Question

Use the t-distribution to find a confidence interval for a mean μ given the relevant sample results. Give the best point estimate for μ, the margin of error, and the confidence interval. Assume the results come from a random sample from a population that is approximately normally distributed.

A 99% confidence interval for μ using the sample results x¯=49.7, s=12.6, and n=10

Round your answer for the point estimate to one decimal place, and your answers for the margin of error and the confidence interval to two decimal places.

Point estimate =

Margin of Error=

The 99% confidence interval is:

Answer 1

Solution :

Given that,

n = 10

   = 49.7

s = 12.6

Note that, Population standard deviation() is unknown..So we use t distribution.

Our aim is to construct 99% confidence interval.   

c = 0.99

= 1 - c = 1 - 0.99 = 0.01

  /2 = 0.01 2 = 0.005

Also, d.f = n - 1 = 9

    =    =  0.005,9 = 3.2

( use t table or t calculator to find this value..)

The margin of error is given by

E =  /2,d.f. * ( / n )

= 3.2* ( 12.6/ 10 )

E = 12.75

Now , confidence interval for mean() is given by:

( - E ) <   <  ( + E)

( 49.7 - 12.75 )   <   <  ( 49.7 + 12.75 )

36.95 <   < 62.45

Required.99% confidence interval is ( 36.95 , 62.45 )

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