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

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

6.46          41.7316          

80.42         553.0336      

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.

(a) common ratio in the given sequence is, 3

(b) push ups will do kendall do on the 20th week

step-by-step explanation:

geometric sequence is a sequence of numbers that follows a pattern were the next term is found by multiplying by a constant called the common ratio(r)

(a)

given the sequence: 6 , 18 , 54 , 162 ,

this sequence is geometric sequence where each term is 3 times the previous term.

therefore,

common ratio(r) = 3.

(b)

a formula for nth term of geometric sequence is given by;

,   where r is the common ratio between the successive terms and is the first term,

to find how many push ups will kendall do on the 20th week.

here. n = 20th term.

in the given sequence first term

then;

simplify:

therefore, push ups will do kendall do on the 20th week