Question

We randomly chose 20 words from a randomly selected page in the text of interest and counted the number of letters in each word: 5, 5, 2, 11, 1, 5, 3, 8, 5, 4, 7, 2, 9, 4, 8, 10, 4, 5, 6, 6. Suppose the editor was hoping that the book would have a mean length of 6.5 letters. Does this sample indicate that the authors failed to meet this goal?

Conduct a hypothesis test showing all 4 elements of the test

Calculate and interpret a 95% confidence interval for the true mean word length based on this sample.

If you were the editor would you be satisfied with this information or would you want more data and why?

Answer 1

The sample size is n = 20 . The provided sample data along with the data required to compute the sample mean and sample variance are shown in the table below:

	X	X2
	5	25
	5	25
	2	4
	11	121
	1	1
	5	25
	3	9
	8	64
	5	25
	4	16
	7	49
	2	4
	9	81
	4	16
	8	64
	10	100
	4	16
	5	25
	6	36
	6	36
Sum =	110	742

The sample mean is computed as follows:

Also, the sample variance is

Therefore, the sample standard deviation s is

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ = 6.5

Ha: μ ≠ 6.5

This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.05, and the critical value for a two-tailed test is t_c = 2.093

(3) Test Statistics

The t-statistic is computed as follows:

(4) Decision about the null hypothesis

Since it is observed that |t| = 1.666 ≤tc​=2.093, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p = 0.1122 , and since p = 0.1122≥0.05, it is concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean μ is different than 6.5 , at the 0.05 significance level.

The number of degrees of freedom are df = 20 - 1 = 19 and the significance level is α=0.05.

Based on the provided information, the critical t-value for α=0.05 and df = 19 degrees of freedom is t_c = 2.093

The 95% confidence for the population mean μ is computed using the following expression

Therefore, based on the information provided, the 95 % confidence for the population mean μ is

CI = (5.5 - 1.257, 5.5 + 1.257)

CI = (4.243, 6.757)