Answer to Question #275183 in Statistics and Probability for LEA

Question

A random sample of 100 recorded deaths in the country in 2011 showed an average lifespan of 71.8 years with a standard deviation of 8.9 years. Does this seem to indicate that the average lifespan today is greater than 70 years? Use 0.05 as level of significance.

Answer 1

We are given that,

$\bar{x}=71.8,\space s=8.9,\space n=100$

The hypotheses tested are,

$H_0:\mu=70\space vs H_1:\mu\gt 70$

To perform this test, we shall apply the one sample t-test for the population mean as follows.

The test statistic is given as,

$t_c=(\bar{x}-\mu)/(s/\sqrt{n})$

$t_c=(71.8-70)/(8.9/\sqrt{100})=1.8/0.89=2.0225(4dp)$

$t_c$ is compared with the t table value with $(n-1)=100-1=99$ degrees of freedom at $\alpha=0.05$ given by,

$t_{0.05,99}=1.660391$ and the null hypothesis is rejected if $t_c\gt t_{0.05,99}.$

Since $t_c=2.0225\gt t_{0.05,99}=1.660391,$ we reject the null hypothesis and conclude that there is sufficient evidence to indicate that the average lifespan today is greater than 70 years at 5% level of significance.