The owners of Hamburger Hop are concerned that many of their customers are starting to purchase their burgers from Cheesy Chops because of its new burger, which has fewer grams of fat than Hamburger Hop. A random sample of 120 single burgers from Hamburger Hop found a mean of 32 grams of fat and a standard deviation of 2.4 grams. A random sample of 120 single burgers from Cheesy Chops found a mean of 29 grams of fat and a standard deviation of 1.8 grams. Which of the following formulas gives a 98% confidence interval for the difference in mean grams of fat between a Hamburger Hop single burger and a Cheesy Chop single burger?

A3 plus or minus 2.36 times square root of 2.4 over 120 plus 1.8 over 120

B 3 plus or minus 2.36 times square root of 2.4 squared over 120 plus 1.8 squared over 120

C 3 plus or minus 2.36 times square root of 2.4 squared over 119 plus 1.8 squared over 119

D 3 plus or minus 2.054 times square root of 2.4 squared over 119 plus 1.8 squared over 119

E3 plus or minus 2.054 times square root of 2.4 squared over 120 plus 1.8 squared over 120

Option B is correct.

Confidence interval = 3 plus or minus 2.36 times square root of 2.4 squared over 120 plus 1.8 squared over 120

Step-by-step explanation:

Hamburger hop = N₁ (32, 2.4²)

Cheesy chops = N₂ (29, 1.8²)

The combined distribution is the difference between the means of Hamburger hop and Cheesy Chops

N = (N₁ - N₂)

For two distributions (N₁ and N₂) with means p and q and variance x and y,

The new mean of a combined distribution (aN₁ + bN₂) is (ap + bq)

And the new variance = a²x + b²y

So, for our distribution,

N₁ (32, 2.4²)

N₂ (29, 1.8²)

The new mean of N = (N₁ - N₂) will be (32 - 29) = 3

The new variance of N = (N₁ - N₂) will be

(1²)(2.4²) + (-1)²(1.8²) = 2.4² + 1.8²

And the standard deviation = √[2.4² + 1.8²]

And the confidence interval is given as

Confidence interval = (Sample mean) ± (Margin of error)

Sample mean = 3

Margin of error = (critical value) × (standard deviation of the distribution of sample means)

Critical value for a 98% confidence interval is 2.36

(standard deviation of the distribution of sample means) = (standard deviation/√n)

n = sample size = 120

standard deviation = √[2.4² + 1.8²]

(standard deviation of the distribution of sample means) = {√[2.4² + 1.8²]/√120}

= √[(2.4² + 1.8²)/120]

Margin of error = 2.36 × √[(2.4² + 1.8²)/120]

Confidence interval = (Sample mean) ± (Margin of error)

Confidence interval = 3 ± {2.36 × √[(2.4² + 1.8²)/120]}

Confidence interval = 3 plus or minus 2.36 times square root of 2.4 squared over 120 plus 1.8 squared over 120

Hope this Helps!!!

The person who said B is wrong !!!

I just took the test and it was WRONG ! (dont gimme one star just cause this isnt an answer, im just tryna help)