Question

Assume that a simple random sample has been selected and test the given claim. Identify the null and alternative​ hypotheses, test​ statistic, P-value, and state the final conclusion that addresses the original claim. Listed below are brain volumes in cm^3 of unrelated subjects used in a study. Use a 0.05 significance level to test the claim that the population of brain volumes has a mean equal to 1099.8cm ^3.

964
1028
1273
1080
1070
1173
1067
1347
1099
1203

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
	964	929296
	1028	1056784
	1273	1620529
	1080	1166400
	1070	1144900
	1173	1375929
	1067	1138489
	1347	1814409
	1099	1207801
	1203	1447209
Sum =	11304	12901746

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 = 1099.8

Ha: μ ≠ 1099.8

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 = 2.262

(3) Test Statistics

The t-statistic is computed as follows:

(4) Decision about the null hypothesis

Since it is observed that |t| = 0.825 < t_c = 2.262, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p = 0.4305, and since p = 0.4305 > 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 1099.8, at the 0.05 significance level.