Big houses: The U.S. Census Bureau reported that the mean area of U.S. homes built in 2018 was 2559 square feet. Assume that a simple random sample of 23 homes built in 2020 had a mean area of 2695 square feet, with a standard deviation of 175 square feet. Assume the population of areas is normally distributed. Can you conclude that the mean area of homes built in 2020 is not equal to that of homes built in 2018 ? Use the $\alpha =0.01$ level of significance.
$$\begin{array}{l}{H}_0:\mu =2559\\ H_1:\mu \ne 2559\end{array}$$

Part: 0 / 3

Part 1 of 3
(a) Find the $P$-value. Use the TI-84 calculator and round your answer to at least four decimal places.

The $P$-value is $◻$.

Step 1 ：State the null hypothesis $H_0:\mu =2559$, and the alternative hypothesis $H_1:\mu \ne 2559$.

Step 2 ：Given values are: sample mean = 2695, population mean = 2559, sample standard deviation = 175, sample size = 23, and significance level $\alpha =0.01$.

Step 3 ：Calculate the t-score using the formula $(sample\mathrm{\_}mean-population\mathrm{\_}mean)/(sample\mathrm{\_}std\mathrm{\_}dev/\sqrt{sample\mathrm{\_}size})$. The calculated t-score is approximately 3.727.

Step 4 ：Calculate the p-value for a two-tailed test. The p-value is twice the area to the right of the absolute value of the t-score in a t-distribution with degrees of freedom equal to $sample\mathrm{\_}size-1$. The calculated p-value is approximately 0.0012.

Step 5 ：Since the p-value is less than the significance level $\alpha$, we reject the null hypothesis.

Step 6 ：Final Answer: The p-value is $\boxed{{\displaystyle 0.0012}}$.