Question

A random sample of 100 recorded deaths in the United States during the past year showed an average life span of 71.8 years. Assuming a population standard deviation of 8.9 years, does this seem to indicate that the mean life span today is greater than 70 years

Answer 1

This indicates that the chances of the mean life span today is greater than 70 years is 0.5871

Explanation:

The past year showed an average life span of 71.8 years.

`\mu = 71.8 years`

Population standard deviation = 8.9 years

`\sigma = 8.9 years`

We are supposed to find the mean life span today is greater than 70 years

i.e. P(X>70)

Formula :
`Z = (x-\mu)/(\sigma)`

`Z=(70-71.8)/(8.9)`

Z=−0.202

P(X<70)=0.4129

P(X>70)=1-P(X<70)=1-0.4129=0.5871

Hence This indicates that the chances of the mean life span today is greater than 70 years is 0.5871

Answer 2

We conclude that the mean life span today is greater than 70 years.

Explanation:

We are given that a random sample of 100 recorded deaths in the United States during the past year showed an average life span of 71.8 years. Assuming a population standard deviation of 8.9 years.

Let Null Hypothesis,
`\mu`
`\leq` 70 years {means that the mean life span today is less than or equal to 70 years}

Alternate Hypothesis,
`\mu` > 70 years {means that the mean life span today is greater than 70 years}

The test statistics that will be used here is One-sample z test statistics;

T.S. =
`(\bar X - \mu)/((\sigma)/(√(n) ) )` ~ N(0,1)

where,
`\bar X` = sample mean = 71.8 years

`\sigma` = population standard deviation = 8.9 years

n = sample size = 100

So, test statistics =
`(71.8 - 70)/((8.9)/(√(100) ) )`

= 2.022

Now, since we are not given with significance level so we assume it to be 5%. At 5% significance level, the z table gives critical value of 1.96. Since our test statistics is more than the critical value of z so we have sufficient evidence to reject null hypothesis as it fall in the rejection region.

Therefore, we conclude that the mean life span today is greater than 70 years.