Your research will find out whether the population mean number of kilowatt-hours per day of electricity consumed is significantly more than 21 for a certain house. Assume that the distribution is normal-shaped. Use the grouped data summary from a large sample of 25 randomly-selected days shown below and use the known population standard deviation of $1.692 \mathrm{kWh} /$ day to test the claim using a $1 \%$ level o significance. \begin{tabular}{|c|c|} \hline \begin{tabular}{c} Electricity consumed \\ in kWh/day \end{tabular} & Frequency \\ \hline 20 & 2 \\ \hline 21 & 5 \\ \hline 22 & 3 \\ \hline 23 & 4 \\ \hline 24 & 5 \\ \hline 25 & 6 \\ \hline \end{tabular} What are the correct hypotheses? \[ \begin{array}{l} H_{0}: \mu \quad \checkmark \checkmark \sigma^{\infty} \nRightarrow \sim \sigma^{\circ} 21 \quad \checkmark \sigma^{\circ} \mathrm{kWh} / \mathrm{day} \\ H_{a}: \mu \quad \vee \infty^{\infty}>\vee \vee \infty^{\circ} \mathrm{kWh} / \text { day } \\ \end{array} \] Based on the hypotheses, find the following to 3 decimal places. Test statistic: p-value: The correct decision is: Select an answer What would be the correct summary? Is there evidence or not to reject or not that the population mean number of $\mathrm{kWh} /$ day of electricity consumed at this house is significantly more than 21 ? The correct summary would be: Select an answer
Solution
Step 1 ：The hypotheses for this problem are as follows: Null Hypothesis (H0): μ ≤ 21 kWh/day Alternative Hypothesis (Ha): μ > 21 kWh/day The null hypothesis states that the population mean number of kilowatt-hours per day of electricity consumed is not significantly more than 21, while the alternative hypothesis states that it is significantly more than 21. To calculate the sample mean, we multiply each value by its frequency, sum these products, and then divide by the total number of observations (25 in this case). Sample mean (x̄) = (20*2 + 21*5 + 22*3 + 23*4 + 24*5 + 25*6) / 25 = 22.8 kWh/day The test statistic (z) is calculated using the formula: z = (x̄ - μ) / (σ / √n), where μ is the population mean, σ is the population standard deviation, and n is the sample size. z = (22.8 - 21) / (1.692 / √25) = 5.327 The p-value is the probability of observing a test statistic as extreme as the one calculated, assuming the null hypothesis is true. Since we are conducting a one-tailed test (because the alternative hypothesis is μ > 21), we look up the z-score in a standard normal distribution table or use a calculator to find the area to the right of the z-score. The p-value associated with a z-score of 5.327 is virtually 0. The decision rule for a one-tailed test at the 1% significance level is to reject the null hypothesis if the p-value is less than 0.01. Since our p-value is virtually 0, we reject the null hypothesis. In conclusion, there is strong evidence to reject the claim that the population mean number of kilowatt-hours per day of electricity consumed at this house is not significantly more than 21. The data suggests that the mean consumption is indeed significantly more than 21 kWh/day.
