Question

Show your work. Carry out all calculations to at least 3 significant digits.

A company selling licenses of new e-commerce software advertised that firms using this software could obtain, on average during the first year, a minimum yield (in cost savings) of 20 percent on average on their software investment. To disprove the claim, we checked with 10 different firms which used the software.

These firms reported the following cost-saving yields (in percent) during the first year of their operations: {17.0, 19.2, 21.5, 18.6, 22.1, 14.9, 18.4, 20.1, 19.4, 18.9}.

a. Do we have significant evidence to show that the software company’s claim of a minimum of 20 percent in cost savings was not valid? Test using ? = 0.05.

b. Compute a 95% confidence interval for the average cost-saving yield estimate.

Answer 1

The sample size is n = 10. 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
	17.0	289
	19.2	368.64
	21.5	462.25
	18.6	345.96
	22.1	488.41
	14.9	222.01
	18.4	338.56
	20.1	404.01
	19.4	376.36
	18.9	357.21
Sum =	190.1	3652.41

The sample mean is computed as follows:

  

Also, the sample variance is

Therefore, the sample standard deviation s is

The provided sample proportion is p =19.01% and the sample standard deviation is s = 2.071, and the sample size is n = 10.

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: p >= 20%

Ha: p < 20%

This corresponds to a left-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 right-tailed test is t_c = -1.833

(3) Test Statistics

The t-statistic is computed as follows:

  

(4) Decision about the null hypothesis

Since it is observed that t =−1.512 > t_c​=-1.833, it is then concluded that the null hypothesis is rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is not enough evidence to claim that the population proportion is greater than 20%, at the 0.05 significance level.

Confidence Interval

The 95% confidence interval is 17.528 < p < 20.492

a. Do we have significant evidence to show that the software company’s claim of a minimum of 20 percent in cost savings was not valid?

Yes, the claim was not valid

b. Compute a 95% confidence interval for the average cost-saving yield estimate.

17.528 < p < 20.492