Answer :
SOLUTION
To solve this we will apply the formula
[tex]C.I=p\pm Z\frac{\alpha}{2}\sqrt{\frac{p(1-p)}{n}}[/tex]The required probability p becomes
[tex]\begin{gathered} p=\frac{88}{120}=0.73 \\ n=sample\text{ size}=120 \\ Z\frac{\alpha}{2}=Z_{0.95}=1.96 \end{gathered}[/tex]Note that the Z score was gotten using a calculator.
Plugging in the values we have
[tex]\begin{gathered} C.I=p\pm Z\frac{\alpha}{2}\sqrt{\frac{p(1-p)}{n}} \\ C.I=0.73\pm1.96\sqrt{\frac{0.73(1-0.73)}{120}} \\ C.I=0.73\pm1.96\sqrt{\frac{0.73(0.27)}{120}} \\ C.I=0.73\pm1.96\times0.0405277 \\ C.I=0.73\pm0.07943 \end{gathered}[/tex]So, we have
[tex]\begin{gathered} 0.73+0.07943 \\ =0.8094344\text{ = 80.94\%} \\ and\text{ } \\ 0.73-0.07943 \\ =0.65057=65.06\% \end{gathered}[/tex]So we have the interval as (65.06%, 80.94%).
Since the confidence interval does not fall around 90%, but less than 90%
Hence it is not reasonable to claim that 90% have one
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