Mario buys tickets for either 47 or 57 dollars. He purchases 14 tickets and spends 738 dollars. How many tickets did he purchase
1 answer:
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Answer:
a) The standard error would be of 8.58 pounds.
b) The margin of error is 14.11 pounds.
c) The 90% confidence interval for the population mean is between 232.89 pounds and 261.11 pounds
Step-by-step explanation:
a.What is the standard error?
The standard error is
In which is the standard deviation of the population and n is the size of the sample. So
The standard error would be of 8.58 pounds.
b.What is the margin of error at 90% confidence?
We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:
Now, we have to find z in the Ztable as such z has a pvalue of .
So it is z with a pvalue of , so
Now, find the margin of error M as such
The margin of error is 14.11 pounds.
c. Using my sample of 30, what would be the 90% confidence interval for the population mean?
Lower bound: Sample mean subtracted by the margin of error.
247 - 14.11 = 232.89 pounds
Upper bound
247 + 14.11 = 261.11 pounds
The 90% confidence interval for the population mean is between 232.89 pounds and 261.11 pounds
When we factor out something, we divide the factor by the equation. For example if we were to factor 3 out of 6, It will be 3(2). We got this answer by dividing 3 and 6. Using this concept:
![\frac{-1}{4}[( \frac{-1}{2} / \frac{-1}{4}) - ( \frac{5}{4} y / \frac{-1}{4})]](https://tex.z-dn.net/?f=%20%5Cfrac%7B-1%7D%7B4%7D%5B%28%20%5Cfrac%7B-1%7D%7B2%7D%20%2F%20%5Cfrac%7B-1%7D%7B4%7D%29%20-%20%28%20%5Cfrac%7B5%7D%7B4%7D%20y%20%2F%20%20%5Cfrac%7B-1%7D%7B4%7D%29%5D%20%20)
Simplifying:
Hope this helps!
Simplifying:
Hope this helps!