Question

In previous years, the average number of sheets recycled per bin was 59.3 sheets, but they believe this number may have increase with the greater awareness of recycling around campus. They count through 79 randomly selected bins from the many recycle paper bins that are emptied every month and find that the average number of sheets of paper in the bins is 62.4 sheets. They also find that the standard deviation of their sample is 9.86 sheets.
What is the value of the test-statistic for this scenario? Round your answer to 3 decimal places.

What are the degrees of freedom for this t-test?

Answer 1

Answer:

There is enough evidence to support the claim that the average number of sheets recycled per bin was more than 59.3 sheets.          

Step-by-step explanation:

We are given the following in the question:  

Population mean, μ = 59.3

Sample mean, $\bar{x}$ = 62.4

Sample size, n = 79

Alpha, α = 0.05

Sample standard deviation, s = 9.86

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 59.3\text{ sheets}\\H_A: \mu > 59.3\text{ sheets}$

We use one-tailed t test to perform this hypothesis.

Formula:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }$

Putting all the values, we have

$t_{stat} = \displaystyle\frac{62.4 - 59.3}{\frac{9.86}{\sqrt{79}} } = 2.7945$

Degree of freedom = n - 1 = 78

Now, $t_{critical} \text{ at 0.05 level of significance, 78 degree of freedom } = 1.6646$

Since,                        

$t_{stat} > t_{critical}$

We fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.

Conclusion:

Thus, there is enough evidence to support the claim that the average number of sheets recycled per bin was more than 59.3 sheets.