(9^6 * 7^-9)^-4. Thanks!

Mathematics

1 answer:

Anuta_ua [19.1K]1 year ago

7 0

Answer:

$\frac{7^{36}}{9^{24}}$

Step-by-step explanation:

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A marketing researcher wants to find a 96% confidence interval for the mean amount those visitors spend per person per day while

ki77a [65]

Answer:

$n=(\frac{2.054(12)}{4})^2 =37.97 \approx 38$

So then the minimum sample to ensure the condition given is n= 38

Step-by-step explanation:

Notation

$\bar X$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

$\sigma=12$ represent the population standard deviation

n represent the sample size

ME = 4 the margin of error desired

Solution to the problem

When we create a confidence interval for the mean the margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

And on this case we have that ME =4 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

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

The critical value for 96% of confidence interval now can be founded using the normal distribution. The significance is $\alpha=1-0.96 =0.04$ . And in excel we can use this formula to find it:"=-NORM.INV(0.02;0;1)", and we got $z_{\alpha/2}=2.054$ , replacing into formula (b) we got: