Answer to Question #261915 in Statistics and Probability for Jenny

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?

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

We shall use the student's t distribution to perform this test since the population variances are unknown as described as follows,

We can summarize the information above as,

$n_1=35,\space \bar{x}_1=92.4,\space s_1=7.3$

$n_2=42,\space \bar{x}_2=133.8,\space s_2=9.9$

a.

The hypotheses to be tested are,

$H_0:\mu_1=\mu_2$

$Against$

$H_1:\mu_1\lt\mu_2$

b.

Since it is a lower tailed one sided test, we only one critical value given as,

$t_{\alpha, v}$ where $\alpha =0.1$ is the desired significance level and $v$ is the degrees of freedom given by the formula,

$v=((s_1^2/n_1+s_2^2/n_2)^2)/((s_1^2/n_1)^2/(n_1-1)+(s_2^2/n_2)/(n_2-1))$

$v=((53.29/35+98.01/42)^2)/((53.29/29)^2/34+(98.01/42)^2/41)=73.98\approx74$

Therefore, $t_{0.1,74}= -1.293097$

Thus the critical value is $t_{0.1,74}=-1.293097$

c.

The test statistic is given by the formula below,

$t^*=(\bar{x}_1-\bar{x}_2)/\sqrt{s_1^2/n_1+s_2^2/n_2}=(92.4-133.8)/\sqrt{53.29/35+98.01/42})$

$t^*=-41.4/1.9637=-21.083(3\space dp)$

Thus the test value/statistic is $t^*=-21.083$

d.

The null hypothesis is rejected if $t^*\lt t_{0.1,74}$

Since $t^*=-21.083\lt t_{0.1,74}=-1.293097$, we reject the null hypothesis.

e.

Since the null hypothesis is rejected, we can conclude that there is sufficient evidence to support the claim that aluminum bottles chills at a lesser time than glass bottles at 10% level of significance.