Question

The New York City Department of Health and Mental Hygiene conducts regular inspections of restaurants. Each restaurant receives an inspection score, with lower scores indicating a more satisfactory inspection. Following are the scores for the most recent inspection for random samples of 25 restaurants in the boroughs of Manhattan and Queens.

Manhattan: 2, 22, 23, 0, 40, 12, 37, 43, 27, 15, 24, 8, 38, 4, 17, 21, 11, 18, 13, 30, 27, 19, 38, 21, 4

Queens: 27, 18, 14, 20, 35, 8, 42, 29, 0, 12, 25, 0, 19, 6, 13, 22, 5, 11, 0, 10, 19, 39, 2, 8, 19

Using a significance level of 0.05, can you conclude a significant difference in mean inspection scores between the two boroughs?

Answer 1

Given that,
mean(x)=20.56
standard deviation , s.d1=12.165
number(n1)=25
y(mean)=16.12
standard deviation, s.d2 =11.7
number(n2)=25
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.064
since our test is two-tailed
reject Ho, if to < -2.064 OR if to > 2.064
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =20.56-16.12/sqrt((147.98723/25)+(136.89/25))
to =1.3153
| to | =1.3153
critical value
the value of |t α| with min (n1-1, n2-1) i.e 24 d.f is 2.064
we got |to| = 1.3153 & | t α | = 2.064
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 1.3153 ) = 0.201
hence value of p0.05 < 0.201,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 1.3153
critical value: -2.064 , 2.064
decision: do not reject Ho
p-value: 0.201
we do not have enough evidence to support the claim that significant difference in mean inspection scores between the two boroughs.