A random sample of 20 U.S.adults finds they have an average height of 170.12 cm with sample standard deviation 9.8 cm.Find the TS, p-value, and decision for testing whether the average height of an American adult differs from 175 cm using a 0.05 level of significance.
A) TS = -2.23; p-value = 0.0380; fail to reject H₀
B) TS = -2.23; p-value = 0.0190; reject H₀
C) TS = 2.23; p-value = 0.0380; reject H₀
D) TS = -2.23; p-value = 0.0380; reject H₀

Question

Quizplus · Accepted Answer

The test statistic (TS) is calculated using the formula for a t-test: $TS = \frac{\bar{x} - \mu}{(s/\sqrt{n})}$, where $\bar{x}$ is the sample mean, $\mu$ is the population mean under the null hypothesis, $s$ is the sample standard deviation, and $n$ is the sample size. Plugging in the given values, $TS = \frac{170.12 - 175}{(9.8/\sqrt{20})}$, which results in approximately -2.23. Since this is a two-tailed test (we're checking if the average height differs, which could be either higher or lower), we look at the two-tailed p-value for the t-distribution with $n-1 = 19$ degrees of freedom. The p-value of 0.0380 is less than the significance level of 0.05, leading us to reject the null hypothesis, indicating that the average height of an American adult is significantly different from 175 cm.

A Random Sample of 20 U