Question

John works in the local waste station and notes that on any given day 1500 tons of waste, with a standard deviation of 500 tons is being received in these types of facilities across the city. John however, has become more quantitatively aware and finds that on average at his tip that he receives 1700 tons of waste per day. John would like to know if the amount of waste he receives is significantly more than the other tips.

a) To test Johns question what test would you use and why? Justify your answer.

b) What level of significance would be applicable for this test and why?

c) Conduct the hypothesis test using the critical value approach.

d) Construct a table using a significance level of 0.5, 1.0, 2.5, 5 and 10% and investigate how the conclusion would change as a result of changing the significance value.

（plz show all working）

Answer 1

Solution:-

a)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u < 1500
Alternative hypothesis: u > 1500

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small.

b)

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample z-test.

Analyze sample data. Using sample data, we compute the standard error (SE), z statistic test statistic (z).

SE = s / sqrt(n)

S.E = 500
z = (x - u) / SE

z = 0.40

c)

z_critical = + 1.645

Rejection region is z > + 1.645

where s is the standard deviation of the sample, x is the sample mean, u is the hypothesized population mean, and n is the sample size.

The observed sample mean produced a z statistic test statistic of + 1.645.

Interpret results. Since the z-value (0.40) does not lies in the rejection region, hence we cannot reject the null hypothesis.

d)

Significance level	z-critical	Decision
0.5%	+ 2.576	Do not reject H0
1.0%	+ 2.327	Do not reject H0
2.5%	+ 1.96	Do not reject H0
5%	+ 1.645	Do not reject H0
10%	+ 1.282	Do not reject H0