Answer:
The sample statistics are attached.
First, we need to determine the hypothesis.
The null hypothesis should be:
.
The alternative hypothesis should be: ![H_{a}: u_{africa}](https://tex.z-dn.net/?f=H_%7Ba%7D%3A%20u_%7Bafrica%7D%20%20%3Cu_%7Basia%7D)
The next step is about calculating the<em> t statistics </em>based on the samples. This <em>t test </em>will give the probability value or p-value, which determines if there's enough to reject the null hypothesis. The formula we need to use to find the t-value is: ![t=\frac{x_{1}-x_{2} }{\sqrt{\frac{s_{1} ^{2} }{n_{1} } +\frac{s_{2} ^{2} }{n_{2} } } }](https://tex.z-dn.net/?f=t%3D%5Cfrac%7Bx_%7B1%7D-x_%7B2%7D%20%20%7D%7B%5Csqrt%7B%5Cfrac%7Bs_%7B1%7D%20%5E%7B2%7D%20%7D%7Bn_%7B1%7D%20%7D%20%2B%5Cfrac%7Bs_%7B2%7D%20%5E%7B2%7D%20%7D%7Bn_%7B2%7D%20%7D%20%7D%20%7D)
Replacing all the values into the formula, we have:
![t=\frac{52.4-58.1}{\sqrt{\frac{(7.4)^{2} }{45}+\frac{(8.8)^{2} }{35} } } \\t=\frac{-5.7}{\sqrt{1.2+2.2} } =\frac{-5.7}{1.8} =-3.2](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B52.4-58.1%7D%7B%5Csqrt%7B%5Cfrac%7B%287.4%29%5E%7B2%7D%20%7D%7B45%7D%2B%5Cfrac%7B%288.8%29%5E%7B2%7D%20%7D%7B35%7D%20%20%7D%20%7D%20%5C%5Ct%3D%5Cfrac%7B-5.7%7D%7B%5Csqrt%7B1.2%2B2.2%7D%20%7D%20%3D%5Cfrac%7B-5.7%7D%7B1.8%7D%20%3D-3.2)
Therefore, the t-value is -3.2.
Now, we using a level of significance of 0.01, and a degree of freedom (df) of 34, we use the t-table to find the p-value for this results. (df = N -1; in this case, we take the smaller sample, which is 35, giving us 34 of df).
Therefore, according to the table attached, <em>the p-value is less than 0.01,</em> which is less than our level of significance. This result means that the null hypothesis is rejected. In other words, there's enough evidence to say that <em>the life expectancy of people in Africa is less than the life of expectancy of people in Asia.</em>
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