Find the margin of error at a 90% confidence level.

Question

In a survey, 22 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $46 and standard deviation of $9. Find the margin of error at a 90% confidence level.

Give your answer to two decimal places.

Answer 1

Hi Kaitlyn,

Margin of error is always critical value*standard error, just so you know that for future reference. This is another close call on whether to use z or t--we've got an approximate normal distribution, but low n, no known population standard deviation. The calculation isn't that cumbersome, so I'll provide both types.

First step is to compute standard error:

SE=s/sqrt(n)

s=standard deviation

n=sample size

Here:

s=9

n=22

SE=9/sqrt(22)

SE=1.92

Now, the z-critical value for 90% confidence is 1.645, which might be worth memorizing, just for future reference. Therefore,

z*=1.645

SE=1.92

MoE_z=3.16

Standard error remains the same regardless of which critical value type we use, so:

SE=1.92

For t, we need degrees of freedom, which can be calculated by:

df=n-1

df=22-1

df=21

Go to the t-table, look up 21 degrees of freedom in the row, 90% confidence in the column. This gives:

t*=1.721

Now, we have:

t*=1.721

SE=1.92

MoE_t= (1.721*1.92)

MoE_t=3.30

Not too much difference in critical values. I hope this helps.