Problem
A simple random sample of 20 pages from a dictionary is obtained. The numbers of words defined on those pages are found, with the results $n=20, \bar{x}=58.5$ words, $s=16.2$ words. Given that this dictionary has 1487 pages with defined words, the claim that there are more than 70,000 defined words is equivalent to the claim that the mean number of words per page is greater than 47.1 words. Use a 0.10 significance level to test the claim that the mean number of words per page is greater than 47.1 words. What does the result suggest about the claim that there are more than 70,000 defined words? Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim. Assume that the population is normally distributed. What are the null and alternative hypotheses? A. \[ \begin{array}{l} H_{0}: \mu=47.1 \text { words } \\ H_{1}: \mu<47.1 \text { words } \end{array} \] B. \[ \begin{array}{l} H_{0}: \mu=47.1 \text { words } \\ H_{1}: \mu>47.1 \text { words } \end{array} \] C. \[ \begin{array}{l} H_{0}: \mu=47.1 \text { words } \\ H_{1}: \mu \neq 47.1 \text { words } \end{array} \] D. \[ \begin{array}{l} H_{0}: \mu>47.1 \text { words } \\ H_{1}: \mu<47.1 \text { words } \end{array} \] Determine the test statistic. 3.15 (Round to two decimal places as needed.) Determine the P-value. (Round to three decimal places as needed.)
Solution
Step 1 ：The null and alternative hypotheses are: \[H_{0}: \mu=47.1 \text { words }\] \[H_{1}: \mu>47.1 \text { words }\]
Step 2 ：The test statistic is calculated using the formula: \[t = \frac{\bar{x} - \mu_{0}}{s/\sqrt{n}}\] Substituting the given values, we get: \[t = \frac{58.5 - 47.1}{16.2/\sqrt{20}} \approx 3.15\]
Step 3 ：The P-value is the probability of observing a test statistic as extreme as, or more extreme than, the observed test statistic, assuming the null hypothesis is true. The P-value is calculated using the test statistic and is approximately 0.003.
Step 4 ：Final Answer: The test statistic is \(\boxed{3.15}\). The P-value is \(\boxed{0.003}\).
