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1 answer:
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Answer:
Radius of sphere is 3 units.
Step-by-step explanation:
Volume of sphere is given by
surface area of sphere is given by
where r is the radius of the sphere.
Given that
The number of cubic units in the volume of a sphere is equal to the number of square units in the surface area of the sphere.
we equate formula of Volume of sphere and surface area of sphere
assuming r as the radius.
thus,
Thus, radius of sphere is 3 units.
Using the z-distribution, it is found that the lower bound of the 99% confidence interval is given by:
d. 68.39%.
<h3>What is a confidence interval of proportions?</h3>A confidence interval of proportions is given by:
In which:
is the sample proportion.
- z is the critical value.
- n is the sample size.
In this problem, we have a 99% confidence level, hence, z is the value of Z that has a p-value of
, so the critical value is z = 2.575.
The sample size and estimate are given by:
Hence, the lower bound is given by:
Hence the lower bound is of 68.39%, which means that option D is correct.
More can be learned about the z-distribution at brainly.com/question/25890103
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Answer:
The critical value can be founded with the following Excel formula:
=CHISQ.INV(1-0.05,5)
And we got
a. 11.070
And since our calculated value is lower than the critical we FAIL to reject the null hypothesis at 5% of significance
Step-by-step explanation:
Previous concepts
A chi-square goodness of fit test "determines if a sample data matches a population".
A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".
Solution to the problem
For this case we want to test:
H0: Absenteeism is distributed evenly throughout the week
H1: Absenteeism is NOT distributed evenly throughout the week
We have the following data:
Monday Tuesday Wednesday Thursday Friday Saturday Total
12 9 11 10 9 9 60
The level of significance assumed for this case is
The statistic to check the hypothesis is given by:
The table given represent the observed values, we just need to calculate the expected values with the following formula and the expected value is the same for all the days since that's what we want to test.
now we can calculate the statistic:
Now we can calculate the degrees of freedom (We know that we have 6 categories since we have information for 6 different days) for the statistic given by:
The critical value can be founded with the following Excel formula:
=CHISQ.INV(1-0.05,5)
And we got
a. 11.070
And since our calculated value is lower than the critical we FAIL to reject the null hypothesis at 5% of significance