A Florida newspaper claims that the average number of confirmed Zika cases in sm

Question

A Florida newspaper claims that the average number of confirmed Zika cases in small cities in the Miami-Dade county exceeds 20. We want to test this claim. Provide the null and alternative hypotheses. A random sample of 6 small cities in the Miami-Dade County in Florida yield the following Data:

City Cases

Aventura 22

Doral 22

HIaleah 23

Medley 20

Pinecrest 32

Surfside 20

Describe the experiment, the uncertain quantity, and the possible values of the uncertain quantity for this problem. Compute the test statistic for the hypothesis test, determine the probability - value; and at 90% confidence, test the hypothese.

Explanation / Answer

Given that,
population mean(u)=20
sample mean, x =23.1667
standard deviation, s =4.4907
number (n)=6
null, Ho: =20
alternate, H1: >20
level of significance, = 0.05
from standard normal table,right tailed t /2 =2.015
since our test is right-tailed
reject Ho, if to > 2.015
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =23.1667-20/(4.4907/sqrt(6))
to =1.727
| to | =1.727
critical value
the value of |t | with n-1 = 5 d.f is 2.015
we got |to| =1.727 & | t | =2.015
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value :right tail - Ha : ( p > 1.7273 ) = 0.07235
hence value of p0.05 < 0.07235,here we do not reject Ho

ANSWERS
---------------
null, Ho: =20
alternate, H1: >20
test statistic: 1.727
critical value: 2.015
decision: do not reject Ho
p-value: 0.07235

2.
Confidence Interval
CI = x ± t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x)=23.1667
Standard deviation( sd )=4.4907
Sample Size(n)=6
Confidence Interval = [ 23.1667 ± t a/2 ( 4.4907/ Sqrt ( 6) ) ]
= [ 23.1667 - 2.015 * (1.833) , 23.1667 + 2.015 * (1.833) ]
= [ 19.473,26.861 ]

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