Question

You want to know about the dining habits of college students in the US. Since you do not have resources to survey all college students in the nation, you take a random sample of 1501. You learn that, on average, they eat out 4 times per week. Also, these sample data are generally distributed about the average of 4 times, by 1.5 times per week. 1. What is the average times per week that college students across the nation eat out

Answer 1

Answer:

the average times per week that college students across the nation eat out lies within these 95% confidence interval

      $3.938 < \mu < 4.062$

Step-by-step explanation:

From the question we are told that

    The sample size is n =  1501

    The sample mean  is  $\= x = 4$

      The standard deviation is  $\sigma = 1.5$

From the question we are told the confidence level is  95% , hence the level of significance is    

      $\alpha = (100 - 95 ) \%$

=>   $\alpha = 0.05$

Generally from the normal distribution table the critical value  of  $\frac{\alpha }{2}$ is  

   $Z_{\frac{\alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as  

      $E = Z_{\frac{\alpha }{2} } * \frac{\sigma }{\sqrt{n} }$

=>   $E = 1.96 * \frac{1.5 }{\sqrt{1501} }$

=>   $E = 0.06196$

Generally the estimate for average times per week that college students across the nation eat out at a 95% confidence is  mathematically represented as

      $\= x -E < \mu < \=x +E$

      $4 -0.06196 < \mu < 4 + 0.06196$

=>   $3.938 < \mu < 4.062$

Hence the true average times per week that college students across the nation eat out lies within this interval