<u>Answer:</u>
<u>1. A. You allow the passenger to board his flight when the passenger has a weapon.</u>
<u>2. B. You select the passenger for further inspection when the passenger has no weapon.</u>
<u>Explanation:</u>
1. Remember, a Type I error in simple words means that the assumption "the passenger has a weapon" (null hypothesis) is <em>actually true,</em> but the airport security screener <em>incorrectly concluded it is false. </em>In other words, he assumed the passenger had no weapon and allowed the passenger to board his flight <u>when he actually did have one.</u>
<em>2. While, </em><em>a </em><em>Type II error </em><em>means that </em>the assumption "the passenger has a weapon" (null hypothesis) is <em>actually false, </em>but the airport security screener <em>incorrectly concluded it is true. </em>In other words, he assumed the passenger had a weapon and selected the passenger for further inspection <u>when he actually didn't have one.</u>
Answer:
The mean for the second week is $2 less than the first and in percentage it is 22% less.
Step-by-step explanation:
The mean is given by the sum of all individual values divided by the number of values. For the first week the sum is:
sum1 = 6.5 + 8 + 7.25 + 13.5 + 9.75
sum1 = 45
Since she spent 10 less in the second week the sum is:
sum2 = sum1 - 10 = 45 - 10 = 35
The mean for each week is:
mean1 = sum1/5 = 45/5 = 9
mean2 = sum2/5 = 35/5 = 7
difference = mean1 - mean2 = 9-7 = 2
difference(%) = [2/9]*100 = 0.22*100 = 22%
The mean for the second week is $2 less than the first and in percentage it is 22% less.
Step-by-step explanation:
50-50+15d =125-50
15d = 75
15d/15=75/15
d=5