Question

Summary for LTime Anderson-Darling Normality Test A-Squared P-Value 0.43 0.299 Mean 0.98549 StDev 0.29727 0.08837 Variance Skew ness 0.260626 0.301601 100 Kurtosis 0.30103 Minimum 1st Quartile 0.77815 0.95424 1.20412 Median 3rd Q uartile 0.3 0.6 0.9 1.2 1.5 Maximum 1.74819 95% Confidence Interval for Mean 0.92651 1.04448 95% Confidence Interval for Median 0.90309 1.04139 95% Confidence Interval for StDev 95% Confidence Intervals 0.26101 0.34533 0.900 0.925 0.950 1.000

1. The values analyze of LTime follow a normal distribution
a. True
b. False

2. The 75% of values are below of:
a. 0.954
b. 1.204
c. 1.748
d. 1.220

3. A second sampling is made from LTime process. The value of mean was
0.91, is correct to think that the average has changed significantly?
a. True
b. False

Answer 1

1. In Anderson-darling test the hypothesis is

H₀: The sample follows normal distribution

against

H₁: H₀ is not true

and we reject H₀ if the p-value is less than $\alpha$ (we choose it 0.05 in general)

Here p-value is 0.299 which is less than 0.05

Hence we failed to reject the null hypothesis and conclude that the sample follows normal distribution.option

Thus the answer is a. True.

2. Under third quartile (Q₃) there lies 75% of the data. Here Q₃ is 1.20412.

Therefore, the answer is b. 1.204.

3. The 95% confidence interval for mean is (0.92651, 1.04448). For the second sampling the mean is 0.91, which is outside of the confidence interval. Thus the mean has changed significantly.

Hence, the answer is a. True.