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navik [9.2K]
3 years ago
6

Please please help It’s my last assignment

Mathematics
1 answer:
spin [16.1K]3 years ago
7 0
A. Congruent by SSS

Let me know if is wrong or right, please!
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2 years ago
How many weeks of data must be randomly sampled to estimate the mean weekly sales of a new line of athletic footwear? We want 98
Irina18 [472]

Answer:

n=(\frac{2.33(1400)}{400})^2 =66.50 \approx 67

So the answer for this case would be n=67 rounded up

Step-by-step explanation:

Information given

\bar X represent the sample mean for the sample  

\mu population mean

\sigma=1400 represent the sample standard deviation

n represent the sample size  

Solution to the problem

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 =400 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 98% of confidence interval now can be founded using the normal distribution. And the critical value would be z_{\alpha/2}=2.33, replacing into formula (b) we got:

n=(\frac{2.33(1400)}{400})^2 =66.50 \approx 67

So the answer for this case would be n=67 rounded up

8 0
3 years ago
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