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kipiarov [429]
3 years ago
13

Beer shelf life is a problem for brewers and distributors because when beer is stored at room temperature, its flavor deteriorat

es. When the average furfuryl ether content reaches 6 μg per liter, a typical consumer begins to taste an unpleasant chemical flavor. At α = .05, would the following sample of 12 randomly chosen bottles stored for a month convince you that the mean furfuryl ether content exceeds the taste threshhold? 8.92 6.99 5.54 5.73 6.38 5.51 6.45 7.50 8.48 5.56 6.90 6.46
Mathematics
1 answer:
mr_godi [17]3 years ago
6 0

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.

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