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We draw a random sample of size 25 from a normal population with variance 2.4. If the sample mean is 12.5, what is a 99% confidence interval for the population mean?

Respuesta :

abidemiokin

Answer:

11.2≤[tex]\mu[/tex]12.8

Step-by-step explanation:

Confidence interval for the population mean is expressed by the formula;

CI = xbar ± Z(S/√n) where;

xbar is the sample mean = 12.5

Z is the z score at 99% confidence = 2.576

S is the standard deviation = √variance

S = √2.4 = 1.5492

n is the sample size = 25

Substituting the given values into the formula given above,

CI = 12.5 ± 2.576(1.5492/√25)

CI = 12.5 ± 2.576(0.30984)

CI = 12.5 ± 0.7981

CI = (12.5-0.7981, 12.5+0.7981)

CI = (11.2019, 12.7981)

Hence the 99% confidence interval for the population mean is 11.2≤[tex]\mu[/tex]12.8 (to 1 decimal place)

Cricetus

A 99% confidence interval for the population mean will be "11.2 [tex]\leq[/tex] 12.8".

Statistics

According to the question,

Sample mean, [tex]\bar x[/tex] = 12.5

Z score at 99%, Z = 2.576

Standard deviation, S = √Variance

                                    = √2.4

                                    = 1.5492

Sample size, n = 25

We know the formula,

Confidence interval, CI = [tex]\bar x \ \pm[/tex] Z ([tex]\frac{S}{\sqrt{n} }[/tex])

By substituting the given values, we get

                                        = 12.5 [tex]\pm[/tex] 2.576 ([tex]\frac{1.5492}{\sqrt{25} }[/tex])

                                        = 12.5 [tex]\pm[/tex] 2.576 (0.30984)

                                        = 12.5 [tex]\pm[/tex] 0.7981

Now,

                                   Cl = (12.5 - 0.7981, 12.5 + 0.7981)

                                        = (11.2019, 12.7981) or,

                                        = (11.2, 12.8)

Thus the above answer is appropriate.        

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