Solve percent problems (if anyone has done this test can you please screen shot it or just type the answers) Thank you very much
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Answer:
We conclude that not more than 500 gallons of wastewater per hour, on average, is discharged into a neighboring lake.
Step-by-step explanation:
We are given that an industrial plant can discharge not more than 500 gallons of wastewater per hour, on average, into a neighboring lake.
Four one-hour periods are selected randomly over a period of one week. The following are observed:
1384, 683, 1534, 405
Let = <u><em>population average gallons of wastewater discharged per hour</em></u>
So, Null Hypothesis, :
500 gallons {means that not more than 500 gallons of wastewater per hour, on the average, is discharged into a neighboring lake}
Alternate Hypothesis, :
> 500 gallons {means that more than 500 gallons of wastewater per hour, on the average, is discharged into a neighboring lake}
The test statistics that will be used here is <u>One-sample t-test statistics</u> because we don't know about population standard deviation;
T.S. = ~
where, = sample mean =
= 1001.5 gallons
s = sample standard deviation = = 543.79
n = sample of periods = 4
So, <u><em>the test statistics</em></u> = ~
= 1.844
The value of t-test statistics is 1.844.
Since in the question we are not given with the level of significance so we assume it to be 5%. Now, at 0.05 level of significance, the t table gives a critical value of 2.353 at 3 degrees of freedom for the right-tailed test.
Since the value of our test statistics is less than the critical value of z as 1.844 < 2.353, so we have <u><em>insufficient evidence to reject our null hypothesis</em></u> as it will fall in the rejection region.
Therefore, we conclude that not more than 500 gallons of wastewater per hour, on average, is discharged into a neighboring lake.