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Sagot :
(a) We can conclude that there is no sufficient evidence that there is a significant difference in true average [tex]CO_2[/tex] loss at [tex]$18^{\circ}C[/tex] for the traditional pour and the slanted pour.
(b) We can conclude that there is sufficient evidence that there is a significant difference in true average [tex]\mathrm{CO}_2[/tex] loss at [tex]12^{\circ} \mathrm{C}[/tex] for the traditional pour and the slanted pour.
(a):
The Null and Alternative hypothesis is as follows:
Null hypothesis, [tex]H_0[/tex] : As we can see there is no significant difference in true average CO_2 loss at [tex]18^{\circ}C[/tex] for the traditional pour and the slanted pour.
Hence, we can write it as: [tex]H_0: \mu_1=\mu_2[/tex]
Alternative hypothesis,[tex]H_a[/tex]: As we can see, there is a significant difference in true average CO_2 loss at [tex]18^{\circ} C[/tex] for the traditional pour and the slanted pour.
Hence, we can write it as: [tex]$H_0: \mu_1 \neq \mu_2$[/tex]
The level of significance is [tex]\alpha=0.01[/tex]
The sample information is as follows:
Mean S_d n
Traditional at [tex]18^0C[/tex] 4.0 0.5 4
Slanted at [tex]18^0C[/tex] 3.7 0.3 4
Calculating the value of pooled proportion.
[tex]$\begin{aligned}s_p^2 & =\frac{\left(n_1-1\right) s_1^2+\left(n_2-1\right) s_2^2}{n_1+n_2-2} \\& =\frac{(4-1)(0.5)^2+(4-1)(0.3)^2}{4+4-2} \\& =\frac{0.75+0.27}{8-2} \\& =\frac{1.02}{6} \\& =0.17\end{aligned}$[/tex]
Calculating the value of test statistic.
[tex]$\begin{aligned}t & =\frac{\bar{x}_1-\bar{x}_2}{\sqrt{s_p^2\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}} \\& =\frac{4-3.7}{\sqrt{0.17\left(\frac{1}{4}+\frac{1}{4}\right)}} \\& =\frac{0.30}{0.2915} \\& =1.028991511 \\& \approx 1.03\end{aligned}$[/tex]
Calculating the degrees of freedom, the value will be:
[tex]$\begin{aligned}d f & =n_1+n_2-2 \\& =4+4-2 \\& =8-2 \\& =6\end{aligned}$[/tex]
Based on the usual t table values, the critical value of t for the two tail test at 5% significance and 6 degrees of freedom is 2.447.
Now we will compare the calculated value with the critical value.
Here, the calculated value less than the critical value so fails to reject the Null hypothesis.
Hence, we can conclude that there is no sufficient evidence that there is significant difference of true average CO_2 loss at [tex]$18^{\circ}C[/tex] for the traditional pour and the slanted pour.
(b):
The Null and Alternative hypothesis is as follows:
Null hypothesis, [tex]H_0[/tex] : As we can see that, there is no significant difference of true average [tex]CO_2[/tex] loss at [tex]12^{\circ} C[/tex] for the traditional pour and the slanted pour.
Hence, we can write it as: [tex]H_0: \mu_1=\mu_2[/tex]
Alternative hypothesis, [tex]H_a[/tex] : As we can see that, there is significant difference of true average [tex]CO_2[/tex] loss at [tex]12^{\circ} C[/tex] for the traditional pour and the slanted pour.
Hence, we can write it as: [tex]H_0: \mu_1 \neq \mu_2[/tex]
The level of significance is, [tex]\alpha=0.01[/tex]
The sample information is as follows:
Mean S_d n
Traditional at [tex]18^0C[/tex] 3.3 0.2 4
Slanted at [tex]18^0C[/tex] 2.0 0.3 4
Calculating the value of pooled proportion.
[tex]$\begin{aligned}s_p^2 & =\frac{\left(n_1-1\right) s_1^2+\left(n_2-1\right) s_2^2}{n_1+n_2-2} \\& =\frac{(4-1)(0.2)^2+(4-1)(0.3)^2}{4+4-2} \\& =\frac{0.75+0.27}{8-2} \\& =\frac{0.39}{6} \\& =0.065\end{aligned}$[/tex]
Calculating the value of test statistic.
[tex]$\begin{aligned}t & =\frac{\bar{x}_1-\bar{x}_2}{\sqrt{s_p^2\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}} \\& =\frac{3.3-2.0}{\sqrt{0.065\left(\frac{1}{4}+\frac{1}{4}\right)}} \\& =\frac{1.3}{0.18028} \\& =7.211102551 \\& \approx 7.21\end{aligned}$[/tex]
Now calculate the value of the degrees of freedom:
[tex]$\begin{aligned}d f & =n_1+n_2-2 \\& =4+4-2 \\& =8-2 \\& =6\end{aligned}$[/tex]
From the standard t table values, we can observe that the critical value of t for the two-tail test at the 5% level of significance and 6 degrees of freedom is 2.447.
Now comparing the calculated value with the critical value.
Here, the calculated value is greater than the critical value so reject the Null hypothesis.
Hence, conclude that there is sufficient evidence that there is a significant difference in true average [tex]\mathrm{CO}_2[/tex] loss at [tex]12^{\circ} \mathrm{C}[/tex] for the traditional pour and the slanted pour.
For similar question on Null and Alternative hypothesis
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