Question #338810

A random sample of 100 recorded deaths in the United States during the past year showed an average life span of 71.8 years, with a standard deviation of 8.9 years. Does this seem to indicate that the average life span today is greater than 70 years? Use a 0.05 level of significance. The computed value is


1
Expert's answer
2022-05-16T15:18:15-0400

The following null and alternative hypotheses need to be tested:

H0:μ70H_0:\mu\le70

Ha:μ>70H_a:\mu>70

This corresponds to a right-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

Based on the information provided, the significance level is α=0.05,\alpha = 0.05, and the critical value for a right-tailed test is zc=1.6449.z_c = 1.6449.

The rejection region for this right-tailed test is R={z:z>1.6449}.R = \{z: z> 1.6449\}.

The z-statistic is computed as follows:


z=xˉμσ/n=71.8708.9/1002.0225z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{71.8-70}{8.9/\sqrt{100}}\approx2.0225

Since it is observed that z=2.0225>1.6449=zc,z = 2.0225>1.6449= z_c , it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is p=P(Z>2.0225)=0.021562,p=P(Z>2.0225)= 0.021562, and since p=0.021562<0.05=α,p= 0.021562<0.05=\alpha, it is concluded that the null hypothesis is rejected.

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

is greater than 70, at the α=0.05\alpha = 0.05 significance level.


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