Question

Many hotels have begun a conservation program that encourages guests to re-use towels rather than have them washed on a daily basis. A recent study examined whether one method of encouragement might work better than another. Different signs explaining the conservation program were placed in the bathrooms of the hotel rooms, with random assignment determining which rooms received which sign. One sign mentioned the importance of environmental protection, whereas another sign claimed that 75% of the hotel’s guests choose to participate in the program. The researchers suspected that the latter sign, by appealing to a social norm, would produce a higher proportion of hotel guests who agree to re-use their towels. Researchers used the hotel staff (a mid-sized, mid-priced hotel in the Southwest that was part of a well-known national hotel chain) to record whether guests staying for multiple nights agreed to reuse their towel after the first night.

(a) Identify the observational units, explanatory variable, and response variable in this study.

(b) State the null and alternative hypotheses in symbols, and be sure to define the parameter in the context of this study.

The following table displays the observed data in this study:

	Social Norm	Environmental protection	Total
Guest opted to re-use towels	98	74	172
Guest did not opt to re-use towels	124	137	261
Total	222	211	433

(c) Calculate the conditional proportions of re-use in each group.

(e) Use a two-sample z-test to test the hypotheses that you stated in (a). Report the test statistic and p-

value.

(f) Report your test decision at the α = 0.10, 0.05, and 0.01 significance levels. Also summarize what

these test decisions reveal about the strength of evidence for the researchers’ conjecture.

(g) Produce and interpret a 90% confidence interval for the difference in probabilities of re-using towels

between these two signs.

Answer 1

Given that,
sample one, x1 =98, n1 =222, p1= x1/n1=0.441
sample two, x2 =74, n2 =211, p2= x2/n2=0.351
finding a p^ value for proportion p^=(x1 + x2 ) / (n1+n2)
p^=0.397
q^ Value For Proportion= 1-p^=0.603
null, Ho: p1 = p2
alternate, produce a higher proportion of hotel guests who agree to re-use their towels, H1: p1 > p2
we use test statistic (z) = (p1-p2)/√(p^q^(1/n1+1/n2))
zo =(0.441-0.351)/sqrt((0.397*0.603(1/222+1/211))
zo =1.929
| zo | =1.929
critical value
the value of |z α| at los 0.05% is 1.645
we got |zo| =1.929 & | z α | =1.645
p-value: right tail - Ha : ( p > 1.9285 ) = 0.02689
hence value of p0.05 > 0.02689,here we reject Ho
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(a) study of examined whether one method of encouragement might work better than another
(b) null, Ho: p1 = p2 , alternate, produce a higher proportion of hotel guests who agree to re-use their towel H1: p1 > p2,

proportion of hotel guests who agree to re-use their towels by following social norm sign,
(c) p1 (social norm) = 0.441, p2 (environmental protection) = 0.351
(d & e) test statistic: 1.929, p-value: 0.02689
(f) level of significance, α = 0.05
from standard normal table,right tailed z α/2 =1.645
since our test is right-tailed
reject Ho, if zo > 1.645
make decision
hence value of | zo | > | z α| and here we reject Ho
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level of significance, α = 0.1
from standard normal table,right tailed z α/2 =1.282
since our test is right-tailed
reject Ho, if zo > 1.282
critical value
the value of |z α| at los 0.1% is 1.282
hence value of | zo | > | z α| and here we reject Ho
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level of significance, α = 0.01
from standard normal table,right tailed z α/2 =2.326
since our test is right-tailed
reject Ho, if zo > 2.326
make decision
hence value of |zo | < | z α | and here we do not reject Ho
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(g)
CI = (p1-p2) ± sqrt( p1 * (1-p1)/n1 + p2 * (1-p2)/n2 )
where,
p1, p2 = proportion of both sample observation
n1,n2 = size of both group
a = 1 - (confidence Level/100)
Za/2 = Z-table value
CI = confidence interval
CI = [ (0.4414-0.3507) ± 2.58 * 0.0468]
= [ -0.03 , 0.2115 ]