Question

The contents of seven similar containers of sulfuric acid are 9.8, 10.2, 10.4, 9.8,10.0, 10.2, and 9.6 liters. Find a 95% confidence interval for the mean contents of all such containers, assuming an approximately normal distribution. (Round your answers into two decimal places.)

Answer 1

The 95 confidence interval is
`9.738  <  \mu < 10.262`

Explanation:

The sample size is n = 7

The sample data is 9.8, 10.2, 10.4, 9.8,10.0, 10.2, and 9.6 liters

Generally the sample mean is mathematically represented as

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

=>
`\= x = ( 9.8 + 10.2 + 10.4 +\cdots + 9.6)/(7 )`

=>
`\= x = 10`

Generally the standard deviation is mathematically represented as

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

=>
`\sigma = \sqrt{( ( 9.8 - 10)^2 + ( 10.2 - 10)^2+ \cdots + ( 9.6 - 10)^2)/(7-1) }`

=>
`\sigma =0.283`

Note : We are making use of t distribution because n is small i.e n < 30

Generally the degree of freedom is mathematically represented as

`df = n - 1`

=>
`df = 7 - 1`

=>
`df = 6`

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 at a degree of freedom of is

`t_{(\alpha )/(2), 6 } =  2.447`

Generally the margin of error is mathematically represented as

`E = t_{(\alpha )/(2) , 6} *  (\sigma )/(√(n) )`

=>
`E = 2.447  *  (0.283)/(√(7) )`

=>
`E =  0.262`

Generally 95% confidence interval is mathematically represented as

`\= x -E <  \mu <  \=x  +E`

=>
`10 -0.262 <  \mu <10 + 0.262`

=>
`9.738  <  \mu < 10.262`