Question

A random sample of the ACT scores of 400 students at Big State University provided a sample mean score of 22.48 with a sample standard deviation of 5.76. Find the p-value when testing the claim that u, the population mean ACT score, is greater than 22.
.024
.048
.096
.192
.256

Answer 1

Answer on Question #42184 – Math - Statistics and Probability

A random sample of the ACT scores of 400 students at Big State University provided a sample mean score of 22.48 with a sample standard deviation of 5.76. Find the p-value when testing the claim that $\mu$, the population mean ACT score, is greater than 22.

.024 .048 .096 .192 .256

Solution

Evidence: $n = 400$, $\bar{x} = 22.48$, $s = 5.76$. If the sample size is large ($n=400$), then apply normal approximation and compute the statistic:

$$z = \frac {\bar {x} - \mu}{\frac {s}{\sqrt {n}}} = \frac {22.48 - 22}{\frac {5.76}{\sqrt {400}}} = 1.6666.$$

Compute $p - value = P(z > 1.6666) = 1 - P(z \leq 1.6666) = 0.047797 \approx 0.048$

(via Excel “=1-NORMSDIST(1,6666)”). We can see p-value<$\alpha = 0.05$. In similar way we can test other values of $\mu$.

Answer: 0.048.

www.AssignmentExpert.com