Respuesta :
Complete Question
Find the range for the population mean value with 95% and 65% confidence intervals for each set of data.
[tex]\= x_1= 3.611 \ cm[/tex], [tex]\sigma_1=0.02\ c m[/tex], [tex]n_1=24[/tex],
[tex]\=x_2=3.632 \ cm[/tex], [tex]\sigma_2=0.06 \ cm[/tex], [tex]n_2= 17[/tex]
Answer:
The range for the population mean value with 95% confidence intervals for the first set of data is
[tex]2.602 <\mu < 2.619[/tex]
The range for the population mean value with 95% confidence intervals for the second set of data is
[tex]2.601 <\mu < 2.661[/tex]
The range for the population mean value with 65% confidence intervals for the first set of data is
[tex]2.605 <\mu< 2.617[/tex]
The range for the population mean value with 65% confidence intervals for the second set of data is
[tex]2.611 <\mu < 2.6526[/tex]
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
[tex]df = n_1 -1[/tex]
=> [tex]df = 24 -1[/tex]
=> [tex]df = 23[/tex]
From the question we are told the confidence level is 95% , hence the level of significance is
[tex]\alpha = (100 - 95 ) \%[/tex]
=> [tex]\alpha = 0.05[/tex]
Generally from the t distribution table the critical value of at is
[tex]t_{\frac{\alpha }{2} , df = 23} = 2.068[/tex]
Generally the margin of error is mathematically represented as
[tex]E_1 = t_{\frac{\alpha }{2} , df = 23} * \frac{\sigma_1 }{\sqrt{n_1} }[/tex]
=> [tex]E_1 = 2.068 * \frac{0.02}{\sqrt{24} }[/tex]
=> [tex]E_1 = 0.00844 [/tex]
Generally 95% confidence interval is mathematically represented as
[tex]\= x_1 -E_1 < \mu < \=x_1 +E_1[/tex]
[tex]2.611 - 0.00844 < \mu < 2.611 + 0.00844[/tex]
=> [tex]2.602 <\mu < 2.619[/tex]
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
[tex]df_2 = n_2 -1[/tex]
=> [tex]df_2 = 17 -1[/tex]
=> [tex]df_2 = 16[/tex]
From the question we are told the confidence level is 95% , hence the level of significance is
[tex]\alpha = (100 - 95 ) \%[/tex]
=> [tex]\alpha = 0.05[/tex]
Generally from the t distribution table the critical value of at is
[tex]t_{\frac{\alpha }{2} , df = 16} = 2.12 [/tex]
Generally the margin of error is mathematically represented as
[tex]E_1 = t_{\frac{\alpha }{2} , df = 23} * \frac{\sigma_2 }{\sqrt{n_2} }[/tex]
=> [tex]E_2 =2.12 * \frac{0.06}{\sqrt{17} }[/tex]
=> [tex]E_2 = 0.03085 [/tex]
Generally 95% confidence interval is mathematically represented as
[tex]\= x_2 -E_2 < \mu < \=x_2 +E_2[/tex]
[tex]2.632 - 0.03085<\mu < 2.632 + 0.03085[/tex]
=> [tex]2.601 <\mu < 2.661[/tex]
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
[tex]df = n_1 -1[/tex]
=> [tex]df = 24 -1[/tex]
=> [tex]df = 23[/tex]
From the question we are told the confidence level is 95% , hence the level of significance is
[tex]\alpha = (100 - 65 ) \%[/tex]
=> [tex]\alpha = 0.35[/tex]
Generally from the t distribution table the critical value of at is
[tex]t_{\frac{\alpha }{2} , df = 23} = 1.3995 [/tex]
Generally the margin of error is mathematically represented as
[tex]E_3 = t_{\frac{\alpha }{2} , df = 23} * \frac{\sigma_1 }{\sqrt{n_1} }[/tex]
=> [tex]E_3 = 1.3995 * \frac{0.02}{\sqrt{24} }[/tex]
=> [tex]E_3 = 0.00571 [/tex]
Generally 95% confidence interval is mathematically represented as
[tex]\= x_1 -E_3 < \mu < \=x_1 +E_3 [/tex]
[tex]2.611 - 0.00571 <\mu < 2.611 + 0.00571[/tex]
=> [tex]2.605 <\mu< 2.617[/tex]
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
[tex]df_2 = n_2 -1[/tex]
=> [tex]df_2 = 17 -1[/tex]
=> [tex]df_2 = 16[/tex]
From the question we are told the confidence level is 95% , hence the level of significance is
[tex]\alpha = (100 - 65 ) \%[/tex]
=> [tex]\alpha = 0.35[/tex]
Generally from the t distribution table the critical value of at is
[tex]t_{\frac{\alpha }{2} , df = 16} = 1.41930 [/tex]
Generally the margin of error is mathematically represented as
[tex]E_4 = t_{\frac{\alpha }{2} , df = 23} * \frac{\sigma_2 }{\sqrt{n_2} }[/tex]
=> [tex]E_4 =1.41930 * \frac{0.06}{\sqrt{17} }[/tex]
=> [tex]E_4 = 0.020653 [/tex]
Generally 95% confidence interval is mathematically represented as
[tex]\= x_2 -E_4 < \mu < \=x_2 +E_4[/tex]
[tex]2.632 - 0.020653<\mu < 2.632 + 0.020653[/tex]
=> [tex]2.611 <\mu < 2.6526[/tex]
