Question 914629: 1.In a survey of 150 randomly selected teenagers, 75 stated that they drank alcoholic beverage, at least once during the last 2 weeks.
(a)Calculate the point estimate for the population proportion of teenagers who drank alcoholic beverage at least once during the last 2 weeks.
(b)Construct the 95% confidence interval for the true proportion of teenagers who drank alcoholic beverage at least once during the last 2 weeks.
(a) Given that n = 81, x= 55 and σ = 9, find the
(i) 87% confidence interval
(ii) 92% confidence interval
Answer by AlgebraLady88(44)   (Show Source):
You can put this solution on YOUR website! (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.
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