Question

You measure 40 watermelons' weights, and find they have a mean weight of 67 ounces. Assume the population standard deviation is 11.4 ounces. Based on this, what is the maximal margin of error associated with a 99% confidence interval for the true population mean watermelon weight.

Answer 1

Answer:

$35.36,44.64$

Step-by-step explanation:

Sample size $n=40$

Mean weight $\=x =67$

Standard deviation $\sigma=11.4$

Confidence Interval $CI=0.99$

\alpha==0.01

Therefore

$Z_{\frac{\alpha}{2}}=Z_[0.005]$

From table

$Z_{\frac{\alpha}{2}}=2.576$

Generally, the equation for Margin of error is mathematically given by

$M.E=Z_{\frac{\alpha}{2}}*(\frac{\sigma}{sqrt{n}}$

$M.E=2.576*(\frac{11.4}{\sqrt{40}}$

$M.E=4.64$

Therefore Estimated mean is

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

$40-4.64<\mu< 40+4.64$

$35.36<\mu < 44.64$

$35.36,44.64$