Question

In a survey, 24 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $42 and standard deviation of $2. Construct a confidence interval at a 98% confidence level.

Answer 1

Answer:

The 98% confidence interval for the mean amount spent on their child's last birthday gift is between $40.98 and $43.02.

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 24 - 1 = 23

98% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 23 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.98}{2} = 0.99$. So we have T = 2.5

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 2.5\frac{2}{\sqrt{24}} = 1.02$

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 42 - 1.02 = $40.98.

The upper end of the interval is the sample mean added to M. So it is 42 + 1.02 = $43.02.

The 98% confidence interval for the mean amount spent on their child's last birthday gift is between $40.98 and $43.02.