HypTest Mu - Wastewater discharge A wastewater treatment plant must monitor its

Question

HypTest Mu - Wastewater discharge A wastewater treatment plant must monitor its discharge continuously in order to reduce the chances of a significant environmental event. The Environmental Protection Agency (EPA) has certified a plant for a discharge level of 1552 gallons per hour. On a randomly selected day, discharge rates were sampled at 6 different times. This data - rounded to the nearest integer - is reported in Table 1. Note: assume that discharge rate is a variable that is appx. normally distributed. This condition is essential if this "small sample" estimation procedure is to generate valid inferences. Table 1: Wastewater discharge rates (gal/hr) 1561 1568 1543 1571 1547 1559 Assuming that this sample is representative of the entire population or discharge rates, an EPA official would like to test the hypothesis that discharge rates, on average, do not differ from the certified level of 1552 gallons/hour. Let = 0.02 A)The appropriate null/alternative hypothesis pair for this study is: Ho: < 1552; Ha: 1552 Ho: < 1552; Ha: > 1552 Ho: 1552; Ha: 1552 Ho: x = 1552; Ha: x > 1552 Ho: 1552; Ha: < 1552 Ho: = 1552; Ha: > 1552 Ho: = 1552; Ha: 1552 Ho: x 1552; Ha: x > 1552 Ho: 1552; Ha: > 1552 Note: you should carry at least 5 decimal precision for any intermediate calculations then round your answer as indicated in the problem B) The point estimate for the true average wastewater discharge rate for this treatment plant is: gal/hr Note: round your answer to the nearest hundredth C) The point estimate for the true standard deviation for the wastewater discharge rate of this treatment plant is: Note: round your answer to the nearest hundredth D) The test statistic for this data set is: Note: Round your answer to the nearest hundredth. E) The statistical decision and corresponding English interpretation are: FTR Ho: we cannot conclude that waste water discharge rates are, on average, different than 1552 gal/hr. Reject Ho: conclude that waste water discharge rates are, on average, different than 1552 gal/hr. In fact, we can conclude that the discharge rate is more than 1552 gal/hr FTR Ho: we can conclude that waste water discharge rates are, on average, different than 1552 gal/hr. In fact, we can conclude that the discharge rate is more than 1552 gal/hr Reject Ho: we can conclude that waste water discharge rates are, on average, different than 1552 gal/hr. However, we cannot say if the discharge rate is more or less than 1552 gal/hr Reject Ho: we can conclude that waste water discharge rates are, on average, different than 1552 gal/hr. In fact, we can conclude that the discharge rate is less than 1552 gal/hr Reject Ha: we can conclude that waste water discharge rates are, on average, different than 1552 gal/hr. In fact, we can conclude that the discharge rate is more than 1552 gal/hr FTR Ho: we can conclude that waste water discharge rates are, on average, different than 1552 gal/hr. In fact, we can conclude that the discharge rate is less than 1552 gal/hr FTR Ho: we cannot conclude that waste water discharge rates are, on average, different than 1552 gal/hr. In fact, we can conclude that the discharge rate is more than 1552 gal/hr Note: It might be helpful to calculate the p-value (using the calculator or some computer program) or to find the range in which the p-value lies (using the appropriate table) and THEN answer part E.

Explanation / Answer

Given that,
population mean(u)=1552
sample mean, x =1558.1667
standard deviation, s =11.1788
number (n)=6
null, Ho: =1552
alternate, H1: !=1552
level of significance, = 0.02
from standard normal table, two tailed t /2 =3.365
since our test is two-tailed
reject Ho, if to < -3.365 OR if to > 3.365
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =1558.1667-1552/(11.1788/sqrt(6))
to =1.351
| to | =1.351
critical value
the value of |t | with n-1 = 5 d.f is 3.365
we got |to| =1.351 & | t | =3.365
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != 1.3512 ) = 0.2345
hence value of p0.02 < 0.2345,here we do not reject Ho
ANSWERS
---------------
null, Ho: =1552
alternate, H1: !=1552
test statistic: 1.351
critical value: -3.365 , 3.365
decision: do not reject Ho
p-value: 0.2345

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