Question

In a survey, 20 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 $6. Construct a confidence interval at a 90% confidence level.

Answer 1

Answer:

The confidence interval at a 90% confidence level is ($43.79, $48.21).

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.9}{2} = 0.05$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$.

So it is z with a pvalue of $1-0.05 = 0.95$, so $z = 1.645$

Now, find M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.645*\frac{6}{\sqrt{20}} = 2.21$

The lower end of the interval is the mean subtracted by M. So it is 46 - 2.21 = 43.79

The upper end of the interval is the mean added to M. So it is 46 + 2.21 = 48.21.

The confidence interval at a 90% confidence level is ($43.79, $48.21).