Question

An industrial plant claims to discharge no more than 1000 gallons of wastewater per​ hour, on the​ average, into a neighboring lake. An environmental action group decides to monitor the​ plant, in case this limit is being exceeded. A random sample of four hours is selected over a period of a week. The observations​ (gallons of wastewater discharged per​ hour) are

1291​,2252,2652, 2344 Complete parts a through d below.

a. Find the sample​ mean,x overbar, standard​ deviation, s, and standard​ error, se.

b. to test H₀:To test Upper H₀: mu μ=1000 vs. Upper H Subscript Ha​: mu μ greater than >1000​,find the test statistic.

t= ( ) round to two decimal places.

c. Determine the​ P-value and decide if there is enough evidence to reject the null hypothesis at the 0.05 significance level.

the P-value is ( ) round to three decimal places

Answer 1

	Values ( X )	Σ ( Xi - X̅ )2
	1291	711914.0625
	2252	13747.5625
	2652	267547.5625
	2344	43785.5625
Total	8539	1036994.75

Mean X̅ = Σ Xi / n
X̅ = 8539 / 4 = 2134.75

Sample Standard deviation SX = √ ( (Xi - X̅ )2 / n - 1 )
SX = √ ( 1036994.75 / 4 -1 ) = 587.9327

Standard Error = S/√(n) = 293.9664

Part b)

Test Statistic :-
t = ( X̅ - µ ) / (S / √(n) )
t = ( 2134.75 - 1000 ) / ( 587.9327 / √(4) )
t = 3.86

Part c)

P - value = P ( t > 3.8601 ) = 0.015

Reject null hypothesis if P value < α = 0.05 level of significance
P - value = 0.0154 < 0.05, hence we reject null hypothesis
Conclusion :- Reject null hypothesis