Chapter 5 Lecture Notes (5.7) One-Sample Inference body of water available for S

Question

Chapter 5 Lecture Notes (5.7) One-Sample Inference body of water available for STAT-4013 Statistical MethodsI Example: The amount of sewage and industrial pollutants dumped into a affects the health of the water by reducing the amount of dissolved oxygen a aquatic life. Over a 2-month period, 15 measurements were taken from a rive oxyg ro sowage tratment plant The amount of ded h en in the 15 water specimens yielded a mean of 4.88 parts per million ( 0.22 ppm. The current research asserts that the mean dissolved oxygen level must ppm for fish to survive. Is there significant evidence tha differs from 5.0 ppm? ppm) and standard deviation 5.0 bet t the mean dissolved oxygen content

Explanation / Answer

Given that,
population mean(u)=5
sample mean, x =4.88
standard deviation, s =0.22
number (n)=15
null, Ho: =5
alternate, H1: !=5
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.14
since our test is two-tailed
reject Ho, if to < -2.14 OR if to > 2.14
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =4.88-5/(0.22/sqrt(15))
to =-2.113
| to | =2.113
critical value
the value of |t | with n-1 = 14 d.f is 2.14
we got |to| =2.113 & | t | =2.14
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -2.1125 ) = 0.0531
hence value of p0.05 < 0.0531,here we do not reject Ho
ANSWERS
---------------
null, Ho: =5
alternate, H1: !=5
test statistic: -2.113
critical value: -2.14 , 2.14
decision: do not reject Ho
p-value: 0.0531

we do not have enough evidence to support the claim that mean dissolved oxygen content differs from 5 ppm

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