Question

A marketing specialist wants to estimate the average amount spent by visitors to an online retailer's newly-designed website. From the data in a preliminary study she guesses that the standard deviation of the amount spent is about 12 dollars.

How large a sample should she take to estimate the mean amount spent to within 4 dollars with 98% confidence? (Round your answer up to the next largest integer).

Answer 1

Answer:

n=49

Step-by-step explanation:

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Assuming the X follows a normal distribution

$X \sim N(\mu, \sigma=12)$

We know that the margin of error for a confidence interval is given by:

$Me=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$   (1)

The next step would be find the value of $\z_{\alpha/2}$, $\alpha=1-0.98=0.02$ and $\alpha/2=0.01$

Using the normal standard table, excel or a calculator we see that:

$z_{\alpha/2}=\pm 2.33$

If we solve for n from formula (1) we got:

$\sqrt{n}=\frac{z_{\alpha/2} \sigma}{Me}$

$n=(\frac{z_{\alpha/2} \sigma}{Me})^2$

And we have everything to replace into the formula:

$n=(\frac{2.33(12)}{4})^2 =48.86$

And if we round up the answer we see that the value of n to ensure the margin of error required $Me=4$ dollars is n=49.

Answer 2

She takes to estimate the mean amount spent to be within 4 dollars with 98% confidence is 49.

What is the margin of error?

The probability or the chances of error while choosing or calculating a sample in a survey is called the margin of error.

A marketing specialist wants to estimate the average amount spent by visitors to an online retailer's newly-designed website.

From the data in a preliminary study, she guesses that the standard deviation of the amount spent is about 12 dollars.

Assuming the X follows a normal distribution then we have

$X \sim N(\mu, \sigma = 12)$

We know that the margin of error for a confidence interval is given as

$ME= z_{\alpha /2} = \dfrac{\sigma }{\sqrt{n}}$  ...1

Then the value of α will be

α = 1 - 0.98 = 0.02

Then

α/2 = 0.01

Then the value of the z-score is 2.33, then we have

$n = (\dfrac{z_{\alpha / 2} \times\sigma }{ME })^2\\\\\\n = \dfrac{2.33 \times 12}{4}^2\\\\\\n = 48.86 \approx 49$

The margin of error required for 4 dollars is n = 49.

More about the margin of error link is given below.

https://brainly.com/question/6979326