for a sample of 40 firms in the kaiser health coverage survey the mean monthly c

Question

for a sample of 40 firms in the kaiser health coverage survey the mean monthly cost of the premium for a hmo plan was 405.02. the sample standard deviation of premium costs was $112.08. assume that the population has a normal distribution. using a 90% confidence level, find the margin of error Question 10 of 11 For a sample of 40 firms in the Kaiser health coverage a normal distribution. Using a 90% confidence level, fin O A. 8.232 O B. 15.343 O c. 44.212 D. 29.843 Reset Selection Mark for Review What's This? Previous Next Save Exit ere to search

Explanation / Answer

given that,
sample mean, x =405.02
standard deviation, s =112.08
sample size, n =40
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 112.08/ sqrt ( 40) )
= 17.7214
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.1
from standard normal table, two tailed value of |t /2| with n-1 = 39 d.f is 1.68488
margin of error = 1.68488 * 17.7214
= 29.85844 ~ 29.843 which is nearest to the ccomputed

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