91Ó°ÊÓ

When we talk about hypothesis testing in statistics, we always begin with the null hypothesis. The null hypothesis, often symbolized as \( H_0 \), is a statement that indicates no effect or no difference. In the context of your exercise, it suggests that the mean water consumption of households remains at the original level, which is 1245 cubic feet per month.
This hypothesis acts as a default or baseline statement. You assume it to be true unless evidence strongly indicates otherwise. It's a bit like a courtroom where you are innocent until proven guilty. Hypothesis tests are employed to make informed decisions about accepting or rejecting this null hypothesis based on data.
By rejecting the null hypothesis, as seen in the problem, you provide evidence that the mean consumption has likely changed, possibly due to external factors such as the campaign for water conservation.

Once you've established your hypotheses, the next step is to calculate a test statistic known as the z-score. The z-score is a measure that tells you how many standard deviations a data point is from the mean. In hypothesis testing, it helps you understand whether your sample mean is a rare event under the assumption that the null hypothesis is true.
The formula used in this problem is: \[ Z = \frac{\bar{x} - \mu}{ \frac{\sigma}{\sqrt{n}}}\]

• \( \bar{x} \) is the sample mean
• \( \mu \) is the population mean
• \( \sigma \) is the population standard deviation
• \( n \) is the sample size

After plugging in the numbers from the problem, a z-score of -2.8 is computed. This value is important as it quantifies the distance between the sample mean (1175 cubic feet) and the population mean (1245 cubic feet) in terms of standard deviations.

The p-value in hypothesis testing tells you the probability of observing a test statistic as extreme as, or more extreme than, the observed statistic, given that the null hypothesis is true. Essentially, it helps you understand whether your observed sample data is consistent with the null hypothesis.
In the context of your problem, a p-value of 0.0026 was computed from the z-score of -2.8. Since this p-value is considerably lower than the significance level (\( \alpha = 0.025 \)), it suggests that the observed data is highly unlikely under the null hypothesis. This leads us to reject the null hypothesis.
A low p-value implies strong evidence against the null hypothesis, reinforcing the likelihood of the sample mean being different from the established population mean due to the city's campaign.

The critical value is a threshold in hypothesis testing beyond which you reject the null hypothesis. It corresponds to the significance level, \( \alpha \), which is predetermined to decide on the test's strictness.
For a one-tailed test at \( \alpha = 0.025 \), the critical z-value is calculated to be -1.96. This value serves as a cutoff point. If our calculated z-score falls beyond this critical value—in the tail of the distribution—then we reject the null hypothesis.
In this exercise, since our computed z-score of -2.8 is indeed less than -1.96, it indicates that the sample mean is in the rejection region. Thus, you confidently reject the null hypothesis, further confirming that the city's water-saving campaign likely led to reduced water consumption.