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Rate (Per Day) Frequency Below .100

Rate (per day) Frequency
Below .100 12
.100-below .150 20
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.200-below .250 15
.250 or more 13
: An article, "A probabilistic Analysis of Dissolved Oxygen-Biochemical Oxygen Demand Relationship in Streams," reports data on the rate of oxygenation in streams at 20 degrees Celsius in a certain region. The sample mean and standard deviation were computed as; xbar = .173 and Sx = .066 respectively. Based on the accompanying frequency distribution (on the left), can it be concluded that the oxygenation rate is normally distributed variable. Conduct a chi-square test at alpha = .05

a. State the null and alternate hypothesis of the test

b. Briefly described the approach you need to use to calculate expected values to perform the Chi-Square contrast

c. What is the conclusion, do you reject or accept the null (also be sure to address the questions on the Answer Sheet as well)

1 Answer

2 votes

Final answer:

The null hypothesis (H0) states that the oxygenation rate is normally distributed, while the alternate hypothesis (H1) states it is not. Calculating expected values involves integrating the normal distribution for each rate bin. To conclude, compare the chi-square test statistic with the critical value to determine whether to reject or accept H0.

Step-by-step explanation:

Chi-square Test for Normality

To answer the question regarding the normality of the oxygenation rate, one must perform a chi-square goodness-of-fit test. The approach to calculating the expected frequencies for the oxygenation rate involves using the sample mean and standard deviation in combination with the properties of the normal distribution.

a. Null and Alternate Hypotheses

The null hypothesis (H0) would be that the oxygenation rate is normally distributed. The alternate hypothesis (H1) would claim otherwise.

b. Expected Values Calculation Approach

To calculate the expected values, we integrate the normal distribution (with the given mean and standard deviation) over each bin's range to find the probability of a rate falling into that range. Then, multiply the sample size by each probability to get the expected frequency for each bin.

c. Conclusion

To conclude whether to reject or accept the null hypothesis, compare the calculated chi-square test statistic to the critical value from the chi-square distribution table based on the degrees of freedom and significance level. If the test statistic exceeds the critical value, we reject H0, otherwise, we fail to reject H0.

User Otto Nascarella
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