Respuesta :
Answer:
Step-by-step explanation:
From the question we are told that
The first sample size is [tex]n_1 = 1000[/tex]
The second sample size is [tex]n_2 = 6000[/tex]
The number that had significant outside activity in the sample with ALL is [tex]k_1 = 700[/tex]
The number that had significant outside activity in the sample without ALL is [tex]k_2 = 5000[/tex]
Considering question a
The percentage of children with ALL have significant social activity outside the home when younger is mathematically represented as
[tex]\^ p_1 = \frac{700}{1000} * 100[/tex]
=> [tex]\^ p_ 1 = 0.7 = 70\%[/tex]
Considering question b
The percentage of children without ALL have significant social activity outside the home when younger is mathematically represented as
[tex]\^ p_2 = \frac{5000}{6000}[/tex]
=> [tex]\^ p_ 2 = 0.83[/tex]
Generally the sample odds ratio for significant social activity outside the home when younger, comparing the groups with and without ALL is mathematically represented as
[tex]r = \frac{\* p _1}{ \^ p_2 }[/tex]
=> [tex]r = \frac{0.7}{ 0.83 }[/tex]
=> [tex]r = 0.141[/tex]
Considering question c
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 normal distribution table the critical value of [tex]\frac{\alpha }{2}[/tex] is
[tex]Z_{\frac{\alpha }{2} } = 1.96[/tex]
Generally the lower limit of the 95% confidence interval for the population odds ratio for significant social activity outside the home when younger, comparing the groups with and without ALL is mathematically represented as
[tex]a = e^{ln ( r ) - Z_{\frac{\alpha }{2}} \sqrt{ [ \frac{1}{ k_1 } ] + [ \frac{1}{ c_1 } ] + [\frac{1}{k_2} ] + [\frac{1}{ c_2 } ] } }[/tex]
Here [tex]c_1 \ and \ c_2[/tex] are the non-significant values i.e people that did not play outside when they were young in both samples
The values are
[tex]c_1 = 1000 - 700 = 300[/tex]
and [tex]c_2 = 6000 - 5000[/tex]
=> [tex]c_2 = 1000[/tex]
=> [tex]a = e^{ln ( 0.141 ) - 1.96 \sqrt{ [ \frac{1}{ 700 } ] + [ \frac{1}{ 1000} ] + [\frac{1}{5000} ] + [\frac{1}{ 300 } ] } }[/tex]
=> [tex]a = 0.1212[/tex]
Generally the upper limit of the 95% confidence interval for the population odds ratio for significant social activity outside the home when younger, comparing the groups with and without ALL is mathematically represented as
[tex]b = e^{ln ( 0.141 ) + 1.96 \sqrt{ [ \frac{1}{ 700 } ] + [ \frac{1}{ 1000} ] + [\frac{1}{5000} ] + [\frac{1}{ 300 } ] } }[/tex]
[tex]b = 0.1640[/tex]
Generally the 95% confidence interval for the population odds ratio for significant social activity outside the home when younger, comparing the groups with and without ALL is
[tex]95\% CI = [ 0.1212 , 0.1640 ][/tex]
Generally looking and the confidence interval obtained we see that it is less that 1 hence this means that there is a greater odd of developing ALL in groups with insignificant social activity compared to groups with significant social activity