Question

as part of your work for an environmental group, you want to see if the mean amount of waste generated per adult in your community is less than the national average of 5 pounds per day. you take a simple random sample of 27 adults in your community and find that they average 4.2 pounds with a standard deviation of 1.2 pounds. suppose you know the amount of waste generated per day follows a normal distribution. test at 0.01 significance. round answers to 4 decimal places.
a. ( h_0 ): select an answer?
b. ( h_1 ): select an answer?
c. test statistic:
d. p - value:
e. select the decision rule: select an answer
f. there select an answer enough evidence to conclude select an answer

Explanation:

Step1: State the hypotheses

The null hypothesis \(H_{0}\) is that the mean amount of waste generated per adult in the community is equal to the national average. The alternative hypothesis \(H_{1}\) is that the mean amount of waste generated per adult in the community is less than the national average.
\(H_{0}:\mu = 5\)
\(H_{1}:\mu<5\)

Step2: Calculate the test statistic

The formula for the \(t -\)test statistic in a one - sample \(t -\)test is \(t=\frac{\bar{x}-\mu}{s/\sqrt{n}}\)
where \(\bar{x} = 4.2\), \(\mu = 5\), \(s = 1.2\), and \(n = 27\)

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Step3: Calculate the degrees of freedom and the p - value

The degrees of freedom \(df=n - 1=27-1 = 26\)
For a one - tailed \(t -\)test with \(df = 26\) and \(t=-3.4641\), using a \(t -\)distribution table or a calculator (e.g., in R: pt(-3.4641,26)), the p - value is \(P(T< - 3.4641)\)
The p - value \(p\approx0.0009\)

Step4: Decision rule

The significance level \(\alpha = 0.01\). The decision rule is: Reject \(H_{0}\) if \(p - value<\alpha\)

Step5: Make a decision

Since \(p=0.0009<0.01=\alpha\), we reject the null hypothesis.

Answer:

a. \(H_{0}:\mu = 5\)
b. \(H_{1}:\mu<5\)
c. Test Statistic: \(-3.4641\)
d. P - value: \(0.0009\)
e. Decision Rule: Reject \(H_{0}\) if \(p - value<0.01\)
f. There is enough evidence to conclude that the mean amount of waste generated per adult in the community is less than the national average of 5 pounds per day.