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. 9.7% 7.4% 5.9% 7.7% 7.5% 8.9% What is the value of the test statistic, if the alternative hypothesis is the mean titanium content is greater than 9.5%?

Tstatistic = - 3.068

Step-by-step explanation:

Given the data: 9.7% 7.4% 5.9% 7.7% 7.5% 8.9%

Mean titanium > 9.5%

Test statistic :

(m - μ) / (s/√n)

From the data: obtain the sample mean and standard deviation

From the data given ;

Calculate the same mean and standard deviation :

Using calculator :

Sample mean (m) = ΣX / n

Sample size (n) = 6

m = 47.1 / 6 = 7.85

Standard deviation (s) :

Using calculator = 1.317

Hence,

(m - μ) / (s/√n)

Tstatistic = (7.85 - 9.5) / (1.317/√6)

Tstatistic = - 1.65 / 0.5376629

Tstatistic = - 3.068

d

step-by-step explanation:

d

step-by-step explanation:

first you will have to convert 3/4 into a decimal

which is .75

then you will have to divide 1.5 by .75

which is 2

no you have to divide