Question

A recent study of 26 city residents showed that the time they had lived at their present address has a mean of 10.3 years with the standard deviation of 3 years. Find the 95% confidence interval of the true mean. Assume that the variable is approximately normally distributed.

Answer 1

The 95% confidence interval is
`9.15<  \mu < 11.45`

Explanation:

From the question we are told that

The sample size is n = 26

The mean is
`\= x = 10.3`

The standard deviation is
`s = 3`

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 *  (3)/(√(26) )`

=>
`E = 1.1532`

Generally 95% confidence interval is mathematically represented as

`\= x -E <  \mu <  \=x  +E`

=>
`10.3  -1.1532 <  \mu < 10.3  +  1.1532`

=>
`9.15<  \mu < 11.45`