Question

The research department of a particular solar manufacturer believes that the mean energy output of their solar panels is 227 Watts. A random sample of 25 solar panels were tested and the average energy output was 225.5 Watts. It is known that the standard deviation of energy output for the panels the manufacturer produces is 2.5 Watts. A test is carried out to determine if the average energy output of the manufactured panels is different from 227 Watts. What is the p-value for this test? a. 0.4987 b. 0.0013 c. 0.0026 d. 0.9974

Answer 1

This would be a one sample t-test. The null would be $\dpi{80} H_0 : \bar x = \mu$ and the alternate would be $\dpi{80} H_A : \bar x \ne \mu$ . The t statistic would be $\dpi{80} t = \frac{\bar x - \mu}{se(\mu)}$ or $\dpi{80} t = \frac{\bar x - \mu}{\sigma / \sqrt{n}}$ or $\dpi{80} t = \frac{225.5 - 227}{2.5 / \sqrt{25}}$ or $\dpi{80} t = \frac{-1.5}{0.5}$ or $\dpi{80} t = - 3$ .

The p-value is basically the significance level below which a given test statistic would fail to reject the null. We have the p-value for two-sided test as $\dpi{80} P(x < -3) + P(x > 3) = 0.0031 + 0.0031 = \underline{0.0062}$ , for $\dpi{80} x \sim t (df=n-1)$ or $\dpi{80} x \sim t (df=24)$ .

The correct option might be option (c), considering 0.0026 is a typo for 0.0062. But if that is not the case, none of the above is correct.

Answer 2

You are almost right. What you have done to find tcrit is actually to find zcrit. When you have zcrit as -3 the area of the two tails (p) is in fact 0.0026. (You used t table not z table)