Question 914629
(a) The point estimate would be 75/150= 0.5

(b) 95% confidence interval for the true proportion of teenagers who drank alcoholic beverage at least once during the last 2 weeks :

Formula would be confidence interval= point estimate +/- margin of error.

To calculate margin of error, we will use the formula :  z * (sqrt [p(1-p])/n)

A 95 % level of confidence gives us a z score of 1.96 .  The other confidence levels most often used are the 90% and 99% levels. However, the 95% confidence level is the most popular.

We plug in the numbers to get 1.96 * sqrt [ (0.5 * 0.50)/ 150 ]
which gives us 0.080. So the intervals will be 0.5 +/- 0.080

(c) Given that n = 81, x= 55 and σ = 9, find the
(i) 87% confidence interval
(ii) 92% confidence interval 

First, we find the standard error=  standard deviation/ sqrt n = 9/ sqrt 81 =1
(i)Then , we find the 
     alpha (α): α = 1 - (confidence level / 100)
                  = 1 -  87/100
                  = 1 - 0.87 
                  = 0.13

    Find the critical probability (p*): p* = 1 - α/2
                                           = 1 - 0.13/2 
                                           = 0.935

    To express the critical value as a t score:

        Find the degrees of freedom (DF)= sample size -1
                                          81-1= 80

      Therefore, the critical value is the t-score having 80 degrees of freedom and a critical probability equal to 0.935. We can use a T Distribution Calculator online to find that the critical value is 1.530

Now, we can find the margin of error:

ME= critical value * standard error
  = 1.530 * 1
  = 1.530

The range of the confidence interval is defined by the sample mean +/- margin of error.

Therefore, the 87 % confidence level says that the mean falls within the interval  55 +/- 1.530.  

Note: We can express the critical value as a t-score or a z-score. When the sample size is small (less than 30), it is preferred that we use the critical value expressed as a t-score, but if the sample size is larger, using either one yields similar results. Above, I could have expressed the critical value as a z score. I would have equated the critical probability with the cumulative probability and used a normal distribution calculator to yield a critical value z-score of 1.514. Above, I used the t-score of 1.530.