Mathematics, 28.10.2020 16:40, andrewblack033

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

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answered: grugfhb

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

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

2.602

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

2.601

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

2.605

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

2.611

Step-by-step 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

$t_{\frac{\alpha }{2} , df = 23} = 2.068$

Generally the margin of error is mathematically represented as

$E_1 = t_{\frac{\alpha }{2} , df = 23} * \frac{\sigma_1 }{\sqrt{n_1} }$

=> $E_1 = 2.068 * \frac{0.02}{\sqrt{24} }$

=> 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

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

$t_{\frac{\alpha }{2} , df = 16} = 2.12$

Generally the margin of error is mathematically represented as

$E_1 = t_{\frac{\alpha }{2} , df = 23} * \frac{\sigma_2 }{\sqrt{n_2} }$

=> $E_2 =2.12 * \frac{0.06}{\sqrt{17} }$

=> 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

=> 2.601

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

$t_{\frac{\alpha }{2} , df = 23} = 1.3995$

Generally the margin of error is mathematically represented as

$E_3 = t_{\frac{\alpha }{2} , df = 23} * \frac{\sigma_1 }{\sqrt{n_1} }$

=> $E_3 = 1.3995 * \frac{0.02}{\sqrt{24} }$

=> 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

=> 2.605

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

$t_{\frac{\alpha }{2} , df = 16} = 1.41930$

Generally the margin of error is mathematically represented as

$E_4 = t_{\frac{\alpha }{2} , df = 23} * \frac{\sigma_2 }{\sqrt{n_2} }$

=> $E_4 =1.41930 * \frac{0.06}{\sqrt{17} }$

=> 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

=> 2.611

answered: Guest

answer:

\boxed {\frac{1}{5}}

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

solution-

60 * \frac{1}{5} = 12

90 * \frac{1}{5} = 18

75 * \frac{1}{5} = 15

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

hence, the answer would be \boxed {\frac{1}{5}}

What is the scale factor of axyz to auvw?

answered: Guest

the answer is c hope it

step-by-step explanation:

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