Page 382 - DMTH404_STATISTICS

Statistics

Notes 27.2 T test

Acme Corporation manufactures light bulbs. The CEO claims that an average Acme light bulb
lasts 300 days. A researcher randomly selects 15 bulbs for testing. The sampled bulbs last an
average of 290 days, with a standard deviation of 50 days. If the CEO’s claim were true, what is
the probability that 15 randomly selected bulbs would have an average life of no more than 290
days?

Note There are two ways to solve this problem, using the T Distribution Calculator.
Both approaches are presented below. Solution A is the traditional approach. It requires
you to compute the t score, based on data presented in the problem description. Then, you
use the T Distribution Calculator to find the probability. Solution B is easier. You simply
enter the problem data into the T Distribution Calculator. The calculator computes a t
score “behind the scenes”, and displays the probability. Both approaches come up with
exactly the same answer.

Solution A
The first thing we need to do is compute the t score, based on the following equation:
t = [ x – m ] / [ s / sqrt( n ) ]

t = ( 290 – 300 ) / [ 50 / sqrt( 15) ] = –10 / 12.909945 = – 0.7745966
where x is the sample mean, ì is the population mean, s is the standard deviation of the sample,
and n is the sample size.

Now, we are ready to use the T Distribution Calculator. Since we know the t score, we select
“T score” from the Random Variable dropdown box. Then, we enter the following data:

 The degrees of freedom are equal to 15 – 1 = 14.
 The t score is equal to – 0.7745966.
The calculator displays the cumulative probability: 0.226. Hence, if the true bulb life were 300
days, there is a 22.6% chance that the average bulb life for 15 randomly selected bulbs would be
less than or equal to 290 days.
Solution B:
This time, we will work directly with the raw data from the problem. We will not compute the
t score; the T Distribution Calculator will do that work for us. Since we will work with the raw
data, we select “Sample mean” from the Random Variable dropdown box. Then, we enter the
following data:
 The degrees of freedom are equal to 15 - 1 = 14.

 Assuming the CEO’s claim is true, the population mean equals 300.
 The sample mean equals 290.
 The standard deviation of the sample is 50.
The calculator displays the cumulative probability: 0.226. Hence, there is a 22.6% chance that the
average sampled light bulb will burn out within 290 days.

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