Question 333318
you can apply the binomial method directly which can be tedious and difficult
or "if" the assumptions are met, can apply a normal approximation to the binomial.
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The enable the use of the normal approximation, the following need to be met
1. {{{n[1]*p[1]>5}}} and {{{n[2]*p[2]>5}}} and {{{n[1]*(1-p[1])>5}}} and {{{n[2]*(1-p[2])>5}}}
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you provided p1 = 0.48 n1 = 100 p2 = 0.052 n2 =100 
NOTE: not sure if your p2 is correct.  Nevertheless, it looks like the requirements above are met.
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90% confidence interval for difference in the population proportions yields Z=1.645
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Upper confidence interval value:  {{{(p[1]-p[2]) + Z*Sqrt(p[1]*(1-p[1])/n[1] +p[2]*(1-p[2])/n[2])}}}
Lower confidence interval value:  {{{(p[1]-p[2]) - Z*Sqrt(p[1]*(1-p[1])/n[1] +p[2]*(1-p[2])/n[2])}}}
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confidence interval = {{{(0.48-0.052)-1.645*sqrt(0.48*0.52/100+0.052*0.948/100)}}} to {{{(0.48-0.052)+1.645*sqrt(0.48*0.52/100+0.052*0.948/100)}}}
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you can do the math