Question

A random sample of 16 ATM transactions at the Last National Bank of Flat Rock revealed a mean transaction time of 2.8 minutes with a standard deviation of 1.2 minutes. The width (in minutes) of the 95 percent confidence interval for the true mean transaction time is plus/minus (to three decimal places):

Answer 1

The width is
`W = 1.2786`

Explanation:

From the question we are told that

The sample size is n = 16

The mean time is
`\= x = 2.8 \ minute`

The standard deviation is
`s = 1.2 \ minutes`

Generally the degree of freedom is mathematically represented as

`df = n - 1`

=>
`df = 16 - 1`

=>
`df = 15`

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 t distribution table the critical value of
`(\alpha )/(2)` at a degree of freedom of df = 15 is

`t_{(\alpha )/(2) , 15 } =  2.131`

Generally the margin of error is mathematically represented as

`E = t_{(\alpha )/(2), 15 } *  (\sigma )/(√(n) )`

=>
`E = 2.131  *  ( 1.2 )/(√(16) )`

=>
`E = 0.6393`

Generally the width the confidence 95% confidence interval

`W = 2 * E`

=>
`W = 2 * 0.6393`

=>
`W = 1.2786`