The confidence interval for the difference p1 - p2 of the population proportion is (0.063, 0.329)
<h3>How to determine the confidence interval?</h3>
The given parameters are:
n₁ = 93; n₂ = 80
p₁ = 0.814; p₂ = 0.618
The critical value at 95% confidence interval is
z = ±1.96
So, we have:
Substitute known values in the above equation
Evaluate
This gives
CI = 0.196 ± 0.133
Expand
CI = (0.196 - 0.133, 0.196 + 0.133)
Evaluate
CI = (0.063, 0.329)
Hence, the confidence interval is (0.063, 0.329)
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Answer:
IF its a pretest the grade doesnt count im pretty sure.
Answer:
Step-by-step explanation:
25.
x + 2x -75 = 90
3x = 90+75
3x = 165
x = 55
Measure of A = 55
Measure of B = 2(55) - 75 = 110 - 75 = 35
28.
2x+10-x+55 = 90
x = 90-65
x = 25
measure of A = 2(25) +10 = 60
measure of B = -25+55 = 30
31.
2x+3+3x - 223 = 180
5x -220 = 180
5x = 400
x = 80
measure of A = 163
measure of B = 17
32.
-4x+40+x+50 = 180
-3x +90 = 180
-3x = 90
x= -30
measure of A = -4(-30) +40 = 160
measure of B = -30+50 = 20
Answer:
a) 0.0523 = 5.23%
b) 0.9477 = 94.77%
c) 0.2242 = 22.42%
d) YES
Step-by-step explanation:
This situation could be modeled with a binomial distribution where
The probability of getting exactly k “successes” in n trials is given by
Where p is the probability of “success”.
In this case we can assume that “success” is the fact that the household interviewed is tuned to Found.
So, p=0.19 and n = 14 (the households interviewed).
a)
The probability that none of the households are tuned to Found is P(X=0)
b)
The probability that at least one household is tuned to Found is
1- P(X=0) = 1-0.0523 = 0.9477 = 94.77%
c)
The probability that at most one household is tuned to Found is P(X=0)+P(X=1)
d)
According to this sample, the probability that more than one household is tuned to Found would be 100%-22.42% = 77.58%, so it does appear that the 19% share value is wrong.
