Question

A study compared the cost of restaurant meals when people pay individually versus splitting the bill as a group. In the experiment half of the people were told they they would each be responsible for individual meals costs and the other half were told the cost would be split equally among the six people at the table. The 24 people paying individually had a mean cost of 37.29 Israeli shekels with a standard deviation of 12.54, while the 24 people splitting the bill had a higher mean cost of 50.92 Israeli shekels with a standard deviation of 14.33. The raw data can be found in SplitBill and both distributions are reasonably bell-shaped. Use this information to find a 99% confidence interval for the difference in mean meal cost between these two situations.

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Round your answers to two decimal places.

The 99% confidence interval is

Answer 1

given that,
mean(x)=37.29
standard deviation , σ1 =12.54
population size(n1)=24
y(mean)=50.92
standard deviation, σ2 =14.33
population size(n2)=24
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((157.2516/24)+(205.3489/24))
= 3.8869
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, α = 0.01
from standard normal table, two tailed z α/2 =2.576
since our test is two-tailed
value of z table is 2.576
margin of error = 2.576 * 3.8869
= 10.0128
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (37.29-50.92) ± 10.0128 ]
= [-23.6428 , -3.6172]
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