Question

10. Using sample 0.10, 0.11, 0.11, 0.12, 0.13, 0.12,0.11, 0.12, 0.11, 0.11 test the hypothesis MX = 0.10 against the left-hand alternative MX normality, choose ax-5%. 0.10. Assume

Answer 1

Sample size, n =10 <30, small sample. So, we use t-test.

t-test for the population mean(MX):

Null hypothesis (H0): Mean, MX =0.10

Alternative Hypothesis (H1): Mean, MX<0.10

Sample mean, $\bar{X}$ =$\Sigma Xi/n$ =1.14/10 =0.114​​​​​​

Standard deviation, s =$\sqrt{\Sigma (Xi-\bar{X})^2/(n-1)}$ =0.0084​​​​​​

Test statistic, t =$(\bar{X} - MX) /(s/\sqrt{n})$ =$(0.114- 0.10) /(0.0084/\sqrt{10})$ =5.27

At degrees of freedom, df =n - 1 =10 - 1 =9 and $\alpha$ =5% for left-tailed test, the critical value of t is: t_crit = -1.833

Conclusion:

We failed to reject the null hypothesis at 5% significance level because the test statistic of 5.27 does not fall in the rejection region which is shown in the following image:

S2 七-1.833

Thus, there is no sufficient evidence to claim MX<0.10