Word Problem

Question

Jennifer suspected that her dog ate more than the average dog (of the same breed) did. While the average dog eats 900g of food a day (SD = 450), she monitored her dog for a month and found that her dog ate 1,100g of food a day (SD = 530). Can she conclude that her dog eats more than the average dog? Why?  

Answer 1

This is an ancient question. Please remove.

For the practice then:

Assume normal distribution (this is actually a bad assumption as one can not have negative intake and zero is only 2 sd from the mean. )

Step one: create test Hypothesis. H₀ : Daily food intake is ≤ 900g/day . One sided test so it includes null.

H_a: Daily food intake > 900g/day

Step two: As population parameters are known a Z statistic will be used. Calculate the Z statistic for for the test. Z = (x - μ)/σ when we are testing individuals instead of means. μ = 900g ; σ = 450g x = 1100 ; Z = (1100 - 900)/450 = .444

Step three: calculate inverse Z distribution (i.e. P(x>Z)) (note: multiple methods are available from tables, to hand calculators , to on line calculators, I used the calculator here {https://www.calculator.net/z-score-calculator.html?c1raw=1100&c1mean=900&c1sd=450&calctype=zscore&x=Calculate} ). P(x>Z) = .3284

Step four: Draw conclusion. As P(x>Z) is greater than .05 we can not reject the null hypothesis at the 95% confidence level.