Confidence Interval Assignment

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Question 1: Find the mean, mod, median, and standard deviation of the following data:

5, 6, 7, 8, 9,8,7,8

Based on these results, check whether the value of 10 is usual?

Solution

The mean is calculated by using the formula given below

x ¯=(∑_i=1ⁿx_i)/n = (5+6+7+8+9+8+7+8)/8=58/8=7.25

The median is calculated by using the formula given below (after arranging the data in ascending order of magnitude)

Median=((n/2)^th+(n/2+1)^th)/2

=((8/2)^th+(8/2+1)^th)/2

=((4)^th+(5)^th)/2

=(7+8)/2=7.5

Therefore, the required median is 7.5

s=√((∑_i=1ⁿ(x_i-x ¯ )² )/(n-1))=1.282

The mode is defined as the most repeated value in the dataset. The mode is 8 as it is repeated to a maximum of 3 times

The standard deviation is calculated by using the formula given below

Here, 10 is not in the sample and therefore, it is not considered as an usual value

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Question 2. Pre-Employment Drug Screening Results are shown in the following Table:

Solution:

i. If 1 of the 200 test subjects is randomly selected, find the probability that the subject had a positive test result, given that the subject actually uses drugs. That is,

findP (positive test result | subject uses drugs) = 8/10 = 0.8

ii. If 1 of the 200 test subjects is randomly selected, find the probability that the subject actually uses drugs, given that he or she had a positive test result. That is,

findP (subject uses drugs |positive test result) = 8/18 = 0.444

Question 3. This is observation from previous years about the impact of students working while they are enrolled in classes, due to students too much work, they are spending less time on their classes. First, the observer need to find out, on average, how many hours a week students are working. They know from previous studies that the standard deviation of this variable is about 5 hours. A survey of 200 students provides a sample mean of 7.10 hours worked. What is a 95% confidence interval based on this sample?

Solution:

The 95% confidence interval is calculated by using the formula given below

(x ¯-Z_(1-∝/2)*σ/√n,x ¯+Z_(1-∝/2)*σ/√n)=(7.1-1.96*5/√200,7.1+1.96*5/√200)
=(7.1-1.96*0.354,7.1+1.96*0.354)
= (7.1 - 0.693, 7.1 + 0.693)
= (6.407, 7.793)

The 95% confidence interval for the sample is (6.407, 7.793)

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