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Use the given degree of confidence and sample data to construct a confidence interval
for the population proportion p.
Of 83 adults selected randomly from one town, 64 have health insurance.
Find a 90% confidence interval for the true proportion of all adults in the town who have
health insurance.

Respuesta :

Answer:

0.726 < p < 0.874

Step-by-step explanation:

p = 64/83 = about 0.8

confidence interval = p +- z [tex]\sqrt{\frac{p (1-p)}{n} }[/tex]

= 0.8 +- (1.64 x 0.04472)

= 0.8 +- 0.073343

= (0.726656 and 0.873343)

0.726 < p < 0.874

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