Question

1. The waiting times (in minutes) of customers at a certain bank is given below. 5.5 6.5 6.7 7.2 6.4 7.3 (a) Construct a 95% confidence interval for the mean waiting time of all customers at the bank. (b) Based on the confidence interval in (a), will the bank manager be satisfied if she wishes the average waiting time in her bank to be less than 7 minutes

Answer 1

The 95% confidence interval is

`5.92  <   \mu  <  7.284`

Explanation:

From the question we are told that

The data is 6.5 6.7 7.2 6.4 7.3

Generally the sample mean is mathematically represented as

`\= x = (\sum x )/(n)`

=>
`\= x = ( 5.5 + 6.5 + \cdot 7.3 )/( 6)`

=>
`\= x = 6.6`

Generally the standard deviation is mathematically represented as

`\sigma = \sqrt{ ( \sum (x_i - \= x )^2 )/( n-1) }`

=>
`\sigma = \sqrt{ ( (5.5 - 6.6 )^2 +(6.5 - 6.6 )^2 + \cdot + (7.3 - 6.6 )^2 )/( 6-1) }`

=>
`\sigma = 0.6512`

Gnerally the degree of freedom is mathematically represented as

`df = n- 1`

=>
`df = 6 -1`

=>
`df = 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 t distribution table the critical value of
`(\alpha )/(2)` is

`t_{(\alpha )/(2) , 5 } = 2.571`

Generally the margin of error is mathematically represented as

`E = t_{(\alpha )/(2), 5  } *  (\sigma )/(√(n) )`

=>
`E =  2.571  *  (0.6512)/(√(6) )`

=>
`E = 0.6835`

Generally 95% confidence interval is mathematically represented as

`\= x  -E <   \mu  <  \= x  +E`

=>
`6.6  - 0.6835 <   \mu  <  6.6  +  0.6835`

=>
`5.92  <   \mu  <  7.284`