Young people may feel they are carrying the weight of the world on their shoulders when in reality they are too often carrying an excessively heavy backpack. the article "effectiveness of a school-based backpack health promotion program" reported the following data for a smaple of 131 sixth graders: for backpack weight (lb) \bar{x} = 13.83, s = 5.05 ; for backpack weight as a percentage of body weight, a 95% ci for the population mean was (13.62; 15.89)a.) calculate and interpret a 99% ci for population mean backpack weight. b.) obtain a 99% ci for population mena weight as a percentage of body weight. c.) the american academy of orthopedic surgeons recommends that backpack weight be at most 10% of body weight. what does your calculation of part (b) suggest and why?

A.  [12.92;14.97]lb

B.  [13.25;16.26]%

C. check explanation

Step-by-step explanation:

Hello!

To properly resolve any statistical problem you need to first identify your study variables and summarize the given data.

A.

Study variable:

X: "backpack weight of a sixth-grader" (lb)

sample n= 131

sample mean: x[bar]= 13.83 lb

sample standard deviation: S= 5.05 lb

Now you need to calculate a CI for the population mean of the backpack weight of sixth graders (μ) with a confidence level of 99%

Since you need to estimate the population mean, and you have a sample large enough (n≥30), although you don't know the distribution of this population, you can approximate the distribution of the sample mean to normal (applying CLT) and use the statistic Z= (x[bar] - μ)/(S/√n) ≈ N(0;1)

The formula for the CI is

x[bar] ± * (S/√n)

The confidence level to use is 1-α = 0.99

α = 0.01 ⇒ α/2 = 0.005

⇒ = -2.58

Now you calculate the interval

13.83 ± 2.58 * 5.05/√131 ⇒ [12.92;14.97]lb

So with a confidence level of 99%, you'll expect that the real mean for the backpack weight of sixth graders is contained by the interval [12.92;14.97]lb

B.

It seems that for this item the information was measured from the same sample taken in item A.

Study variable

X: "Backpack weight as percentage of body weight" (lb)

Assuming is the same sample: n= 131

It was calculated the 95% CI [13.62;15.89]lb

And you need to calculate a 99% CI

For this item, you use the same statistic as before

x[bar] ± * (S/√n)

To calculate the asked interval you need to know the confidence level, the sample mean, sample standard deviation and sample size.

To know the sample mean and standard deviation for the backpack weight as percentage of body weight, you'll need to deduce it from the given interval.

Since this interval is for the population mean, it is constructed arround the sample mean. That means, that the sample mean is at the center of the interval.

x[bar] = (Li+Ls)/2 ⇒ x[bar] = (13.62+15.89)/2 = 14.76%

The standard deviation needs a little more calculation,

amplitude a= Ls - Li ⇒ a= 15.89-13.62= 2.27

semiamplitude d= a/2 ⇒ d=2.27/2= 1.135lb ≅ 1.14

⇒ = 1.96

d= * (S/√n) ⇒ 1.14= 1.96 * (S/√131)

⇒ S= 6.657% ≅ 6.66%

Now the new CI has a confidence level of 99%

The Z value you need to use is = 2.58

x[bar] ± * (S/√n) ⇒ 14.76 ± 2.58 * (6.66/√131)

⇒ [13.25;16.26]%

With a confidence level of 99% you can expect that the interval [13.25;16.26]% contains the population mean of the backpack weight as percentage of body weight of sixth graders.

C.

"The American Academy of Orthopedic Surgeons recommends that backpack weight is at most 10% of body weight." As you can see in the calculated interval [13.25;16.26]%, the weight carried by the sixth graders is above the recommended backpack weight as percentage of body weight. This means that the sixth graders carry more weight than recommended.

I hope you have a SUPER day!

391.6 is the correct answer