91Ó°ÊÓ

The p-value is an important concept in hypothesis testing, allowing us to understand whether the results of our test are statistically significant. It tells us how likely we are to observe a test statistic as extreme as the one calculated, assuming the null hypothesis is true. In this exercise, the test statistic (Z) was found to be 2. This means we need to find the probability of observing a Z value of 2 or more in a standard normal distribution.
The standard normal distribution is symmetrical and centered around zero. By using a Z-table or statistical software, we can find the area to the right of Z=2 which provides the p-value. In our case, the p-value was determined to be approximately 0.0228, indicating the probability of observing a sample mean as extreme as 75.2 when assuming the population mean is 72, given a standard deviation of 6 and sample size of 16.

The significance level, denoted as \(\alpha\), is a threshold set by the researcher. It determines the criteria for rejecting the null hypothesis. Common significance levels used in testing are 0.05, 0.01, and 0.10, but they can vary depending on the study. Essentially, if the p-value is less than or equal to the significance level, we reject the null hypothesis.
In our exercise, we tested two significance levels: 0.01 and 0.025. When the p-value of 0.0228 was compared to these levels, it was greater than 0.01, indicating insufficient evidence to reject the null hypothesis at this level. However, it was less than 0.025, allowing us to reject the null hypothesis with a significance level of 0.025, indicating a stronger assurance of a true effect.

Normal distribution is a cornerstone concept in statistics, providing the foundation for many statistical tests including the one in this exercise. It is characterized by a bell-shaped curve that is symmetric around the mean. The shape of the curve is determined by its mean (center) and standard deviation (spread).
In hypothesis testing, especially when the sample size is large, the distribution of the sample mean tends to be normally distributed as per the Central Limit Theorem, regardless of the distribution of the original data. In the given task, the population was stated to be normally distributed with a mean of 72 and a standard deviation of 6, allowing us to use the normal distribution accurately to calculate our p-value and test statistic.

The test statistic is a standardized value that helps assess the null hypothesis by comparing the observed sample data against the expected distribution under the null hypothesis. For a normally distributed population, the Z-test is commonly used. The Z-test transforms the sample mean into a Z-score through the formula: \Z = \frac{{\bar{X} - \mu_0}}{{\sigma / \sqrt{n}}}\.
In our problem, with the sample mean \(\bar{X}\) being 75.2, the hypothesized population mean \(\mu_0\) as 72, standard deviation \(\sigma\) as 6, and sample size \(n\) as 16, the test statistic was computed to be \Z = 2\. This Z value indicates how many standard deviations the sample mean is away from the hypothesized population mean, guiding the decision on whether the null hypothesis can be rejected when compared against the p-value and significance levels.