2. A certain TV can be purchased from the manufacturer for $160. A certain online retailer has a standard markup of 30%, and a c
ertain superstore has a standard markup of 40%. (a) What is the price of the TV when purchased online? What is the price of the TV when purchased at the superstore? (b) What is the difference between the prices in Part (a)? How does this relate to the difference in the markup?
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1 answer:
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Answer:
99% confidence interval for the population proportion of passing test scores is [0.5986 , 0.9414].
Step-by-step explanation:
We are given that a high school principal wishes to estimate how well his students are doing in math.
Using 40 randomly chosen tests, he finds that 77% of them received a passing grade.
Firstly, the pivotal quantity for 99% confidence interval for the population proportion is given by;
P.Q. = ~ N(0,1)
where, = sample proportion of students received a passing grade = 77%
n = sample of tests = 40
p = population proportion
<em>Here for constructing 99% confidence interval we have used One-sample z proportion test statistics.</em>
So, 99% confidence interval for the population proportion, p is ;
P(-2.5758 < N(0,1) < 2.5758) = 0.99 {As the critical value of z at 0.5%
level of significance are -2.5758 & 2.5758}
P(-2.5758 < < 2.5758) = 0.99
P( <
<
) = 0.99
P( < p <
) = 0.99
<u>99% confidence interval for p</u> = [ ,
]
= [ ,
]
= [0.5986 , 0.9414]
Therefore, 99% confidence interval for the population proportion of passing test scores is [0.5986 , 0.9414].
Lower bound of interval = 0.5986
Upper bound of interval = 0.9414