Answer and Step-by-step explanation:
we have the following data:
Point estimate = sample mean = \ bar x = 12.39
Population standard deviation = \ sigma = 3.7
Sample size = n = 177
a) the margin of error with a 90% confidence interval
α = 1 - 90%
alpha = 1 - 0.90 = 0.10
alpha / 2 = 0.05
Z \ alpha / 2 = Z0.05 = 1,645
Margin of error = E = Z \ alpha / 2 * (\ sigma / \ sqrtn)
we replace:
E = 1.645 * (3.7 / \ sqrt177)
Outcome:
E = 0.46
b) margin of error with a 99% confidence interval
α = 1-99%
alpha = 1 - 0.99 = 0.01
alpha / 2 = 0.005
Z \ alpha / 2 = Z0.005 = 2,576
Margin of error = E = Z \ alpha / 2 * (\ sigma / \ sqrtn)
we replace:
E = 2,576 * (3.7 / \ sqrt177)
Outcome:
E = 0.72
c) A larger confidence interval value will increase the margin of error.
Answer:
From my knowledge and my research the answer is B! Thank you
Step-by-step explanation:
Answer:
167 Pages Left To Read
Step-by-step explanation:
15*5=75
75*5=375
375+ 25*10=625
792-625=167
I’m guessing it’s the first one
