Answer:
Explanation:
We would set up the hypothesis test.
For the null hypothesis,
µ = 98
For the alternative hypothesis,
µ ≠ 98
This is a 2 tailed test.
Since no population standard deviation is given, the distribution is a student's t.
Since n = 270
Degrees of freedom, df = n - 1 = 270 - 1 = 269
t = (x - µ)/(s/√n)
Where
x = sample mean = 97
µ = population mean = 98
s = samples standard deviation = 7
t = (97 - 98)/(7/√270) = - 2.35
We would determine the p value using the t test calculator. It becomes
p = 0.02
Since alpha, 0.1 > than the p value, 0.02, then we would reject the null hypothesis. Therefore, At a 10% level of significance, there sufficient evidence to support the claim that the technique performs differently than the traditional method.