13. Using Excel - Estimating the difference between two population means You are

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13. Using Excel - Estimating the difference between two population means You are a marketing expert for a company that is producing a new kind of energy bar. You want to provide evidence that consuming the new type of bar 1 hour before engaging in strenuous exercise of long duration-such as running a marathon-will improve performance You decide to conduct a test at a half-marathon race. You give the bar to a randomly selected group of 33 participants (the treatment group). You randomly select a second group of 30 participants, to whom you do not give the energy bar (the control group) To answer the questions that follow, download an Excel® spreadsheet containing the race times in minutes for each of the two groups by clicking on the following words in bold: Download Excel File In the following sample Excel spreadsheet, enter the appropriate formulas to estimate the difference in mean times between the treatment and control groups (Average Time for Control group - Average Time for Treatment group). Note that only the first 20 rows are shown here, but all values are included in the downloaded Excel page. Use a 95% confidence level Note: The function T.INV.2T in Excel 2010 is equivalent to TINV for earlier versions of Excel. Similarly, the function STDEV.S in Excel 2010 is equivalent to STDEV for earlier versions of Excel Interval Estimate of Differences in Population Means: (Population standard deviations unknown) 1 Control Treatment 2 80.95 3 157.73132.57 4 139.41105.03 155.54 Control Treatment Sample Size Sample Mean Sample Standard Deviation 144 134.96 COUNT(A2:A31) =COUNT( B2 : B34) =AVERAGE(A2:A31) 6 109.84128.43 7 76.76 8 122.57134.06 9 103.42 146.98 10 137.75126.12 11169.91101.28 12 132.58122.17 13 107.14125.93 126.72 = STDEV(B2 : B34) =E7^2/E5 + F7^2/F5 =SQRT(E9) Estimate of Variance Standard Error Confidence Coefficient 0.95 Level of Significance 0.05

Explanation / Answer

TRADITIONAL METHOD
given that,
control table
mean(x)=131.6358
standard deviation , s.d1=27.3632
number(n1)=19
treatment table
y(mean)=130.2342
standard deviation, s.d2 =20.2495
number(n2)=19
I.
stanadard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
stanadard error = sqrt((748.745/19)+(410.042/19))
= 7.81
II.
margin of error = t a/2 * (stanadard error)
where,
t a/2 = t -table value
level of significance, =
from standard normal table, two tailedand
value of |t | with min (n1-1, n2-1) i.e 18 d.f is 2.101
margin of error = 2.101 * 7.81
= 16.408
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (131.6358-130.2342) ± 16.408 ]
= [-15.006 , 17.809]
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DIRECT METHOD
given that,
mean(x)=131.6358
standard deviation , s.d1=27.3632
sample size, n1=19
y(mean)=130.2342
standard deviation, s.d2 =20.2495
sample size,n2 =19
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
where,
x1,x2 = mean of populations
sd1,sd2 = standard deviations
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 131.6358-130.2342) ± t a/2 * sqrt((748.745/19)+(410.042/19)]
= [ (1.402) ± t a/2 * 7.81]
= [-15.006 , 17.809]
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interpretations:
1. we are 95% sure that the interval [-15.006 , 17.809] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population proportion

the average time of participants who received the bar was 131.6358 minutes. The average time of participants who did not receive
he bar was 130.2342 minutes. You can say with 95% confidence that runners who ate the energy bar ran this course, on average,
etween minutes and -15.006 , 17.809 minutes than those who did not eat the energy bar.

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