Question

A study tested whether significant social activities outside the house in young children affected their probability of later developing the white blood cell disease acute lymphoblastic leukemia (ALL). They compared 1000 children with ALL to 6000 children without ALL. Of the ALL kids, 700 had significant social activity outside the home when younger. Of the non-ALL kids, 5000 had significant social activity outside the home.

Required:
a. What percentage of children with ALL have significant social activity outside the home when younger?
b. What is the sample odds ratio for significant social activity outside the home when younger, comparing the groups with and without ALL?
c. What is a 95% confidence interval for the population odds ratio for significant social activity outside the home when younger, comparing the groups with and without ALL?
d. Does the 95% confidence interval from part c. indicate that the amount of social activity is related to ALL? Why or why not?
e. Test whether significant social activities outside the house in young children affected their probability of later developing ALL, using the significance level of 0.05. Show the hypotheses, test statistic, p-value, and the conclusion.

Answer 1

Explanation:

From the question we are told that

The first sample size is
`n_1 = 1000`

The second sample size is
`n_2 = 6000`

The number that had significant outside activity in the sample with ALL is
`k_1 = 700`

The number that had significant outside activity in the sample without ALL is
`k_2 = 5000`

Considering question a

The percentage of children with ALL have significant social activity outside the home when younger is mathematically represented as

`\^ p_1 = (700)/(1000) * 100`

=>
`\^ p_ 1 = 0.7 = 70\%`

Considering question b

The percentage of children without ALL have significant social activity outside the home when younger is mathematically represented as

`\^ p_2 = (5000)/(6000)`

=>
`\^ p_ 2 = 0.83`

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

`r = (\* p _1)/( \^ p_2 )`

=>
`r = (0.7)/( 0.83 )`

=>
`r = 0.141`

Considering question c

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 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

`a = e^{ln ( r ) - Z_{(\alpha )/(2)} \sqrt{ [ (1)/( k_1 ) ] + [ (1)/( c_1 ) ] + [(1)/(k_2) ] + [(1)/( c_2 ) ] } }`

Here
`c_1 \ and \ c_2` are the non-significant values i.e people that did not play outside when they were young in both samples

The values are

`c_1 = 1000 - 700 = 300`

and
`c_2 = 6000 - 5000`

=>
`c_2 = 1000`

=>
`a = e^{ln ( 0.141 ) - 1.96 \sqrt{ [ (1)/( 700 ) ] + [ (1)/( 1000) ] + [(1)/(5000) ] + [(1)/( 300 ) ] } }`

=>
`a = 0.1212`

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

`b = e^{ln ( 0.141 ) + 1.96 \sqrt{ [ (1)/( 700 ) ] + [ (1)/( 1000) ] + [(1)/(5000) ] + [(1)/( 300 ) ] } }`

`b = 0.1640`

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

`95\% CI = [ 0.1212 , 0.1640 ]`

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