Question

# Sixty-four people are asked to measure the width of a large room. The mean measurement is...

Sixty-four people are asked to measure the width of a large room. The mean measurement is 9.78 metres with a variance of 63 squaredcentimetres. Assuming that the measurement errors are random (and normally distributed), the true width of the room is unlikely (i.e., = :05) to be less _____ than or greater than _______ .

(Hont; t-dist test)

Solution:

Confidence interval for population mean() using t distribution

Given that,

n = 64 ....... Sample size

= 9.78 ....... Sample mean

= 63 ........Sample variance

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

= 0.05

/2 = 0.05 2 = 0.025

Also, n = 64 .....given

d.f= n-1 = 63

=    = = 1.998

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

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

9.78 - 1.998*(63/64)    9.78 +1.998*(63/64)

7.7973< < 11.7627

Answer: μ is unlikely ( = :05) to be less than 11.7627 or greater than 7.7973.

#### Earn Coins

Coins can be redeemed for fabulous gifts.