Question

The titanium content of an alloy is being studied in the hope of ultimately increasing the tensile strength. An analysis of six recent heats chosen at random produces the following titanium contents.

5.2%

9.8%

8.6%

7.9%

6.2%

9.2%

What is the value of the test statistic, if the alternative hypothesis is the mean titanium content is greater than 9.5%? Round your final answer to two decimal places.

Answer 1

Mean X̅ = Σ Xi / n
X̅ = 46.9 / 6 = 7.8167
Sample Standard deviation SX = √ ( (Xi - X̅ )2 / n - 1 )
SX = √ ( 15.9283 / 6 -1 ) = 1.7848

Test Statistic :-
t = ( X̅ - µ ) / ( S / √(n))
t = ( 7.8167 - 9.5 ) / ( 1.7848 / √(6) )
t = - 2.31