Given that 0.5 liters is approximately 1.05 pints, how many pints are in one liter? Express your answer as a decimal to the near
2 answers:
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Answer:
1) n=114
2) n=59
3) On this case no, because if we survey just the adults of the nearest college that would be a convenience sample. And when we use "convenience sample" we have some problems associated to bias. This methodology it's not appropiate in order to have a good estimation of the parameter of interest. It's better use a random, cluster or stratified sampling.
Step-by-step explanation:
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".
The population proportion have the following distribution
In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 80% of confidence, our significance level would be given by and
. And the critical value would be given by:
Part 1
The margin of error for the proportion interval is given by this formula:
(a)
And on this case we have that or 6% points, and we are interested in order to find the value of n, if we solve n from equation (a) we got:
(b)
Since we don't have a prior estimate of we can use 0.5 as a good estimate, replacing into equation (b) the values from part a we got:
And rounded up we have that n=114
Part 2
On this case we have a prior estimate for the population proportion and is so replacing the values into equation (b) we got:
And rounded up we have that n=59
Part 3
On this case no, because if we survey just the adults of the nearest college that would be a convenience sample. And when we use "convenience sample" we have some problems associated to bias. This methodology it's not appropiate in order to have a good estimation of the parameter of interest. It's better use a random, cluster or stratified sampling.