Question
A pollution-control inspector suspected that a riverside community was releasing semi treated sewage into a river and this, as a consequence, was changing the level of dissolved oxygen of the river. To check this, he drew 15 randomly selected specimens of river water at a location above the town, and another 15 specimens the town. For the specimens above the town the mean dissolved oxygen reading was 4.92 with a standard deviation of 0.157. For the specimens below the town, the mean dissolved oxygen reading was 4.74 with a standard deviation of 0.32. The lower limit of a 95% confidence interval for the mean differences of the two populations is:
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GPT 3.5
To calculate the lower limit of the 95% confidence interval for the mean difference of the two populations, we can use the formula:
Lower limit = (mean of sample 1 - mean of sample 2) - (critical value * standard error)
First, we need to calculate the critical value. Since the sample size is small and we do not know the population standard deviation, we can use the t-distribution.
The degrees of freedom (DF) is calculated as (n1 + n2 - 2), where n1 and n2 are the sample sizes. In this case, the sample size for each group is 15, so the DF = (15 + 15 - 2) = 28.
To find the critical value, we need to look up the t-score at the 95% confidence level and with 28 degrees of freedom. We can use a t-table or a calculator to find this value. Let's assume it is 2.048 (rounded for simplicity).
Next, we need to calculate the standard error. The formula for the standard error of the mean difference is:
Standard error = sqrt[(s1^2 / n1) + (s2^2 / n2)]
where s1 and s2 are the sample standard deviations, and n1 and n2 are the sample sizes.
Plugging in the values:
Standard error = sqrt[(0.157^2 / 15) + (0.32^2 / 15)]
Calculating this, we get:
Standard error ≈ 0.1009
Now, we can calculate the lower limit of the confidence interval:
Lower limit = (4.92 - 4.74) - (2.048 * 0.1009)
Calculating this, we get:
Lower limit ≈ 0.18
Therefore, the lower limit of the 95% confidence interval for the mean difference of the two populations is approximately 0.18.
Lower limit = (mean of sample 1 - mean of sample 2) - (critical value * standard error)
First, we need to calculate the critical value. Since the sample size is small and we do not know the population standard deviation, we can use the t-distribution.
The degrees of freedom (DF) is calculated as (n1 + n2 - 2), where n1 and n2 are the sample sizes. In this case, the sample size for each group is 15, so the DF = (15 + 15 - 2) = 28.
To find the critical value, we need to look up the t-score at the 95% confidence level and with 28 degrees of freedom. We can use a t-table or a calculator to find this value. Let's assume it is 2.048 (rounded for simplicity).
Next, we need to calculate the standard error. The formula for the standard error of the mean difference is:
Standard error = sqrt[(s1^2 / n1) + (s2^2 / n2)]
where s1 and s2 are the sample standard deviations, and n1 and n2 are the sample sizes.
Plugging in the values:
Standard error = sqrt[(0.157^2 / 15) + (0.32^2 / 15)]
Calculating this, we get:
Standard error ≈ 0.1009
Now, we can calculate the lower limit of the confidence interval:
Lower limit = (4.92 - 4.74) - (2.048 * 0.1009)
Calculating this, we get:
Lower limit ≈ 0.18
Therefore, the lower limit of the 95% confidence interval for the mean difference of the two populations is approximately 0.18.