Question

In a survey, 11 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 $14. Construct a confidence interval at a 90% confidence level.

Answer 1

Answer:

The population standard deviation is not known.

90% Confidence interval by T₁₀-distribution: (38.3, 53.7).

Step-by-step explanation:

The "standard deviation" of $14 comes from a survey. In other words, the true population standard deviation is not known, and the $14 here is an estimate. Thus, find the confidence interval with the Student t-distribution. The sample size is 11. The degree of freedom is thus $11 - 1 = 10$.

Start by finding 1/2 the width of this confidence interval. The confidence level of this interval is 90%. In other words, the area under the bell curve within this interval is 0.90. However, this curve is symmetric. As a result,

• The area to the left of the lower end of the interval shall be $1/2 \cdot (1 - 0.90)= 0.05$.
• The area to the left of the upper end of the interval shall be $0.05 + 0.90 = 0.95$.

Look up the t-score of the upper end on an inverse t-table. Focus on the entry with

• a degree of freedom of 10, and
• a cumulative probability of 0.95.

$t \approx 1.812$.

This value can also be found with technology.

The formula for 1/2 the width of a confidence interval where standard deviation is unknown (only an estimate) is:

$\displaystyle t \cdot \frac{s_{n-1}}{\sqrt{n}}$,

where

• $t$ is the t-score at the upper end of the interval,
• $s_{n-1}$ is the unbiased estimate for the standard deviation, and
• $n$ is the sample size.

For this confidence interval:

• $t \approx 1.812$,
• $s_{n-1} = 14$, and
• $n = 11$.

Hence the width of the 90% confidence interval is

$\displaystyle 1.812 \times \frac{14}{\sqrt{10}} \approx 7.65$.

The confidence interval is centered at the unbiased estimate of the population mean. The 90% confidence interval will be approximately:

$(38.3, 53.7)$.