Answer:
As the calculated value of t =2.1698 is greater than t (0.05,11) = 1.796 reject H0 . It means chosen bottles stored for a month convince you that the mean furfuryl ether content exceeds the taste threshhold.
Step-by-step explanation:
We formulate our null and alternative hypotheses as
H0 u≤ 6 ug Ha : u > 6 ug
The significance level ∝ = 0.05
The test statistic used is
t = X` - u / s/ √n
which if H0 is true, has the students' t test with n-1 = 11 degrees of freedom.
The critical region t > t (0.05,11) = 1.796
We compute the t value from the data
Xi Xi²
8.92 79.5664
6.99 48.8601
5.54 30.6916
5.73 32.8329
6.38 40.7044
5.51 30.3601
6.45 41.6025
7.50 56.25
8.48 71.9104
5.56 30.9136
6.90 47.61
<u>6.46 41.7316 </u>
<u>80.42 553.0336</u><u> </u>
<u />
Now x` = ∑x/ n = 80.42/12 = 6.70
S²= 1/n-1 ( ∑(xi- x`)²= 1/11 ( 553.034 - (80.42)²/12)
= 1/11 (553.034-538.948) = 1.2805
s= 1.1316
Putting the values in the test statistics
t = X` - u / s/ √n = 6.70- 6 / 1.1316 / √12
= 2.1698
The critical region t > t (0.05,11) = 1.796
As the calculated value of t =2.1698 is greater than t (0.05,11) = 1.796 reject H0 . It means chosen bottles stored for a month convince you that the mean furfuryl ether content exceeds the taste threshhold.
