Puppy Food Brand Comparison: ANOVA Weight Gain Analysis
School
University of California, Davis**We aren't endorsed by this school
Course
STA 100
Subject
Statistics
Date
Dec 11, 2024
Pages
2
Uploaded by SuperHumanRainHawk25
STA 100 Discussion 09 SolutionsHandwritten Problems1.Two new brands of puppy food (brands A and B) were compared to a standard brand (brand C). A random sample of75 puppies (25 in each group) was given each brand of food over a period of one month, and the weight gain (in pounds)was measured for each puppy. The results are:BrandABC¯yi10.510.68.2si2.12.24.8ni252525Test whether the average weight gain for the brands of food differ.(a) Find the value ofMS(between).•Solution: First,¯¯y=QQQQQQQIi=1ni¯yin.=262.5+265+20575= 9.77SS(between) =QQQQQQQIi=1ni(¯yi−¯¯y)2=(25)(10.5−9.77)2+ (25)(10.6−9.77)2+ (25)(8.2−9.77)2= 13.3225 + 17.2225 + 61.6225 = 92.1675MS(between) =SS(between)df(between)=92.16752= 46.08(b) Find the value ofMS(within).•Solution:SS(within) =QQQQQQQIi=1(ni−1)s2i= (25−1)(2.1)2+ (25−1)(2.2)2+ (24)(4.8)2= 774.96MS(between) =SS(within)df(within)=774.9672= 10.76(c) State the null and alternative hypotheses, calculate the test-statistic, and find the p-value.•Solution: Mathematically:H0:µ1=µ2=µ3HA:At least oneµidiffersOr you could say in words that:H0: The average weight gain of puppies across all brands of food is the same.HA: The average weight gain of puppies differs for at least one brand of food.FS=MS(between)MS(W ithin)=46.0837510.7633333= 4.2816.At numerator d.f. = 2, and denominator d.f. = 72 (rounded down to 60), we findP{F >4.18}= 0.02, andPr{F >4.98}= 0.01. Thus, our p-value is:0.01<p-value <0.02.1
(d) State the decision and conclusion in terms of the problem, assumingα= 0.05.•Solution: We reject the null, sincep−value <0.05. We conclude that the average weight gain for puppiesdiffers for at least one of the brands of food.(e) What assumption/s of ANOVA may be violated in this example?•Solution: There is a real danger that we are violating the assumption of equal variances. Notice that for BrandC we haes23= 4.82= 23.04which is more than 5 times the variance of Brand As21= 2.12= 4.41. Also, we donot have histograms or boxplots of our data in this problem so it is unclear if we satisfy the assumption of anormality, but weight gain is probably normally distributed.2