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A sociologist wants to determine if the life expectancy of people in africa is less than the life expectancy of people in asia. two samples are randomly selected from each of the two populations. the sample statistics are given below. using \alphaα = 0.01, what is the p-value for this test

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Answer:

The sample statistics are attached.

First, we need to determine the hypothesis.

The null hypothesis should be:
H_(o) : u_(africa) \geq u_(asia).

The alternative hypothesis should be:
H_(a): u_(africa) &nbsp;<u_(asia)

The next step is about calculating the t statistics based on the samples. This t test 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) &nbsp;}{\sqrt{(s_(1) ^(2) )/(n_(1) ) +(s_(2) ^(2) )/(n_(2) ) } }

Replacing all the values into the formula, we have:


t=\frac{52.4-58.1}{\sqrt{((7.4)^(2) )/(45)+((8.8)^(2) )/(35) &nbsp;} } \\t=(-5.7)/(√(1.2+2.2) ) =(-5.7)/(1.8) =-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, the p-value is less than 0.01, 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 the life expectancy of people in Africa is less than the life of expectancy of people in Asia.

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