Question #348563

In a recent survey, a researcher claimed that the average life expectancy of people in a certain country is 72 years. Is his claim correct if a random sample of 25 deaths from his country showed a mean of 71.2 years with standard deviation of 1.2 years? Use 99% confidence level.



1
Expert's answer
2022-06-07T09:22:30-0400

The following null and alternative hypotheses need to be tested:

H0:μ=72H_0:\mu=72

H1:μ72H_1:\mu\not=72

This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation, using the sample standard deviation, will be used.

Based on the information provided, the significance level is α=0.01,\alpha = 0.01, df=n1=24df=n-1=24 and the critical value for a two-tailed test is tc=2.79694.t_c =2.79694.

The rejection region for this two-tailed test is R={t:t>2.79694}.R = \{t:|t|>2.79694\}.

The t-statistic is computed as follows:


t=xˉμs/n=71.2721.2/25=3.33333t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{71.2-72}{1.2/\sqrt{25}}=-3.33333


Since it is observed that t=3.33333>2.79694=tc,|t|=3.33333>2.79694=t_c, it i s then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for two-tailed, df=24df=24 degrees of freedom, t=3.33333t=-3.33333 is p=0.002776,p=0.002776, and since p=0.002776<0.01=α,p= 0.002776<0.01=\alpha, it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean μ\mu

is different than 72, at the α=0.01\alpha = 0.01 significance level.


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