Testing Hypotheses. In Exercises 13–24, assume that a simple random sample has been selected and test the given claim. Unless specified by your instructor, use either the P-value method or the critical value method for testing hypotheses. Identify the null and alternative hypotheses, test statistic, P-value (or range of P-values), or critical value(s), and state the final conclusion that addresses the original claim.

How Many English Words? A simple random sample of 10 pages from Merriam-Webster’s Collegiate Dictionary is obtained. The numbers of words defined on those pages are found, with these results: n = 10, x = 53.3 words, s = 15.7 words. Given that this dictionary has 1459 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 48.0 words. Assume a normally distributed population. Use a 0.01 significance level to test the claim that the mean number of words per page is greater than 48.0 words. What does the result suggest about the claim that there are more than 70,000 defined words?

The sample summary is stated as follows.

Sample size\(n = 10\).

Sample mean\(\bar x = 53.3\)words.

Sample standard deviation\(s = 15.7\)words.

Level of significance\(\alpha = 0.01\).

The claim states that the mean number of words per page is 48.0.

Null hypothesis\({H_0}\): The mean number of words per page is equal to 48.0.

Alternative hypothesis\({H_1}\): The mean number of words per page is greater than 48.0.

For the population mean of words per page as\(\mu \), the hypotheses are stated as follows.

\(\begin{array}{l}{H_0}:\mu = 48.0\\{H_1}:\mu > 48.0\end{array}\)

The test is right-tailed.

For a normally distributed population and a randomly selected sample, use student t-distribution if the population standard deviation is unknown.

The test statistic is given as follows.

\(\begin{array}{c}t = \frac{{\bar x - \mu }}{{\frac{s}{{\sqrt n }}}}\\ = \frac{{53.3 - 48}}{{\frac{{15.7}}{{\sqrt {10} }}}}\\ = 1.0675\end{array}\)

Thus, the test statistic is 1.0675.

The level of significance is\(\alpha = 0.01\).

The degree of freedom is computed as follows.

\(\begin{array}{c}df = n - 1\\ = 10 - 1\\ = 9\end{array}\).

Refer to the t-table for the critical value corresponding to 9 degrees of freedom and the level of significance 0.01 for the one-tailed test, which is \({t_{0.01}} = 2.821\).

The decision rule states the following:

If the test statistic is greater than the critical value, the null hypothesis will be rejected.

If the test statistic is not greater than the critical value, the null hypothesis will fail to be rejected.

Here, the test statistic is 1.067, which is lesser than 2.821. Thus, the null hypothesis is failed to be rejected at a 0.01 level of significance.

As the null hypothesis is failed to be rejected, it can be concluded that there is insufficient evidence to support the claim that the words per page are greater than 48.0.

As there are 1459 pages in the dictionary, the total number of 70000 words is equivalent to 48.0 words per page. Thus, it does not support the claim.