Question

A random sample of 100,000 credit sales in a department store showed an average sale of $82.25. From past data, it is known that the standard deviation of the population is $30.00. The margin of error is 0.186.

Required:
a. What is the 95% confidence interval of the population mean?
b. With a 0.95 probability, determine the sampling error?
c. Determine the standard error of the mean.

Answer 1

Answer:

Step-by-step explanation:

Given that:

Sample Mean $\overline x$ = 82.25

The Standard deviation of the population $\sigma$ = 30.00

The Sample size = 100,000

The margin of error M.O.E = 0.186

At a 95% confidence interval level;

$=\overline x - M.O.E < \mu < \overline x + M.O.E$

= 82.25 - 0.186 < $\mu$ < 82.25 +  0.186

= 82.064 < $\mu$ < 82.436

= (82.064 , 82.436)

The 95% confidence interval of the population mean = (82.064 , 82.436)

Before we can determine the sampling error, we need to find the standard error of the mean.

The standard error of the mean is:

$\mu_ {\bar x} = \mu = 82.25$

$\sigma_{\bar x} = \dfrac{\sigma}{\sqrt{n}}$

$\sigma_{\bar x} = \dfrac{30}{\sqrt{100000}}$

$\sigma_{\bar x} = \dfrac{30}{316.23}$

$\sigma_{\bar x} = 0.096$

At 95% confidence interval, the level of significance = 1 - 0.95 = 0.05

$Z_{0.05/2} = Z_{0.025}$ = 1.96

Now, the sampling error can be determined by using the formula:

$=Z_{\alpha/2} \times \sigma _{\overline x}$

$=Z_{0.025} \times \sigma _{\overline x}$

= 1.96 × 0.096

= 0.18816