Answer:
a. We reject the null hypothesis at the significance level of 0.05
b. The p-value is zero for practical applications
c. (-0.0225, -0.0375)
Step-by-step explanation:
Let the bottles from machine 1 be the first population and the bottles from machine 2 be the second population.
Then we have
,
,
and
,
,
. The pooled estimate is given by
a. We want to test
vs
(two-tailed alternative).
The test statistic is
and the observed value is
. T has a Student's t distribution with 20 + 25 - 2 = 43 df.
The rejection region is given by RR = {t | t < -2.0167 or t > 2.0167} where -2.0167 and 2.0167 are the 2.5th and 97.5th quantiles of the Student's t distribution with 43 df respectively. Because the observed value
falls inside RR, we reject the null hypothesis at the significance level of 0.05
b. The p-value for this test is given by
0 (4.359564e-10) because we have a two-tailed alternative. Here T has a t distribution with 43 df.
c. The 95% confidence interval for the true mean difference is given by (if the samples are independent)
, i.e.,
where
is the 2.5th quantile of the t distribution with (25+20-2) = 43 degrees of freedom. So
, i.e.,
(-0.0225, -0.0375)
Answer:
x= -1/9
Step-by-step explanation:
6+2x=5-7x
-2x. -2x
6=5-9x
-5. -5
1=-9x/-9
-1/9=x
A
f(x) = 2x + 1
f(3) = 2*3 + 1 = 7
B
g(x) = x^2
g(7) = 7^2
g(7)= 49
C
f(g(x)) = 2(g(x) + 1
f(g(x)) = 2 x^2 + 1
f(g(3) = 2* 3^2 + 1
f(g(3) = 19
D
g(f(x)) = (f(x))^2
g(f(x)) = (2x + 1)^2
g(f(3)) = (2*3 + 1)^2 = (6 + 1)^2 = 49
E
f(g(x)) = 2*g(x) + 1
f(g(x)) = 2*x^2 + 1
f(g(3)) = 2*3^2 + 1
f(g(3)) = 2 * 9 + 1
f(g(3)) = 18 + 1
f(g(3) = 19
F
f(g(x)) = 2(g(x)) + 1
f(g(x)) = 2*x^2 + 1
Answer:
The answer is C, the third one.
Answer:
can l
Step-by-step explanation:
can l? can l? can l~