Use the following information to answer the next \(12\) exercises: The U.S. Center for Disease Control reports that the mean life expectancy was \(47.6\) years for whites born in \(1900\) and \(33.0\) years for nonwhites. Suppose that you randomly survey death records for people born in \(1900\) in a certain county. Of the \(124\) whites, the mean life span was \(45.3\) years with a standard deviation of \(12.7\) years. Of the \(82\) nonwhites, the mean life span was \(34.1\) years with a standard deviation of \(15.6\) years. Conduct a hypothesis test to see if the mean life spans in the county were the same for whites and nonwhites.

Find the \(p-\)value.

Short Answer

Expert verified

\(p-\)value approximately \(0\)

Step by step solution

01

Step 1. Given information

  • The Whites born in \(1900\) had a life expectancy of \(47.6\) years
  • The nonwhites had a life expectancy of \(33.0\) years.
  • The average life duration of the \(124\) whites was \(45.3\) years, with a standard deviation of \(12.7\) years.
  • The average life span of the \(82\)nonwhites was \(34.1\) years, with a standard deviation of \(15.6\) years.
02

Step 2. Calculation

To find the \(p-\)value press STAT and twice to select TESTS.

Select the \(2\) sample T test and press ENTER as displayed,

Use the \(\rightarrow\) select Stats and enter the provided sample values.

Once all the sample entries are made press to obtain the results as,

From the above output the p-value is \(5.315896E-85.315896E-8\) which is approximately equal to \(0\).

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