In the U.S., it has been claimed that adults between the ages of 18 and 30 send

Question

In the U.S., it has been claimed that adults between the ages of 18 and 30 send an average of 55.5 text messages per day, with a standard deviation of 22.1. If a random sample of 99 adults in this age group were collected, what would be the mean of the sampling distribution of the sample mean, to one decimal place?

In the U.S., it has been claimed that adults between the ages of 18 and 30 send an average of 58.8 text messages per day, with a standard deviation of 20.2. If a random sample of 123 adults in this age group were collected, what would be the standard deviation of the sampling distribution of the sample mean, to one decimal place?

Explanation / Answer

We know the following about the sampling distribution of the mean. The mean of the sampling distribution (x) is equal to the mean of the population (). And the standard error of the sampling distribution (x) is determined by the standard deviation of the population (), the population size (N), and the sample size (n). These relationships are shown in the equations below:

x =      and      x = [ / sqrt(n) ] * sqrt[ (N - n ) / (N - 1) ]

In the standard error formula, the factor sqrt[ (N - n ) / (N - 1) ] is called the finite population correction or fpc. When the population size is very large relative to the sample size, the fpc is approximately equal to one; and the standard error formula can be approximated by:

x = / sqrt(n).

Based on this mean for first sample would be 55.5

And sd for second sample would be sd=sd/sqrt(n)=1.82

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.