Question

A cell phone company offers two plans to its subscribers. At the time new subscribers sign up, they are asked to provide some demographic information. The mean yearly income for a sample of 40 subscribers to Plan A is $57,000 with a standard deviation of $9,200. For a sample of 30 subscribers to Plan B, the mean income is $61,000 with a standard deviation of $7,100. At the .05 significance level, is it reasonable to conclude the mean income of those selecting Plan B is larger

Answer 1

Yes at 0.05 significance level we can conclude that mean income of those selecting plan B is bigger.

Given mean yearly income for sample of 40 subscribers is $57000 with standard deviation of $9200, mean yearly income for sample of 30 subscribers with standard deviation of $7100.

First we have been given significance level (α) of 0.05. So (1-α)=1-0.05=0.95 and we have to find Z score for this significance level. (1+0.95)/2=0.425

Z=1.44

Now we have to calculate margin of error which will be calculated as under:

$M_{A} =Z*st/\sqrt{n}$

=$1.44*9200/\sqrt{40}$

=1.44*9200/6.32

=1869.10

$M_{B}=Z*st/\sqrt{n}$

=$1.44*7100/\sqrt{30}$

=1.44*7100/3.47=2096.2

We know that margin of error shows the difference between real and calculated values. So if we increase mean of both plans by margin of error of respective plans then mean of plan A becomes=57000+1869.10=58869.10 and mean of plan B becomes =61000+2096.2=630622.2. We can clearly notice that mean of plan B is greater.

Hence it is reasonable to say that the mean of plan B is greater at the significance level of 0.05.

Learn more about margin of error at

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