Question

The building specifications in a certain city require that the sewer pipe used in residential areas have a mean breaking strength of more than 2500 pounds per lineal foot. A manufacturer who would like to supply the city with sewer pipe has submitted a bid and provided the following information: An independent contractor randomly selected seven sections of the manufacturer’s pipe and tested each for breaking strength. The results (pounds per lineal foot) are as follows: 2610, 2750, 2420, 2510, 2540, 2490, and 2680.

1. What type of test should be conducted?

Dependent t test

Independent t test

One sample t test

2. State the null hypothesis in equation format.

3. State the alternative hypothesis in equation format.

4. What is the calculated t-value (to 3 significant digits)?

5. What is the critical t-value (to 3 significant digits)? Use alpha = 0.05.

6. Is the null hypothesis accepted or rejected? Use alpha = 0.05.

7. Is there sufficient evidence to conclude that there is a difference between the mean rate of increase of total phosphorus of the control algal and the water hyacinth? Use alpha = 0.05. Explain in one sentence.

Answer 1

Solution:-

1) One sample t-test.

2)

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

Null hypothesis: u < 2500
Alternative hypothesis: u > 2500

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

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

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = s / sqrt(n)

S.E = 40.693
DF = n - 1

D.F = 7
4)

t = (x - u) / SE

t = -1.76

5)

t_critical = 1.895

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 t statistic test statistic of - 1.76.

Thus the P-value in this analysis is 0.064.

Interpret results. Since the P-value (0.064) is greater than the significance level (0.05), we cannot reject the null hypothesis.

Do not reject the H₀.

From the above test we do not have sufficient evidence in the favor of the claim that the mean breaking strength is more than 2500 pounds.