A statistics student wants to know what percent of the US adult population has r

Question

A statistics student wants to know what percent of the US adult population has run a marathon. He conducts a survey of 730 people and finds that 146 of them have run a marathon. a. Write a 94% confidence interval for the percent of the US population that have completed a marathon, and write a sentence interpreting the confidence interval. b. What is the margin of error of this estimate? c. Assuming the same value of p, how many people would he have to sample to get a 90% confidence interval with a margin of error of less than 2%? d. If the student has no estimate of p or p, what sample size is safest to get that margin of error?

Explanation / Answer

Solution:-

a). Standard error of the mean = SEM = x(N-x)/N3 = 0.015

= (1-CL)/2 = 0.030

Standard normal deviate for = Z = 1.881

Proportion of positive results = P = x/N = 0.200

Lower bound = P - (Z*SEM) = 0.172

Upper bound = P + (Z*SEM) = 0.228

b). Given,

Population size = 730

Success rate = 146

MOE = z * sqrt (p(1-p) / n)

= 1.88 * sqrt (0.2*0.8 / 730)

= 0.0278

c).

At 90% confidence interval

Standard error of the mean = SEM = x(N-x)/N3 = 0.015

= (1-CL)/2 = 0.050

Standard normal deviate for = Z = 1.645

Proportion of positive results = P = x/N = 0.200

Lower bound = P - (Z*SEM) = 0.176

Upper bound = P + (Z*SEM) = 0.224

Minimum sample size = 511

d). If the the student has no estimate of proportion, safest sample size to get this margin of error is 1663.

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