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A simple random sample of 49 8th graders at a large suburban middle school indicated that 89% are involved in some type of after school activity. Find the 95% confidence interval that estimates the proportion of them that are involved in an after school activity.

A. [0.722], [0.978]
B. [0.802], [0.978]
C. [0.702], [0.928]
D. [0.802], [0.778]
E. [0.852], [0.857]
F. None of the above

Respuesta :

ashish4112119

Answer:

B. [0.802], [0.978]

Step-by-step explanation:

n=49

p=.89

confidence level of 95% has a z value of 1.960

z*= 1.960

z*=qnorm(1.95/2)

p+/-z* multiplied by sqrt[ P * ( 1 - P ) / n ]

p+/-z*times(√(p)(1-p)/n)

.89+/-1.960* (square root of (.89)(.11)/49))

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