a fashion designer wants to know how many new dresses women buy each year. Suppose a sample of 849 is drawn with x= 5.3. Assume standard deviation = 1.5.

Question

construct 98% confidence interval. lower and upper bounderies needed

Answer 1

Hi Ashley, 

 

Let's create a "normal" approximated confidence interval. 

 

First we estimate the average number of dresses per year from the population of all women with the sample average of 5.3 from our 849 women. We estimate the standard deviation using our observed standard deviation 1.5.

 

Second we need to calculate the standard error of our sample average. If we use a normal approximation to calculate the standard error of the sample average it will be just the standard deviation divided by the square root of the sample size: 1.5/sqrt(849). Hence, the standard error is approximately 0.0515.

 

Third, we need to ask, "how many standard deviations do we need to go away from the mean to cover 98% of the normal distribution?" More specifically, we want to find the z-scores associated with the 99th percentile and the 1st percentile. This will guarantee us a 98% confidence interval since we are thinking about taking off the top 1% and the bottom 1% of the standard normal distribution. Here is a link to a table that answers that question. Simply see that the 99th percentile is associated with a z-score of about 2.33. Since the standard normal is symmetric, the 1st percentile will be associated with a z-score of -2.33.

 

https://statistics.laerd.com/statistical-guides/img/normal-table-large.png 

 

Now to construct the confidence interval, we simply take our estimate plus and minus our z-score times the standard error:

 

5.3 +/- 2.33*0.0515 = (5.18, 5.42)