Quick anyone know the answer?
1 answer:
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Answer:
The 90% confidence interval for the mean µ of the population of female runners.
( 65.0328 , 66.5672)
Step-by-step explanation:
<u>Step(i)</u>
Given A sample of 12 runners showed a sample mean height of 65.80 inches and a sample standard deviation of 1.95 inches.
Given sample size is n = 12 <30 so small sample
Given sample mean (x⁻) = 65.80 inches
sample standard deviation (S) = 1.95 inches.
<u>Step(ii)</u>
Assume the population is approximately normal.
The 90% confidence interval for the mean µ of the population of female runners.
substitute all above interval
The degrees of freedom γ=n-1 = 12-1=11
From t- table = 1.363 at 90 % 0r 0.10 level of significance
on calculation , we get
(65.80 -0.7672 ,65.80 + 0.7672)
( 65.0328 , 66.5672)
Answer:
b. CHISQ.TEST
Step-by-step explanation:
Notation and previous concepts
A chi-square test is "used to test if the variance of a population is equal to a specified value. This test can be either a two-sided test or a one-sided test. The two-sided version tests against the alternative that the true variance is either less than or greater than the specified value"
The statistic is given by:
In order to calculate the p value we need to have in count the degrees of freedom , on this case . And we can calculate the p value would be given by:
But the correct comand is given by:
b. CHISQ.TEST(Actual range, Expected range)
We just need to put in one column the Observed values and in other the expectec values.