With such a large number of people using text messages as a means of communicati

Question

With such a large number of people using text messages as a means of communication, a company is interested in determining the number of work hours lost due to text messaging. Based on a survey of 43 randomly selected employees (anonymously, of course) the company has determined that the average amount of time spent texting over a one-month period is 198 minutes with a standard deviation of 59 minutes. At a 99% level of confidence, what is the margin of error? (Round your answer to 4 decimal places).

Explanation / Answer

here this can be solved by t distribution; while in some places it is referred that when sample size is greater then 30 we can use z distirbution.

from t distribution:

std error of mean =std deviaiton/(n)1/2 =59/(43)1/2 =8.9974

for (43-1=42) degree of freedom and 99% CI ; t =2.698

hence  margin of error =t*std error =2.698*8.9974 =24.2756

from z distribution:

for 99% CI ; z =2.5758

hence  margin of error =z*std error =23.1758

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