You intend to estimate a population proportion with a confidence interval. The data suggests that the normal distribution is a reasonable approximation for the binomial distribution in this case. While it is an uncommon confidence level, find the critical value that corresponds to a confidence level of 87.3%.

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 87.3% of confidence, our significance level would be given by and . And the critical value would be given by:

And we can use the following excel codes to find the solution:

"=NORM.INV(0.0635,0,1)"

"=NORM.INV(1-0.0635,0,1)"

Step-by-step explanation:

Previous concepts

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  (assuming the normal approximation)

Solution to the problem

The confidence interval for the mean is given by the following formula:  

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 87.3% of confidence, our significance level would be given by and . And the critical value would be given by:

And we can use the following excel codes to find the solution:

"=NORM.INV(0.0635,0,1)"

"=NORM.INV(1-0.0635,0,1)"

h df tyjgtyj ttd y ytkgfty ty

step-by-step explanation:

tyty jry j y yj yttj yhgtyj y ghyj ytthj