Question

The Watch Corporation of Switzerland claims that its watches on average will neither gain nor lose time during a week. A sample of 18 watches provided the following gains (+) or losses (-) in seconds per week.

-0.41	-0.70	-0.30	-0.11	0.34	-0.62	0.46	0.73	-0.30
-0.55	-0.56	-0.72	-0.14	-0.21	-0.23	-0.76	-0.67	0.22

H₀: μ = 0

H₁: μ ≠ 0

Is it reasonable to conclude that the mean gain or loss in time for the watches is 0? Use the 0.05 significance level.

a. At a level of 0.05 significance, we reject H₀: μ = 0. (Negative answer should be indicated by a minus sign. Round the final answers to 3 decimal places.)

If t < or t > , reject H₀.

b. What is the value of the test statistic? (Round the final answer to 3 decimal places.)

Value of the test statistic =           

c. What is your decision regarding H₀?

H₀: μ = 0.

d. What is the p-value?

The p-value is           

.

Answer 1

The sample size is n = 18. 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
	-0.41	0.1681
	-0.7	0.49
	-0.3	0.09
	-0.11	0.0121
	0.34	0.1156
	-0.62	0.3844
	0.46	0.2116
	0.73	0.5329
	-0.3	0.09
	-0.55	0.3025
	-0.56	0.3136
	-0.72	0.5184
	-0.14	0.0196
	-0.21	0.0441
	-0.23	0.0529
	-0.76	0.5776
	-0.67	0.4489
	0.22	0.0484
Sum =	-4.53	4.421

The sample mean is computed as follows:

Also, the sample variance is

Therefore, the sample standard deviation s is

The provided sample mean is −0.252 and the sample standard deviation is s = 0.439 , and the sample size is n = 18

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ = 0

Ha: μ ≠ 0

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 = 2.11

(3) Test Statistics

The t-statistic is computed as follows:

(4) Decision about the null hypothesis

Since it is observed that |t| = 2.435 > t_c = 2.11 it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p = 0.0262, and since p = 0.0262 < 0.05, it is concluded that the null hypothesis is rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean μ is different than 0, at the 0.05 significance level.