Of 94 adults selected randomly from one town, 62 have health insurance. Find a 90% confidence...

Of 94 adults selected randomly from one town, 62 have health insurance. Find a 90% confidence interval for the proportion of all adults in the town who have health insurance.

Solutions

Expert Solution

Solution :

Given that,

n = 94

x = 62

Point estimate = sample proportion = = x / n =62/94=0.660

1 - = 0.34

At 90% confidence level

= 1 - 90%

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645 ( Using z table )

Margin of error = E Z/2 * $\sqrt$ (( * (1 - )) / n)

= 1.645 * $\sqrt$ ((0.66*0.34) /94 )

E = 0.080

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.660-0.080 < p < 0.660+0.080

(0.580,0.740)

orchestra answered 1 year ago

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