Using the z-distribution, it is found that the 90% confidence interval for mean calories in a 30-gram serving of all chocolate chip cookies is (143, 149).
We are given the <em>standard deviation</em> for the population, which is why the <em>z-distribution </em>is used to solve this question.
The information given is:
- Sample mean of .
- Population standard deviation of .
- Sample size of .
The confidence interval is:
The critical value, using a z-distribution calculator, for a <u>95% confidence interval</u> is z = 1.645, hence:
The 90% confidence interval for mean calories in a 30-gram serving of all chocolate chip cookies is (143, 149).
A similar problem is given at brainly.com/question/16807970
Hey there!
In order to compare these fractions, we can give them a common denominator.
Our least common multiple of 3 and 8 is 24. We multiply 2/3 by 8/8 and 5/8 by 3/3 to get
16/24 and 15/24
Therefore, 2/3 is bigger.
Hope this helps!
now, le'ts do the summation using 0.5n+2
(0.5(2.7) + 2)+(0.5(3.0) + 2)+(0.5(3.2) + 2)+(0.5(3.7) + 2)+(0.5(4.4) + 2)+(0.5(4.9) + 2)+(0.5(5.3) + 2)
which gives us 27.6. Now, the paper doesn't state, or at least I don't see it, but that's just the sum of the hours, to get the average we simply divide that by 7, the amount of items, and 27.6 ÷ 7 gives us about 3.94.
does it underestimate or overestimate the one found with the summation? well, the summation gave us 3.9 and 3.94 is just 0.04 or 4 hundredths above it.