Question

Find the range for the population mean value with 95% and 65% confidence intervals for each set of data. 3.611 0.02cm n 24

Answer 1

Complete Question

Find the range for the population mean value with 95% and 65% confidence intervals for each set of data.

`\= x_1= 3.611 \ cm`,
`\sigma_1=0.02\ c m`,
`n_1=24`,

`\=x_2=3.632 \ cm`,
`\sigma_2=0.06 \ cm`,
`n_2= 17`

Answer:

The range for the population mean value with 95% confidence intervals for the first set of data is

`2.602 <\mu <  2.619`

The range for the population mean value with 95% confidence intervals for the second set of data is

`2.601 <\mu <  2.661`

The range for the population mean value with 65% confidence intervals for the first set of data is

`2.605 <\mu<  2.617`

The range for the population mean value with 65% confidence intervals for the second set of data is

`2.611 <\mu <  2.6526`

Explanation:

Generally the range for the population mean value with 95% confidence intervals for the first set of data is mathematically evaluated as follows

Generally the degree of freedom is mathematically evaluated as

`df = n_1 -1`

=>
`df = 24 -1`

=>
`df = 23`

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 is

`t_{(\alpha )/(2)  , df = 23} =  2.068`

Generally the margin of error is mathematically represented as

`E_1 = t_{(\alpha )/(2)  , df = 23}  *  (\sigma_1 )/(√(n_1) )`

=>
`E_1 = 2.068  *  (0.02)/(√(24) )`

=>
`E_1 =  0.00844`

Generally 95% confidence interval is mathematically represented as

`\= x_1 -E_1 <  \mu <  \=x_1  +E_1`

`2.611  - 0.00844 < \mu < 2.611  + 0.00844`

=>
`2.602 <\mu <  2.619`

Generally the range for the population mean value with 95% confidence intervals for the second set of data is mathematically evaluated as follows

Generally the degree of freedom is mathematically evaluated as

`df_2 = n_2 -1`

=>
`df_2 = 17 -1`

=>
`df_2 = 16`

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 is

`t_{(\alpha )/(2)  , df = 16} =  2.12`

Generally the margin of error is mathematically represented as

`E_1 = t_{(\alpha )/(2)  , df = 23}  *  (\sigma_2 )/(√(n_2) )`

=>
`E_2 =2.12  *  (0.06)/(√(17) )`

=>
`E_2 =  0.03085`

Generally 95% confidence interval is mathematically represented as

`\= x_2 -E_2 <  \mu <  \=x_2  +E_2`

`2.632  - 0.03085<\mu < 2.632  + 0.03085`

=>
`2.601 <\mu <  2.661`

Generally the range for the population mean value with 65% confidence intervals for the first set of data is mathematically evaluated as follows

Generally the degree of freedom is mathematically evaluated as

`df = n_1 -1`

=>
`df = 24 -1`

=>
`df = 23`

From the question we are told the confidence level is 95% , hence the level of significance is

`\alpha = (100 - 65 ) \%`

=>
`\alpha = 0.35`

Generally from the t distribution table the critical value of at is

`t_{(\alpha )/(2)  , df = 23} = 1.3995`

Generally the margin of error is mathematically represented as

`E_3 = t_{(\alpha )/(2)  , df = 23}  *  (\sigma_1 )/(√(n_1) )`

=>
`E_3 = 1.3995 *  (0.02)/(√(24) )`

=>
`E_3 =  0.00571`

Generally 95% confidence interval is mathematically represented as

`\= x_1 -E_3 <  \mu <  \=x_1  +E_3`

`2.611  - 0.00571 <\mu <  2.611  + 0.00571`

=>
`2.605 <\mu<  2.617`

Generally the range for the population mean value with 65% confidence intervals for the second set of data is mathematically evaluated as follows

Generally the degree of freedom is mathematically evaluated as

`df_2 = n_2 -1`

=>
`df_2 = 17 -1`

=>
`df_2 = 16`

From the question we are told the confidence level is 95% , hence the level of significance is

`\alpha = (100 - 65 ) \%`

=>
`\alpha = 0.35`

Generally from the t distribution table the critical value of at is

`t_{(\alpha )/(2)  , df = 16} =  1.41930`

Generally the margin of error is mathematically represented as

`E_4 = t_{(\alpha )/(2)  , df = 23}  *  (\sigma_2 )/(√(n_2) )`

=>
`E_4 =1.41930 *  (0.06)/(√(17) )`

=>
`E_4 =  0.020653`

Generally 95% confidence interval is mathematically represented as

`\= x_2 -E_4 <  \mu <  \=x_2  +E_4`

`2.632  - 0.020653<\mu < 2.632  + 0.020653`

=>
`2.611 <\mu <  2.6526`