Question

15) In a recent study of 35 ninth-grade students, the mean number of hours per week that they played video games was 16.6. The standard deviation of the population was 2.8. Find the 95 % confidence interval of the mean of the time playing video games.

Answer 1

Answer:

The 95 % confidence interval of the mean of the time playing video games. is

        $15.67< \mu <17.52$

Step-by-step explanation:

From the question we are told that

      The sample size is $n = 35$

        The sample mean is $\= x = 16.6$

       The standard deviation is  $\sigma = 2.8$

The confidence level is  95%  hence the level of significance is mathematically represented as

     $\alpha = 100 - 95$

     $\alpha = 5$%

      $\alpha = 0.05$

Now the critical value of half of this level of significance obtained from the normal distribution table is  

       $Z_{\frac{\alpha }{2} } = 1.96$

The  reason for the half is that we are considering the two tails of the normal distribution curve which we use to obtain the interval

Now the standard error of the mean is mathematically evaluated as

       $\sigma _{\= x} = \frac{\sigma }{\sqrt{n} }$

substituting values  

     $\sigma _{\= x} = \frac{2.8 }{\sqrt{35} }$

      $\sigma _{\= x} = 0.473$

the 95 % confidence interval of the mean of the time playing video games.

is mathematically evaluated as

    $\= x - (Z_{\frac{\alpha }{2} } * \sigma_{\= x }) < \mu < \= x - (Z_{\frac{\alpha }{2} } * \sigma_{\= x })$

substituting values

    $16.6 - (1.96 * 0.473) < \mu < 16.6 + (1.96 * 0.473)$

     $15.67< \mu <17.52$