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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:
eccentricity; e = 1/7
k = 12
Conic section; Ellipse
Step-by-step explanation:
The first step would be to write the polar equation of the conic section in standard form by multiplying the numerator and denominator by 1/7;
The polar equation of the conic section is now in standard form;
The eccentricity is given by the coefficient of cos theta in which case this would be the value 1/7. Therefore, the eccentricity of this conic section is 1/7.
The eccentricity is clearly between 0 and 1, implying that the conic section is an Ellipse.
The value in the numerator gives the value of k; k = 12