Nutritional information provided by Kentucky Fried Chicken (KFC) claims that eac
ID: 3150203 • Letter: N
Question
Nutritional information provided by Kentucky Fried Chicken (KFC) claims that each small bag of potato wedges contains 4.8 ounces of food and 280 calories. A sample of ten orders from KFC restaurants in New York and New Jersey averaged 358 calories. Assume that the sample is approximately normally distributed. (a) If the sample deviation was s = 54 calories, is there sufficient evidence to indicate that the average number of calories in small bags of KFC potato wedges is greater than advertised? Clearly formulate the null and the alternative hypotheses, select an appropriate test and test at the 1% level of significance. (b) Compute the p-value of the test conducted in part (a). (c) Construct a 99% lower confidence interval for the true mean number of calories in small bags of KFC potato wedges. (d) On the basis of the interval computed in part (c), what would you conclude about the claim that the mean number of calories exceeds 280? How does your conclusion here compare with your conclusion in part (a) where you conducted a formal test of hypothesis?Explanation / Answer
a)
Formulating the null and alternative hypotheses,
Ho: u <= 280
Ha: u > 280
As we can see, this is a right tailed test.
Thus, getting the critical t,
df = n - 1 = 9
tcrit = + 2.821437925
Getting the test statistic, as
X = sample mean = 358
uo = hypothesized mean = 280
n = sample size = 10
s = standard deviation = 54
Thus, t = (X - uo) * sqrt(n) / s = 4.567734398
Also, the p value is, as this is right tailed,
p = 0.000675629
As |t| > 2.821, and P < 0.01, we REJECT THE NULL HYPOTHESIS.
Hence, there is significant evidence that the average number of calories in small bags of KFC potato wedges is greater than advertised. [CONCLUSION]
**************************
b)
Also, the p value is, as this is right tailed,
p = 0.000675629 [ANSWER]
************************
c)
Note that
Lower Bound = X - t(alpha/2) * s / sqrt(n)
where
alpha = (1 - confidence level) = 0.01
X = sample mean = 358
t(alpha) = critical t for the confidence interval = 2.821437925
s = sample standard deviation = 54
n = sample size = 10
df = n - 1 = 9
Thus,
Lower bound = 309.8202814
Hence,
u > 309.8202814 [ANSWER]
******************************
d)
The mean is greater than 309.82 at 99% confidence, so, it significantly exceeds 280.
This is consistent with the test in part a.
********************************
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.