Q. 10.140 In the document "Peak Discharge ... [FREE SOLUTION]

Chapter 10: Q. 10.140 (page 444)

In the document "Peak Discharge from Unlogged and Logged Watersheds," J. Jones and G. Grant compiled (paired) data on peak discharge from storms in two watersheds one unlogged and one logged (\(100%\)clear cut). If there is an effect due to clear cutting one would expect that the runoff would be greater in the logged area than in the unlogged area. The runoffs, in cubic meters per second per square kilometer (\(m^{3}/s/km^{2})\), are provided on the WeissStats site. Use the technology of your choice to do the following.
a. Formulate the null and alternative hypothesis to reflect the expectation expressed above.
b. Perform the required hypothesis test at the \(1%\) significance level.
c. Obtain the interpret a \(99%\) confidence interval for the difference between mean runoffs in the logged and unlogged watersheds.
d. Construct a histogram of the sample data to identify the approximate shape of the paired-difference variable.
e. Based on your result from part (d) do you think that applying the paired \(t-\)procedure in part (b) and (c) is reasonable? Explain your answer.

Short Answer

Expert verified

Part a. The null and alternative hypothesis are \(H_{0}:\mu_{1}=\mu_{2}\) and \(H_{a}:\mu_{1}<\mu_{2}\)

Part b. The data provide sufficient evidence that the mean runoff in the logged area is greater than in the unlogged area at \(1%\) level.

Part c. The \(99%\) confidence interval for the difference between mean runoffs in the logged and unlogged watersheds lies between \(-0.169\) and \(-0.131\) cubic meters per second per square kilometer.

Part d. The histogram of sample is shown in Figure-(1).

Part e. No, the paired t-procedures in parts (b) and (c) are not reasonable.

Step by step solution

Part a. Step 1. Given information

If \(P-value \leq \alpha\), then reject the null hypothesis.

Part a. Step 2. Calculation

Population 1: Unlogged area.

Population 2: Logged area.

Null hypothesis:

\(H_{0}:\mu_{1}=\mu_{2}\)

That is the mean runoff in the logged area is not greater than in the unlogged area.

Alternative hypothesis:

\(H_{a}:\mu_{1}<\mu_{2}\)

That is the mean runoff in the logged area is greater than in the unlogged area.

Thus, the null and alternative hypothesis are \(H_{0}:\mu_{1}=\mu_{2}\) and \(H_{a}:\mu_{1}<\mu_{2}\).

Part b. Step 2. Calculation

Using MINITAB procedure:

From the MINITAB output is,

test statistics \(=-20.50\)

\(P-value=0.000\)

If \(P-value \leq \alpha\), then reject the null hypothesis.

\(P-value \leq \alpha\)

\(0.000 \leq 0.01\)

Here, the \(P-\)value is \(0.000\) which is lesser than the level of significance. Therefore, the null hypothesis is rejected at \(1%\) level significance.

Thus, the data provide sufficient evidence that the mean runoff in the logged area is greater than in the unlogged area at \(1%\) level.

Part c. Step 1. Calculation

Using MINITAB, the value of test statistic and \(p-\)value

From the MINITAB output is,

test statistics \(=-0.169\)

\(P-value=-0.131\)

Thus, the \(99%\) confidence interval for the difference between mean runoffs in the logged and unlogged watersheds lies between \(-0.169\) and \(-0.131\) cubic meters per second per square kilometer.

Part d. Step 1. Calculation

Using MINITAB, the value of test statistic and p-value

From the histogram, It is observed that the distribution of paired difference is skewed to left Moreover, the distribution of the data is not approximately normally distributed.

Thus, the histogram of sample is shown in Figure-(1).

Part e. Step 1. Calculation

No, the paired \(t-\)procedures in parts (b) and (c) are not reasonable because the distribution of data is highly skewed to left with outliers. But some statistician used paired (-procedures because the sample size is very large.

Thus, the No, the paired t-procedures in parts (b) and (c) are not reasonable.