魅影直播

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.