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ikadub [295]
3 years ago
9

a. A simple random sample of size 80 has a mean of 7.31. The population standard deviation is 6.26. Construct a 99% confidence i

nterval for the population mean
Mathematics
1 answer:
Sever21 [200]3 years ago
5 0

Answer:

5.503484≤x≤9.116516

Step-by-step explanation:

The formula for calculating the CI is expressed as;

CI = x±z*s/√n

x is the mean= 7.31

z is the 99% z-score = 2.58

s is the standard deviation = 6.26

n is the sample size = 80

Substitute into the formula

CI = 7.31± 2.58*6.26/√80

CI = 7.31± 2.58*6.26/8.94

CI = 7.31± (2.58*0.7002)

CI = 7.31±1.806516

CI = (7.31-1.806516, 7.31+1.806516)

CI = (5.503484, 9.116516)

Hence the 99% confidence interval for the population mean is

5.503484≤x≤9.116516

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Answer:

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P(GUPUT) = \frac{73}{135}+\frac{59}{135}+\frac{31}{135}-\frac{19}{135}-\frac{9}{135}-\frac{16}{135}+\frac{4}{135}

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