1. You will need your ticker code (NFLX) ( show your ticker code on your homewor

Question

1. You will need your ticker code (NFLX) (show your ticker code on your homework and include a printout of the data!) for stock prices for this question. Use your ticker code to obtain the closing prices for the following three time periods to obtain three data sets:

March 2, 2016 to March 16, 2016                 

Data set A: 95.49, 96.23, 97.36, 97.61, 97.66, 97.86, 97.93, 98.00, 98.13, 99.35, 101.58

March 17, 2016 to March 31, 2016

Data set B: 98.36, 99.59, 99.72, 99.84, 101.06, 101.12, 101.21, 102.19, 102.23, 104.13

           

April 1, 2016 to April 17, 2016                     

Data set C: 102.68, 103.81, 104.35, 104.45, 104.83, 104.94, 105.70, 106.98, 109.65, 110.42, 11.51

To ensure equal sample size per group use only the first 8 observations in each data set. Add 0.1 and 0.2 to each one of the closing prices in data sets B and C, respectively

Imagine you are testing for the effects of two experimental drugs (data sets B and C) relative to a control group (Data set A) on a physiological variable.

a)Conduct an ANOVA on the data set. Show all calculations, table of results, and state your conclusion.

b)Use the Bonferroni-Holm (regardless of whether part “a” is significant or not) to examine all pairwise comparisons. Show all of your calculations and state your conclusions.

Explanation / Answer

(a) following information has been generated using ms-excel

null hypothesis H0: all means are equal

alternate hypothesis H1: atleast one mean is different to other

let level of significance alpha=0.05

since p-value of between treatments is less than alpha=0.05, so we fail to accept H0 and conclude that atleast one treatment is different to other

(b) there are three treatment and there will be k=3 pair of comparison

so bonferroni least significant difference=sqrt(2*mse/r)*t(alpha/k,error df)=sqrt(2*1.1831/8)*t(0.05/3,21)=

=sqrt(2*1.1831/8)*2.6=1.41

so all three pair of difference of means are more than lsd=1.41, so each pair is significant .

observation A B C Anova: Single Factor 1 95.49 98.36 102.68 2 96.23 99.59 103.81 SUMMARY 3 97.36 99.72 104.35 Groups Count Sum Average Variance 4 97.61 99.84 104.45 A 8 778.14 97.2675 0.834621 5 97.66 101.06 104.83 B 8 802.09 100.2613 1.09627 6 97.86 101.12 104.94 C 8 837.74 104.7175 1.618279 7 97.93 101.21 105.7 8 98 101.19 106.98 ANOVA Source of Variation SS df MS F P-value F crit Between treatment 224.8619 2 112.4309 95.03429 3E-11 3.4668 Within treatment(error) 24.84419 21 1.183057 Total 249.7061 23

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