Question

According to a bridal magazine, the average cost of a wedding reception for an American wedding is $8213. Assume that this average is based on a random sample of 450 weddings and that the standard deviation is $2185.
a. What is the point estimate of the corresponding population mean?
b. What is the margin of error for this estimate (use2 standard deviations (divided by Vn) to either side of the mean is used when no specific confidence level is stated).
c. Create and interpret a 98% confidence interval for the corresponding population mean.

Answer 1

Answer:

a. The point estimate of the corresponding population mean is of $8213.

b. The margin of error is of $103.

c. The 98% confidence interval for the corresponding population mean is between $7973.5 and $8452.5, which means that for all weedings, we are 98% sure that the true population mean is in this interval.

Step-by-step explanation:

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

a. What is the point estimate of the corresponding population mean?

The sample mean is of $8213. So, by the Central Limit Theorem, the point estimate of the corresponding population mean is of $8213.

b. What is the margin of error for this estimate (use2 standard deviations (divided by Vn) to either side of the mean is used when no specific confidence level is stated).

Sample of 450, standard deviation of $2185. So

$M = \frac{2185}{\sqrt{450}} = 103$

The margin of error is of $103.

c. Create and interpret a 98% confidence interval for the corresponding population mean.

The confidence interval is given by:

$\mu \pm M*z$

In which $\mu$ is the sample mean and z is the value related to the confidence level.

98% confidence level

So $\alpha = 0.02$, z is the value of Z that has a pvalue of $1 - \frac{0.02}{2} = 0.99$, so $Z = 2.325$.

Then

$\mu - M*z = 8213 - 2.325*103 = 7973.5$

$\mu + M*z = 8213 + 2.325*103 = 8452.5$

The 98% confidence interval for the corresponding population mean is between $7973.5 and $8452.5, which means that for all weedings, we are 98% sure that the true population mean is in this interval.