Among those who take the Graduate Record Examination (GRE), 67 people are randomly selected. This sample...

Among those who take the Graduate Record Examination (GRE), 67 people are randomly selected. This sample group has a mean score of 558 and a standard deviation of 149 on the quantitative portion of the GRE. Find a 99% confidence interval estimate of the population mean.

Solutions

Expert Solution

Solution :

Given that,

Point estimate = sample mean = = 558

Population standard deviation = = 149
Sample size = n =67

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576 ( Using z table )

Margin of error = E = Z/2* ( / $\sqrt$ n)

= 2.576 * ( 149/ $\sqrt$ 67)

= 46.8916

At 99% confidence interval estimate of the population mean is,

- E < < + E

558-46.8916 < < 558+46.8916

511.1084< < 604.8916

(511.1084, 604.8916)

orchestra answered 1 year ago

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