Answer to Question #260984 in Statistics and Probability for Marco

Question

The aluminum bottle, first introduced in 1991 by CCL Container for mainly personal and  household items such as lotions, has become popular with beverage manufacturers.  Besides being lightweight and requiring less packaging, the aluminum bottle is reported  to cool faster and stay cold longer than typical glass bottles. A small brewery tests this  claim and obtains the following information regarding the time (in minutes) required to  chill a bottle of beer from room temperature (75ºF) to serving temperature (45ºF). At  α = 0.10, can it be concluded that aluminum bottles chills at a lesser time than glass  bottles? (12 points) 

Aluminum Glass

Sample Size 35 42

Mean time to chill 92.4 133.8

Sample standard deviation 7.3 9.9

a. State the hypothesis and identify the claim of the researcher.

b. Find the critical value(s).

c. Compute the test value.

d. Make a decision on the null hypothesis.

e. Make a decision on the claim of the researcher. 

Answer 1

Let m₁,m₂ be average times for chilinf for two types of bottles

Null hypothesi is:

$H_0:m_1=m_2;$

Alternative hypothesis

$H_1:m_1<m_2;$

$\alpha=0.1$ - significance level

1) $S_p=\sqrt{\frac{(n_1-1)\cdot s_1^2+(n_2-1)\cdot s_2^2}{n_1+n_2-2}}= \sqrt{\frac{(35-1)\cdot 7.3^2+(42-1)\cdot 9.9^2}{35+42-2}}=8.817$

common standard deviation

2) $t=\frac{\overline{x_1}-\overline{x_2}}{S_p\cdot \sqrt {\frac{1}{n_1}+\frac{1}{n_2}}}=\frac{92.4-133.8}{8.817\cdot \sqrt{\frac{1}{35}+\frac{1}{42}}}=-20.516$

value of t criterion.

If is not true that $m_1<m_2$ than t valuy may not be too big and negative ,threshold value

is $t_{tab}=t_{\alpha,n_1+n_2-2}=t_{0.1,75}=-1.293$ -critical value for t statistics from statistical tables(or qt(0.1,75)=-1.293 with mathcad soft).

3) Conclusion: because t=-20.516<t_tab=-1.293 with reliability 90% H₀ hypothesis should be rejected and average chilling time for aluminium bottles is significantly less than it for glass case with confidence 90%.