Suppose you have selected a random sample of ? displaystyle={14} measurements from a normal distribution. Compare the standard normal ? values with th

Confidence intervals
asked 2021-02-10
Suppose you have selected a random sample of ? \(\displaystyle={14}\) measurements from a normal distribution. Compare the standard normal ? values with the corresponding ? values if you were forming the following confidence intervals.
a) \(95\%\) confidence interval
b) \(80\%\) confidence interval
c) \(98\%\) confidence interval

Answers (1)

Step 1
A random sample of 14 measurements from a normal distribution is selected. That is,
Sample size: \(\displaystyle{n}={14}\)
Let, X be the measurement values.
If some data {x} are normally distributed, the corresponding {z} will be normal with mean 0 and standard deviation 1 where the correspondence between the {x} and {z} is given by \(\displaystyle{z}=\frac{{{x}-\mu}}{\sigma}\). Such z's are called z-scores.
A normal distribution with a mean of 0 and a standard deviation of 1 is called a standard normal distribution. and Z is called as standard normal variate.
That is, \(\displaystyle{Z}~\text{Normal}{\left({0},{1}\right)}\)
It's formula is given as:
Therefore, \(\displaystyle{x}=\mu\pm{z}\sigma\)
Confidence interval for x is: \(\displaystyle{\left(\mu-{z}\sigma,\mu+{z}\sigma\right)}\)
Step 2
(a) \(95\%\) confidence interval:
Using standard normal table ,z value at probability 0.95 corresponding to 1.96.
Therefore , it's confidence interval will be: \(\displaystyle{\left(-{1.96},+{1.96}\right)}\)
A \(95\%\) confidence interval means that if we were to take 100 different samples and compute a \(95\%\) confidence interval for each sample, then approximately 95 of the 100 confidence intervals will contain the true mean value \(\displaystyle{\left(\mu\right)}\)
. (b) \(80\%\) confidence interval:
Using standard normal table ,z value at probability 0.80 corresponding to 1.28.
Therefore, it's confidence interval will be: \(\displaystyle{\left(-{1.28},+{1.28}\right)}\)
(c) \(98\%\) confidence interval:
Using standard normal table , z value at probability 0.98 corresponding to 2.33.
Therefore, it's confidence interval will be: \(\displaystyle{\left(-{2.33},+{2.33}\right)}\)
From the above (a), (b) and (c), it can be observe that:
A higher confidence level is tend to produce a broader confidence interval.
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