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Testing Hypotheses. In Exercises 13鈥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鈥檚 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?

Short Answer

Expert verified

The hypotheses are stated as follows.

H0:=48.0H1:>48.0

The test statistic , and the critical value is .

The null hypothesis is rejected.

Therefore, the mean number of words per page is equal to 48.0. Equivalently, the results also suggest that there cannot be more than 70,000 defined words.

Step by step solution

01

Given information

The sample summary is stated as follows.

Sample size n=10.

Sample mean words x=53.3.

Sample standard deviation words s=15.7.

Level of significance =0.01.

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

02

State the hypotheses

Null hypothesisH0 : The mean number of words per page is equal to 48.0.

Alternative hypothesis H1: The mean number of words per page is greater than 48.0.

For the population mean of words per page as , the hypotheses are stated as follows.

H0:=48.0H1:>48.0

The test is right-tailed.

03

Compute the test statistic

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.

t=x-sn=53.3-4815.710=1.0675

Thus, the test statistic is 1.0675.

04

Compute the critical value

The level of significance is =0.01.

The degree of freedom is computed as follows.

df=n-1=10-1=9

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 .

05

State the decision rule

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.

06

Conclusion

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.

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Most popular questions from this chapter

This exercise contain graphs portraying the decision criterion for a one-mean 2-test. The curve in each graph is the normal curve for the test statistic under the assumption that the null hypothesis is true. For each exercise, determine the

a. rejection region.

c. critical value(s).

b. nonrejection region.

d. significance level.

e. Construct a graph similar to that in Fig. 9.3 on page 361 that depicts your results from parts (a)-(d).

f. Identify the hypothesis test as two tailed, left tailed or right tailed.

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