The 3M company started a new recreation program for its employees in the hope th
ID: 3050449 • Letter: T
Question
The 3M company started a new recreation program for its employees in the hope that a little recreation would improve an employee’s performance at work. To determine whether the high cost of the program is justified, the president of the company wishes to estimate the proportion of the employees who participate in the recreational activities. In a random sample of 200 employees, 60 were found to regularly participate in the recreation program. Give the following information for a 95% confidence interval.
# of successes =
Sample size =
Sample proportion (rounded to nearest hundredth) =
Do you use Z or t? =
Z or t =
Standard Error (Rounded to nearest hundredth)=
Margin of Error (Rounded to nearest hundredth) =
Lower limit (Rounded to nearest hundredth) =
Upper limit (Rounded to nearest hundredth) =
Explanation / Answer
we wil use z =1.96
( please revert for any clariifcation)
x = 60 sample size n = 200 sample proportion p x/n= 0.3000 std error =Se =(p*(1-p)/n) = 0.03 for 95 % CI value of z= 1.9600 margin of error E=z*std error = 0.06 lower confidence bound=sample proportion-margin of error 0.24 Upper confidence bound=sample proportion+margin of error 0.36Related Questions
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