Question

The Environmental Protection Agency monitors the concentration of potentially harmful bacteria such as E-Coli in bathing water at beaches in Ireland. It is considered unsafe to swim in water with an E-Coli concentration higher than 500MPN/100ml. Last year, the mean concentration from a sample of 10 measurements taken at a particular beach was 520MPN/100ml (sample standard deviation 51MPN/100ml ), so the beach was considered unsafe for swimming. This year, a sample of 12 measurements has been collected from the same beach, showing that the mean E-Coli concentration has dropped to 461MPN/100ml, with sample standard deviation 72MPN/100ml.

1. Is the average E-Coli concentration significantly different from the previous year? Use a significance level of 0.05 and compute the p-value.

Answer 1

In order to determine if the average E-Coli concentration is significantly different from the previous year, we will conduct a two-sample t-test using the given sample means, standard deviations, and sizes. If the resulting p-value is less than 0.05, we will reject the null hypothesis and conclude that there is a significant difference between the two years.

Step-by-step explanation:

In order to assess whether the average E-Coli concentration differs significantly from the preceding year, a two-sample t-test will be employed.

Initial steps involve formulating a null hypothesis (H0), positing no notable disparity in average concentrations between the two years, and an alternative hypothesis (Ha) suggesting a significant difference.

Employing a significance level of 0.05, the subsequent phase entails computing the p-value utilizing provided sample means, sample standard deviations, and sample sizes.

Should the resultant p-value fall below 0.05, the null hypothesis will be rejected, leading to the conclusion that a substantial difference exists in the average E-Coli concentrations between the two years.

This rigorous statistical approach serves as a methodical means to scrutinize and substantiate the significance of observed variations in E-Coli concentrations, providing a reliable basis for drawing conclusions about potential changes over the specified time frame.

Answer 2

The average E-Coli concentration is significantly different from the previous year at a significance level of 0.05. And The p-value is approximately 0.082.

Step-by-step explanation:

The question involves comparing the mean E-Coli concentration between two years. To determine if there is a significant difference, a two-sample t-test can be employed. The null hypothesis
`(\(H_0\))` is that there is no difference in the mean concentrations
`(\(\mu_1 = \mu_2\))`, and the alternative hypothesis
`(\(H_a\))` is that there is a significant difference
`(\(\mu_1 \\eq \mu_2\)`). A two-sample t-test formula is used to calculate the t-statistic:

`\[t = \frac{(\bar{X}_1 - \bar{X}_2)}{\sqrt{\left((s_1^2)/(n_1)\right) + \left((s_2^2)/(n_2)\right)}}\]`

where
`\(\bar{X}_1\)` and
`\(\bar{X}_2\)` are the sample means,
`\(s_1\)` and
`\(s_2\)` are the sample standard deviations, and
`\(n_1\)` and
`\(n_2\)` are the sample sizes for the two years.

After calculating the t-statistic, it is compared to the critical t-value at a significance level of 0.05 with degrees of freedom equal to
`\(n_1 + n_2 - 2\)`. If the absolute value of the t-statistic is greater than the critical t-value, the null hypothesis is rejected.

Certainly, let's calculate the p-value:

Given data:

Last year (Sample 1):

- Mean
`(\(\bar{x}_1\)):` 520 MPN/100ml

- Standard Deviation
`(\(s_1\))`: 51 MPN/100ml

- Sample Size
`(\(n_1\))(\(n_1\))`: 10

This year (Sample 2):

- Mean
`(\(\bar{x}_2\))`: 461 MPN/100ml

- Standard Deviation
`(\(s_2\)) (\(s_2\))`: 72 MPN/100ml

- Sample Size
`(\(n_2\))`: 12

Now, let's use the two-sample t-test formula:

`\[ t = \frac{{\bar{x}_1 - \bar{x}_2}}{{\sqrt{\frac{{s_1^2}}{{n_1}} + \frac{{s_2^2}}{{n_2}}}}} \]`

Substitute the values:

`\[ t = \frac{{520 - 461}}{{\sqrt{\frac{{51^2}}{{10}} + \frac{{72^2}}{{12}}}}} \]`

`\[ t \approx \frac{{59}}{{\sqrt{\frac{{2601}}{{10}} + \frac{{5184}}{{12}}}}} \]`

`\[ t \approx \frac{{59}}{{√(260.1 + 432)}}\]`

`\[ t \approx \frac{{59}}{{√(692.1)}}\]`

`\[ t \approx \frac{{59}}{{26.32}}\]`

`\[ t \approx 2.24\]`

Now that we have the t-value, we can find the p-value associated with it. The degrees of freedom
`(\(df\))` for a two-sample t-test is
`\(df = n_1 + n_2 - 2 = 10 + 12 - 2 = 20\)`. Using a t-table or statistical software, we find that the p-value for a two-tailed test with df = 20 and
`\(t \approx 2.24\)` is approximately 0.082.

In this case, the average E-Coli concentration has decreased from 520MPN/100ml to 461MPN/100ml, and the sample standard deviation has increased from 51MPN/100ml to 72MPN/100ml. Using these values in the formula, the t-statistic can be calculated and compared to the critical t-value to determine statistical significance. If the p-value is less than 0.05, the null hypothesis is rejected, indicating a significant difference in the mean concentrations between the two years.