Question 1193195: My apologies, I did not submit the whole problem. I could really use any assistance. Thank you. I appreciate the help. I believe I should be using a t-score but how when I do not know how to get the population mean.
A nutrition expert claims that the average American is overweight. To test his claim, a random sample of 26 Americans was selected, and the difference between each person's actual weight and idea weight was calculated. For this data, we have x¯=15.3 and s=27.5. Is there sufficient evidence to conclude that the expert's claim is true? Carry out a hypothesis test at a 2% significance level. Hint: Pay attention to the variable being measured: the difference between actual weight and ideal. A positive value indicates that the person is overweight.
A. Give the value of the standardized test statistic (give to 3 decimal places):
B. The p-value is (give to 4 decimal places):
Found 2 solutions by math_tutor2020, Theo:
Answer by math_tutor2020(3816)   (Show Source):
You can put this solution on YOUR website!
Part A
Hypothesis:
H0: mu = 0
H1: mu > 0
The mu refers to the population mean of the differences in weight (actual - ideal)
The claim is that mu > 0 to indicate the average American is overweight. In other words, the claim is that the actual weight exceeds the ideal weight on average.
The inequality sign in the alternative hypothesis tells us we're doing a right-tailed test.
As with any hypothesis test, we are testing the null. If the p-value is smaller than alpha, then we reject the null. Otherwise, we fail to reject it.
Determining the p-value requires the test statistic.
First let's visit the given info
n = 26 = sample size
xbar = 15.3 = sample mean
s = 27.5 = sample standard deviation
alpha = 0.02 = 2% significance level
Now to compute the test statistic.
t = (xbar - mu)/( s/sqrt(n) )
t = (15.3 - 0)/( 27.5/sqrt(26) )
t = 2.83690903847162 approximately
t = 2.837
Answer: 2.837
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Part B
I recommend a calculator for this.
You can use a TI (texas instrument) calculator if you have one handy.
If you have a TI calculator, then refer to this documentation page
http://tibasicdev.wikidot.com/tcdf
The tcdf function will give the area under the T curve. The format is
tcdf(lower, upper, v)
where v = degrees of freedom
The instructions on how to access the tcdf function are provided on the page under where it mentions "Menu Location".
2ND DISTR to access the distribution menu
5 to select tcdf(, or use arrows.
So we could say something like
tcdf(2.837, 99, 25)
Notice how n = 26 leads to v = n-1 = 26-1 = 25
Also, recall we're doing a right-tailed test.
Typing that into the calculator would produce roughly 0.00445028 which rounds to 0.0045
If you do not have a TI calculator, then you can go for a free option such as this page
https://stattrek.com/online-calculator/t-distribution.aspx
Type in 25 as the degrees of freedom and 2.837 as the t score. Leave the last box blank. Then hit the "calculate" button.
The value 0.9955 should show up in that box we left empty.
This is the approximate area under the T curve to the left of t = 2.837 due to the notation P(T < 2.837), and this applies only when v = 25
So the area to the right would be 1-0.9955 = 0.0045
The p-value is smaller than alpha, so we reject the null. It appears the nutritionist's claim is correct.
Answer: 0.0045
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! the test statistic is the difference between the person's actual weight and the ideal weight.
if the difference is positive, the person is overweight.
if the difference is negative, the person is underweight.
if the difference is zero, the person is at the ideal weight.
from the data, it appears the average person is overweight by an average of 15.3 pounds.
for this test, the population mean is assumed to be the ideal weight = 0.
sample size = 26
sample mean = 15.3
sample standard deviation = 27.5
population mean is assumed to be 0.
the standard error is equal to the sample standard deviation divided by the square root of the sample size = 27.5 / sqrt(26) = 5.393194 rounded to 6 decimal places.
the formula is t = (x - m) / s.
t is the t-score
x is he sample mean difference from the ideal.
m is the assumed population difference from the ideal.
s is the standard error.
it becomes:
t = (15.3 - 0) / 5.393194 = 2.836909 rounded to 6 decimal places.
the degrees of freedom = sample size minus 1 = 25.
area to the right of a t-score of 2.836909 with 25 degrees of freedom = .00451rounded to 6 decimal places.
area to the right of that t-score = .004451 rounded to 6 decimal places.
critical t-score with 25 degrees of freedom at 2% one-tail significance level equals 2.166587 rounded to 6 decimal places.
the critical t-score is less than the sample t-score, indicating the assumption that the average american is overweight is accepted.
this is also supported by the p-score because the test p-score = .004451 and the critical p-score = .02.
the test p-score is less than the critical p-score, indicating the results are significant as the .02 significance level.
here's what it looks like on a graph.
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