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

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

`3.938  <  \mu < 4.062`

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
`(\alpha )/(2)` is

`Z_{(\alpha )/(2) } =  1.96`

Generally the margin of error is mathematically represented as

`E = Z_{(\alpha )/(2) } *  (\sigma )/(√(n) )`

=>
`E = 1.96  *  (1.5 )/(√(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