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 $32 and standard deviation of $17. Construct a confidence interval at a 98% confidence level.

Answer 1

Answer:

$19.22\:$

Step-by-step explanation:

Tthe population is normally distributed and the sample size is $n=22\:$.

Since the population standard deviation $\sigma$ is unknown and the sample standard deviation $s$, must replace it, the t distribution must be used for the confidence interval.

Hence with degrees of freedom of 21, $t_{\frac{\alpha}{2} }=3.527$.(Read from the t distribution table)

The 98% confidence interval can be constructed  using the formula:

$\bar X-t_{\frac{\alpha}{2}}(\frac{s}{\sqrt{n} } )\:$.

From the question the sample mean is $\bar X=32$dollars and the sample standard deviation is $s=17$ dollars.

We substitute the values into the formula to get

$32-3.527(\frac{17}{\sqrt{22} } )\:$

$19.22\:$

Therefore, we can be 98% confident that the population mean is between is between 19.22 and 44.78 dollars.