Victor Wall

2021-12-26

Let us assume that 1000 different random samples was selected from the same population and for each of them a confidence interval of the population mean was constructed (using exactly the same method) with confidence level of 96%. On average, how many of those 1000 different confidence intervals would contain the true value of the population mean?
What would be the answer if confidence level was 99%?

aquariump9

Step 1
Determine the number of 1000 different confidence intervals would contain the true value of the population mean.
The number of 1000 different confidence intervals would contain the true value of the population mean with the confidence level of 96% is obtained below as follows:
From the information, given that there are 1000 different random samples was selected from the same population.
The required number is,
$1000×\frac{P}{96}\right\}\left\{100\right\}=960$
Thus, on average 960 of 1000 different confidence intervals would contain the true value of the population mean with the confidence level of 96%.
Step 2
Determine the number of 1000 different confidence intervals would contain the true value of the population mean.
The number of 1000 different confidence intervals would contain the true value of the population mean with the confidence level of 99% is obtained below as follows:
From the information, given that there are 1000 different random samples was selected from the same population.
The required number is,
$1000×\frac{99}{100}=990$
Thus, on average 990 of 1000 different confidence intervals would contain the true value of the population mean with the confidence level of 99%.

Bob Huerta

Step 1
a) On average expected number of confidence interval that will contain the true value of population mean
$=np=1000×0.96=960$
b) On average expected number of confidence interval that will contain the true value of population mean
$=np=1000×0.99=990$

karton