reject the null hypothesis in favor of an alternative hypothesis (\( H_a \)). In this setup, the alternative is \( H_a: \mu > 2 \), suggesting that the mean learning time exceeds 2 hours. Overall, rejecting \( H_0 \) would indicate that the company's claim might not hold true based on the sample data.

The sample mean is a statistic used to estimate the population mean. It is calculated by taking the sum of the values in a sample, and dividing it by the number of observations in that sample. In this exercise, the sample mean is calculated from 12 observations or times it took various people to learn the software.

To compute the sample mean, add all the recorded times: 1.75, 2.25, 2.40, 1.90, 1.50, 2.75, 2.15, 2.25, 1.80, 2.20, 3.25, and 2.60. The total sum is 27.15. Then, divide this sum by the number of observations, which is 12. The sample mean is therefore calculated as: \( \bar{x} = \frac{27.15}{12} = 2.2625 \) hours.

Understanding the sample mean helps simplify large datasets into a single representative value, providing a quick outlook of the average performance in learning time across the sample.

The t-distribution is a probability distribution that is used in statistics when the sample size is small and the population standard deviation is unknown. It is slightly different from the normal distribution, being more spread out with heavier tails.

In the exercise at hand, we use the t-distribution because the sample size is relatively small (n = 12). The t-distribution takes into consideration the additional variability inherent in small samples. We compute the t-value as a way of testing the null hypothesis. The t-value formula is \( t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} \), where \( \bar{x} \) is the sample mean, \( \mu_0 \) is the mean stated in the null hypothesis, \( s \) is the sample standard deviation, and \( n \) is the sample size.

Given \( t = \frac{2.2625 - 2}{0.4848/\sqrt{12}} \), the computed t-value is approximately 1.5546. Comparing this t-value with a critical t-value from the t-distribution table will allow for conclusions about the validity of the null hypothesis.

The significance level, denoted as \( \alpha \), is the threshold at which you reject the null hypothesis. It represents the probability of making a Type I error—incorrectly rejecting a true null hypothesis. Common threshold levels are 0.05, 0.01, or 0.10. In this example, a 0.01 or 1% significance level is used, signifying a very strict criterion for rejecting the null hypothesis.

When utilizing a significance level of 0.01 in a one-tailed test, the critical value is derived from the t-distribution. You look up the critical t-value that corresponds to the chosen \( \alpha \) and degrees of freedom in your t-distribution table. Here, the degrees of freedom is 11 (n-1).

The critical t-value was approximately found to be 2.718. To evaluate if you should reject \( H_0 \), you compare the calculated t-value with this critical value. If the t-value is greater, the null is rejected. However, in this case, since the t-value (1.5546) is less than 2.718, there isn't enough evidence to reject the null hypothesis, supporting the company's claim.