Exercise 5. the sample mean and sample standard deviation for the grade point averages of a random sample of 100 college freshmen at a certain university are 2.5 and 0.2 respectively. show that 2.4485 < μ < 2.5515 is a 99% (approximate) confidence interval estimate for the mean μ of grade point averages for the entire freshmen class.

Proved

Step-by-step explanation:

Given that the  sample mean and sample standard deviation for the grade point averages of a random sample of 100 college freshmen at a certain university are 2.5 and 0.2 respectively.

Sample size n =100

Std error = s/sqrt 100 = 0.02

For 99% level critical value is 2.575

Margin of error = 2.575*std error

= 0.0515

Confidence interval 99%

= 2.5±0.0515

=(2.4485, 2.5515)

Hence proved

answer: the answer is simply nothing.

step-by-step explanation: because there is no ending to this question

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