Question

Problem 10

Information about ocean weather can be extracted from radar returns with the aid of a special algorithm. A study is conducted to estimate the difference in wind speed as measured on the ground and via the Seasat satellite. To do so, wind speeds are measured using the two methods simultaneously at 12 specified times. The results follow:

Time	1	2	3	4	5	6	7	8	9	10	11	12
Ground(x)	4.46	3.99	3.73	3.29	4.82	6.71	4.61	3.87	3.17	4.42	3.76	3.30
Satellite(y)	4.08	3.94	5.00	5.20	3.92	6.21	5.95	3.07	4.76	3.25	4.89	4.80

(Adapted from Milton & Arnold, 4th edition, page 367)

Researchers are concerned that the Seasat readings are significantly higher than those taken on the ground. Use the 0.05 level of significance to test their concern. Assume both populations have an approximately normal distribution.

1) As the test statistic of -1.04 is less than the critical value of -1.7959, we fail to reject the null hypothesis. The Seasat wind speed readings are significantly higher than the ground wind speed readings.

2) As the test statistic of -1.04 is greater than the critical value of -1.7959, we reject the null hypothesis. The Seasat wind speed readings are not significantly higher than the ground wind speed readings.

3) As the test statistic of -1.2519 is greater than the critical value of -1.7959, we fail to reject the null hypothesis. The Seasat wind speed readings are not significantly higher than the ground wind speed readings.

4) As the test statistic of -1.2519 is less than the critical value of -1.7959, we reject the null hypothesis. The Seasat wind speed readings are significantly higher than the ground wind speed readings.

Answer 1

Ans:

Time	1	2	3	4	5	6	7	8	9	10	11	12
Ground(x)	4.46	3.99	3.73	3.29	4.82	6.71	4.61	3.87	3.17	4.42	3.76	3.3
Satellite(y)	4.08	3.94	5	5.2	3.92	6.21	5.95	3.07	4.76	3.25	4.89	4.8
d	0.38	0.05	-1.27	-1.91	0.9	0.5	-1.34	0.8	-1.59	1.17	-1.13	-1.5
mean(d)=	-0.412
sd	1.140

n=12

df=12-1=11

alpha=0.05

critical t value=-1.7959

Test statistic:

t=(-0.412-0)/(1.14/SQRT(12))

t=-1.2519

As, test statistic t is not less than critical t value of -1.7959,so we fail to reject the null hypothesis.

Option 3 is correct.

As the test statistic of -1.2519 is greater than the critical value of -1.7959, we fail to reject the null hypothesis. The Seasat wind speed readings are not significantly higher than the ground wind speed readings.