Need help now don't delete plz number 5 only
2 answers:
You might be interested in
Answer:
B. 10000 years old, with a margin of error of 400 years
Step-by-step explanation:
Assuming this complete problem: "12. An archaeologist uses an accelerator mass spectrometer to find the age of a buried branch. At the 68% confidence level, the spectrometer estimates that the branch was 10,000 years old with a following could the spectrometer estimate as the age of the branch at the 95% confidence level?"
The possible options are:
A. 9500 years old, with a margin of error of 500 years
B. 10000 years old, with a margin of error of 400 years
C. 9500 years olds, with a margin of error of 50 years
D. 10000 years old, with a margin of error of 40 year
Solution to the problem
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
represent the sample mean for the sample
population mean (variable of interest)
s represent the sample standard deviation
n represent the sample size
The confidence interval for the mean is given by the following formula:
(1)
The margin of error for this case is given by this formula:
For the first case the confidence was 68% so then and
we can find a critical value on the normal standard distribution that accumulates 0.16 of the area on each tail and this value is
, because P(z<-0.944) = 0.16 and P(Z>.944) = 0.16
The margin of error can be expressed like this:
We can solve for the standard error on this case and we got:
And then for the new confidence interval we need to calculate the new . For the first case the confidence was 95% so then
and
we can find a critical value on the normal standard distribution that accumulates 0.025 of the area on each tail and this value is
, because P(z<-1.96) = 0.025 and P(Z>1.96) = 0.025. So then the new margin of error would be:
The estimation for the mean not changes and from the before and new case is 1000 years. So then the best option for this case would be:
B. 10000 years old, with a margin of error of 400 years
And makes sense since a larger confidence interval means a wider interval.
Answer:
Step-by-step explanation:
Hello!
The objective is to test if the two machines are filling the bottles with a net volume of 16.0 ounces or if at least one of them is different.
The parameter of interest is μ₁ - μ₂, if both machines are filling the same net volume this difference will be zero, if not it will be different.
You have one sample of ten bottles filled by each machine and the net volume of the bottles are measured, determining two variables of interest:
Sample 1 - Machine 1
X₁: Net volume of a plastic bottle filled by the machine 1.
n₁= 10
X[bar]₁= 16.02
S₁= 0.03
Sample 2 - Machine 2
X₂: Net volume of a plastic bottle filled by the machine 2
n₂= 10
X[bar]₂= 16.01
S₂= 0.03
With p-values 0.4209 (for X₁) and 0.6174 (for X₂) for the normality test and using α: 0.05, we can say that both variables of interest have a normal distribution:
X₁~N(μ₁;δ₁²)
X₂~N(μ₂;δ₂²)
Both population variances are unknown you have to conduct a homogeneity of variances test to see if they are equal (if they are you can conduct a pooled t-test) or different (if the variances are different you have to use the Welche's t-test)
H₀: δ₁²/δ₂²=1
H₁: δ₁²/δ₂²≠1
α: 0.05
Using a statistic software I've calculated the test
p-value 0.6168
The p-value is greater than the significance level, so the decision is to not reject the null hypothesis then you can conclude that both population variances for the net volume filled in the plastic bottles by machines 1 and 2 are equal.
To study the difference between the population means you can use
The hypotheses of interest are:
H₀: μ₁ - μ₂ = 0
H₁: μ₁ - μ₂ ≠ 0
α: 0.05
The p-value for this test is: 0.882433
The p-value is greater than the significance level, the decision is to not reject the null hypothesis.
Using a significance level of 5%, there is no significant evidence to reject the null hypothesis. You can conclude that the population average of the net volume of the plastic bottles filled by machine one and by machine 2 are equal.
I hope this helps!
