 Mathematics
Large Sample Estimation of a Population Mean

### Question Description  a) Since we don't know the population standard deviation, only that for the sample, we should use a t-distribution here, with df = n-1 = 7. The critical t-stat for a 98% confidence interval will be:

t_crit = t.inv(p=0.99,df=7) = 2.997951567

The standard error is sample sd / sqrt(n) = 4.1/sqrt(8)

SEM = 1.449568901

So the 99% confidence interval for the sample mean is given by

sample mean +/- t_crit x SEM

= 58.3 +/- 2.997951567 x 1.449568901

= [53.954 to 62.646]

b) Since we don't know the population standard deviation, only that for the sample, we should use a t-distribution here, with df = n-1 = 26. The critical t-stat for a 98% confidence interval will be:

t_crit = t.inv(p=0.99,df=26) = 2.478629824

The standard error is sample sd / sqrt(n) = 4.1/sqrt(27)

SEM = 0.789045368

So the 99% confidence interval for the sample mean is given by

sample mean +/- t_crit x SEM

= 58.3 +/- 2.478629824 x 0.789045368

= [56.344 to 60.256] iainharlow (1041)
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