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r-ruslan [8.4K]
3 years ago
6

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

Mathematics
1 answer:
Colt1911 [192]3 years ago
3 0

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}

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} } } }

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

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>

<em></em>

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3 years ago
We learned in that about 69.7% of 18-20 year olds consumed alcoholic beverages in 2008. We now consider a random sample of fifty
maw [93]

Answer:

(1) The expected number of people who would have consumed alcoholic beverages is 34.9.

(2) The standard deviation of people who would have consumed alcoholic beverages is 10.56.

(3) It is surprising that there were 45 or more people who have consumed alcoholic beverages.

Step-by-step explanation:

Let <em>X</em> = number of adults between 18 to 20 years consumed alcoholic beverages in 2008.

The probability of the random variable <em>X</em> is, <em>p</em> = 0.697.

A random sample of <em>n</em> = 50 adults in the age group 18 - 20 years is selected.

An adult, in the age group 18 - 20 years, consuming alcohol is independent of the others.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 50 and <em>p</em> = 0.697.

The probability mass function of a Binomial random variable <em>X</em> is:

P(X=x)={50\choose x}0.697^{x}(1-0.697)^{50-x};\ x=0,1,2,3...

(1)

Compute the expected value of <em>X</em> as follows:

E(X)=np\\=50\times 0.697\\=34.85\\\approx34.9

Thus, the expected number of people who would have consumed alcoholic beverages is 34.9.

(2)

Compute the standard deviation of <em>X</em> as follows:

SD(X)=\sqrt{np(1-p)}=\sqrt{50\times 0.697\times (1-0.697)}=10.55955\approx10.56

Thus, the standard deviation of people who would have consumed alcoholic beverages is 10.56.

(3)

Compute the probability of <em>X</em> ≥ 45 as follows:

P (<em>X</em> ≥ 45) = P (X = 45) + P (X = 46) + ... + P (X = 50)

                =\sum\limits^{50}_{x=45} {50\choose x}0.697^{x}(1-0.697)^{50-x}\\=0.0005+0.0001+0.00002+0.000003+0+0\\=0.000623\\\approx0.0006

The probability that 45 or more have consumed alcoholic beverages is 0.0006.

An unusual or surprising event is an event that has a very low probability of success, i.e. <em>p</em> < 0.05.

The probability of 45 or more have consumed alcoholic beverages is 0.0006. This probability value is very small.

Thus, it is surprising that there were 45 or more people who have consumed alcoholic beverages.

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