Construct a 95% confidence interval for 'P1-P2'.

The sample statistics listed below are from independent samples. x1= 35 x2= 40 // n1=50 n2=60

interval proportion = - 0.144, 0.204

Step-by-step explanation: from the question.

X1= 35, n1 = 50, x2= 40, n2= 60

P1=x1/n1=35/50 = 7/10 = 0.7,

1-p1 = 1 - 0.7 =0.3

P2 = x2/n2 = 40/60 = 2/3 = 0.67

1 - p2 = 1 - 0.67 = 0.33

95% confidence interval for population proportion is given as

p1-p2 + Zα/2 * √{p1(1-p1)/n1 + p2(1-p2)/n2} This is the upper limit

p1-p2 - Zα/2 * √{p1(1-p1)/n1 + p2(1-p2)/n2} This is the lower limit

p1 = first sample proportion = 0.7

p2 = second sample proportion = 0.67

n1 = first sample size = 50

n2= second sample size = 60

Zα/2= z score for a two tailed test at 5% level of significance = 1.96.

Upper limit

(0.7 - 0.67) + 1.96 * {√(0.7*0.3/50 + 0.67*0.33/60)}

0.03 + 1.96 {√(0.0042 + 0.003685)}

0.03 + 1.96 √0.007885

0.03 + 1.96 ( 0.08879)

0.03 + 0.174

= 0.204.

Upper limit

(0.7 - 0.67) + 1.96 * {√(0.7*0.3/50 + 0.67*0.33/60)}

0.03 + 1.96 {√(0.0042 + 0.003685)}

0.03 + 1.96 √0.007885

0.03 + 1.96 ( 0.08879)

0.03 + 0.174

= 0.204.

Lower limit.

Upper limit

(0.7 - 0.67) + 1.96 * {√(0.7*0.3/50 + 0.67*0.33/60)}

0.03 + 1.96 {√(0.0042 + 0.003685)}

0.03 + 1.96 √0.007885

0.03 + 1.96 ( 0.08879)

0.03 + 0.174

= 0.204.

Upper limit

(0.7 - 0.67) + 1.96 * {√(0.7*0.3/50 + 0.67*0.33/60)}

0.03 + 1.96 {√(0.0042 + 0.003685)}

0.03 + 1.96 √0.007885

0.03 + 1.96 ( 0.08879)

0.03 + 0.174

= 0.204.

Upper limit

(0.7 - 0.67) + 1.96 * {√(0.7*0.3/50 + 0.67*0.33/60)}

0.03 + 1.96 {√(0.0042 + 0.003685)}

0.03 + 1.96 √0.007885

0.03 + 1.96 ( 0.08879)

0.03 + 0.174

= 0.204.

Lower limit

(0.7-0.67) - 1.96 * {√(0.7*0.3/50 + 0.67*0.33/60)}

0.03 - 1.96 {√(0.0042 + 0.003685)}

0.03 - 1.96 √0.007885

0.03 - 1.96 ( 0.08879)

0.03 - 0.174

= - 0.144

interval proportion = - 0.144, 0.204

this is a direct variation proportion. y = kx

y is the total amount due, k is the constant of variation, and x is the number of texts sent.

plug in the 3.20 and the 16 to solve for the "k". this is the constant of proportionality. it is the value that will not change.

3.20 = k(16). solving you get $.20 per text.

the equation would then be y = .20x