00:01
So in this question where the part a is so part a is hypothesis testing.
00:18
So we're comparing the mean of two independent sample let's perform two samples t tests.
00:34
There is two one is null hypothesis null hypothesis next one is alternative hypothesis.
01:00
So null hypothesis that mean two is greater or equal to mean one.
01:08
The mean of oxygen content below the town is greater than and equal to the mean of oxygen content above the town.
01:17
So and this one say that mean two is less than mean one.
01:25
So significance level is 0 .05.
01:35
So calculate the t test.
01:38
So above town t statistic above town is equal to 5.
01:52
So x1 is equal to 5 .06 s1 is equal to 0 .178 and below town x2 is equal to 4 .94 and s2 is equal to 0 .158.
02:20
So standard deviation first find standard pooled standard deviation.
02:57
So pooled deviation the formula is s p is equal to under root n1 minus 1 s1 square plus n2 minus 1 s2 square n1 plus n2 minus 2.
03:21
So the value is approximately is equal to 0 .167.
03:30
So t test is equal to x2 minus x1 bar divided by s p 1 plus n1 plus 1 add 2.
03:44
So this value is equal to minus 2 .023.
03:50
So degree of freedom is equal to so df is equal to n1 plus n2 is equal to 8.
04:06
So rejection in one tail signify the 8 degree of freedom.
04:14
So t is equal to so one tailed test df is equal to 8.
04:26
So t is less than 1 minus 859.
04:32
So this one is degree of freedom t test and s and p standard deviation.
04:53
Next one we find the part b.
05:04
So part b we find the confidence interval.
05:08
Part a we find the t test and part b we find the confidence interval.
05:21
Ci is equal to x1 bar minus x2 bar plus minus t df multiplied by s p.
05:34
So under root 1 plus n1 by n2.
05:40
So value of x is 5 .06 minus x2 4 .94.
05:50
So t value is plus minus 2 .306.
06:04
So 95 percent confidence level.
06:16
So t is equal to divided by and df.
06:20
So is 0 .05 and df is 8.
06:27
So t is equal to 2 .306...