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 $48 and standard deviation of $10. Construct a confidence interval at a 80% confidence level.

Question

Give your answers to one decimal place.

Answer 1

To construct an 80% confidence interval for the mean spending on a child's birthday gift, we can use the formula for the confidence interval:

$$\text{Confidence Interval}=\overline{x}\pm z\left(\frac{s}{\sqrt{n}}\right)$$

Where:

• $\overline{x}$ is the sample mean.
• $z$ is the z-score corresponding to the desired confidence level.
• $s$ is the sample standard deviation.
• $n$ is the sample size.

Given:

• $\overline{x}=48$
• $s=10$
• $n=22$

Step 1: Determine the z-score for 80% confidence level.

For an 80% confidence level, the area in each tail is $(1-0.80)/2=0.10$. Looking up the z-value that corresponds to 0.10 in the right tail, we find:

$$z\approx 1.28$$

Step 2: Calculate the standard error (SE).

$$\text{SE}=\frac{s}{\sqrt{n}}=\frac{10}{\sqrt{22}}\approx \frac{10}{4.690}\approx 2.13$$

Step 3: Calculate the margin of error (ME).

$$\text{ME}=z\cdot \text{SE}\approx 1.28\cdot 2.13\approx 2.73$$

Step 4: Construct the confidence interval.

$$\text{Confidence Interval}=48\pm 2.73$$

Calculating the lower and upper bounds:

$$\text{Lower Bound}=48-2.73\approx 45.3$$ $$\text{Upper Bound}=48+2.73\approx 50.7$$

Final Result:

The 80% confidence interval for the amount spent on a child's last birthday gift is:

$$\boxed{{\displaystyle (45.3,50.7)}}$$