Question

select the correct answer from the drop-down menu.
a proposed null hypothesis states that there is no difference in the water consumption of households in two districts. the sample mean difference is 8.5 liters, and the standard deviation of the distribution of the difference in sample means is 4.5 liters.
the lowest confidence level in which the null hypothesis would be accepted is
reset 68% 95% 99.7%

Explanation:

Step1: Recall Confidence Intervals

For a normal distribution, the confidence intervals are based on the number of standard deviations (z - scores) from the mean. The 68 - 95 - 99.7 rule states that about 68% of the data lies within \( \mu\pm\sigma \), 95% within \( \mu\pm2\sigma \), and 99.7% within \( \mu\pm3\sigma \). Here, the null hypothesis assumes the mean difference is 0 (no difference). The sample mean difference is \( \bar{x}_d = 8.5 \) liters and the standard deviation of the difference in sample means (standard error) \( \sigma_d=4.5 \) liters.

Step2: Calculate the number of standard deviations

We calculate the z - score (number of standard deviations the sample mean difference is from 0) as \( z=\frac{\bar{x}_d - 0}{\sigma_d}=\frac{8.5}{4.5}\approx1.89 \)

Step3: Analyze confidence levels

• For 68% confidence, the critical z - value is \( z = \pm1 \). Since our calculated z - score (1.89) is greater than 1, the sample mean difference lies outside the 68% confidence interval. So we cannot accept the null hypothesis at 68%.
• For 95% confidence, the critical z - value is \( z=\pm1.96 \). Our calculated z - score (1.89) is less than 1.96, so the sample mean difference lies within the 95% confidence interval.
• For 99.7% confidence, the critical z - value is \( z = \pm3 \). The sample mean difference will also lie within this interval, but we are looking for the lowest confidence level.

Answer:

95%