Determine the confidence interval

In summary, students collected water samples from Lake Macatawa to study the impact of salting the roads in winter. They determined the sample mean and adjusted sample variance for the western basin of the lake, and calculated a two-sided confidence interval for the mean sodium content with a confidence level of 0.95. They also collected water samples from the eastern basin and calculated a two-sided confidence interval for the mean sodium content with a confidence level of 0.9, as well as a confidence interval of the form $(-\infty, h]$ with a confidence level of 0.9. Finally, they studied the PCB content in fish from Lake Michigan and calculated the sample mean and a two-sided confidence interval for the mean PCB content
  • #1
mathmari
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Hey! 😊

(a) In winter, the roads around Lake Macatawa are salted. To study the impact of this on Lake Macatawa, students took $32$ water samples from the western basin of the lake and Sodium content (in parts per million, ppm) determined. As a result, the students have receive the following data:
1643538010837.png


(i) Calculate the sample mean.
(ii) Calculate the adjusted sample variance.
(iii) Assume a normal distribution and determine a two-sided confidence interval for the above sample for the mean sodium content with a confidence level of $0.95$. Also enter the used quantile and its (approximate) value.(b) In addition, $35$ water samples were collected from the eastern basin of Lake Macatawa and in each case the sodium content (in ppm) was measured. This resulted in a sample mean of $24.11$ and an adjusted sample variance of $24.44$. Assume a normal distribution.
(i) Determine a two-sided confidence interval for the mean sodium content with a confidence level of $0.9$. Also enter the quantile used and its (approximate) value.
(ii) Determine a confidence interval of the form $(-\infty, h]$ for the mean sodium content with a confidence level of $0.9$. Also state the one used quantile and its (approximate) value.(c) The PCB content was determined from a sample of fish from Lake Michigan (in ppm). It is known that the standard deviation is $0.8$ ppm. Assume a normal distribution. The below were measured:
1643539615613.png


(i) Calculate the sample mean and determine a two-sided confidence interval for the above sample for the mean PCB content with a confidence level of $0.99$. Also enter the used quantile and its
(approximate) value.
(ii) Determine a confidence interval of the form $(-\infty, h]$ for the above sample for the mean PCB content with a confidence level of $0.99$. Give also the quantile used and its (approximate) value.
I have done the following :

(a) (i) We add all elements and divide the result by the number of elements. So the sample mean is equal to \begin{align*}\overline{x}_{32}=& \frac{1}{32}\left(13.0 +18.5 +16.4 +14.8 +19.4+ 17.3 +23.2 +24.9 +20.8 +19.3 +18.8 +23.1 +15.2+ 19.9 +19.1+ 18.1 +25.1+ 16.8+ 20.4 +17.4 +25.2+ 23.1 +15.3+ 19.4 +16.0 +21.7 +15.2+ 21.3+ 21.5 +16.8+ 15.6 +17.6 \right )\\ & =\frac{1}{32}\cdot 610.2=19.06875\end{align*} Is that correct ? :unsure:

(ii) We have that \begin{align*}s_{32}^2&=\frac{1}{32-1}\sum_{i=1}^{32}(x_i-\overline{x}_{32})^2=\frac{1}{31}((13.0-19.06875)^2 +(18.5-19.06875)^2 +(16.4-19.06875)^2 \\ &+(14.8-19.06875)^2 +(19.4-19.06875)^2+ (17.3-19.06875)^2 +(23.2-19.06875)^2 \\ &+(24.9-19.06875)^2 +(20.8-19.06875)^2 +(19.3-19.06875)^2 +(18.8-19.06875)^2 \\ &+(23.1-19.06875)^2 +(15.2-19.06875)^2+ (19.9-19.06875)^2 +(19.1-19.06875)^2\\ &+ (18.1-19.06875)^2 +(25.1-19.06875)^2+ (16.8-19.06875)^2+ (20.4-19.06875)^2 \\ &+(17.4-19.06875)^2 +(25.2-19.06875)^2+ (23.1-19.06875)^2 +(15.3-19.06875)^2\\ &+ (19.4-19.06875)^2 +(16.0-19.06875)^2 +(21.7-19.06875)^2 +(15.2-19.06875)^2\\ &+ (21.3-19.06875)^2+ (21.5-19.06875)^2 +(16.8-19.06875)^2+ (15.6-19.06875)^2 \\ &+(17.6-19.06875)^2 )\\ & = \frac{1}{31}\cdot 328.54875\\ & \approx 10.59835\end{align*} Is that correct ? :unsure:

(iii) Do we use the formula \begin{equation*}\left [M(x)-\frac{\sigma}{\sqrt{n}}q_{1-\frac{a}{2}},M(x)+\frac{\sigma}{\sqrt{n}}q_{1-\frac{a}{2}}\right ] \end{equation*} where $M(x)$is the mean, i.e. $M(x)=19.06875$ ? Is $\sigma$ equal to the square root of the result of (b) ? The value of $q$ is related to the confidence level, right? :unsure:
 
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  • #2
Hey mathmari!

a.i and a.ii look correct to me. (Nod)
In a.iii we do not have $\sigma$, which is unknown. Instead we have $s$, which is the standard deviation of the sample. It is indeed the square root of the result of a.ii.

The value of $q$ is indeed related to the confidence interval. Can we find it? 🤔
 
  • #3
Klaas van Aarsen said:
In a.iii we do not have $\sigma$, which is unknown. Instead we have $s$, which is the standard deviation of the sample. It is indeed the square root of the result of a.ii.

The value of $q$ is indeed related to the confidence interval. Can we find it? 🤔

Ahh ok!

So we use the formula $$\left [M(x)-\frac{\sqrt{V^{\star}}}{\sqrt{n}}t_{n-1,1-a/2}, \ M(x)+\frac{\sqrt{V^{\star}}}{\sqrt{n}}t_{n-1,1-a/2}\right ]$$ right?
Do we have $V^{\star}=s_{32}^2=10.59835$, $M(x)= 19.06875 $ and $t_{n-1,1-a/2}=t_{31,0.975}= 2,037$ ? :unsure:

So we get \begin{equation*}\left [19.06875-\frac{\sqrt{10.59835}}{\sqrt{32}}\cdot 2.037, \ 19.06875+\frac{\sqrt{10.59835}}{\sqrt{32}}\cdot 2.037\right ]\approx \left [17.89646, \ 20.24104\right ]\end{equation*} right? :unsure:
 
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  • #4
mathmari said:
Ahh ok!

So we use the formula $$\left [M(x)-\frac{\sqrt{V^{\star}}}{\sqrt{n}}t_{n-1,1-a/2}, \ M(x)+\frac{\sqrt{V^{\star}}}{\sqrt{n}}t_{n-1,1-a/2}\right ]$$ right?
Do we have $V^{\star}=s_{32}^2=10.59835$, $M(x)= 19.06875 $ and $t_{n-1,1-a/2}=t_{31,0.975}= 2,037$ ?

So we get \begin{equation*}\left [19.06875-\frac{\sqrt{10.59835}}{\sqrt{32}}\cdot 2.037, \ 19.06875+\frac{\sqrt{10.59835}}{\sqrt{32}}\cdot 2.037\right ]\approx \left [17.89646, \ 20.24104\right ]\end{equation*} right?
Looks right to me. (Nod)
 

FAQ: Determine the confidence interval

1. What is a confidence interval?

A confidence interval is a range of values that is likely to include the true population parameter with a certain level of confidence. It is a statistical measure that helps to estimate the precision of a sample statistic.

2. How is a confidence interval determined?

A confidence interval is determined by taking a sample from a population and calculating the sample statistic, such as the mean or proportion. The confidence interval is then calculated by adding and subtracting a margin of error from the sample statistic, using a specific confidence level and the sample's standard deviation.

3. What is the purpose of a confidence interval?

The purpose of a confidence interval is to provide a range of values that is likely to include the true population parameter. It helps to quantify the uncertainty in a sample statistic and provides a measure of the precision of the estimate.

4. What is the significance of the confidence level in a confidence interval?

The confidence level in a confidence interval represents the probability that the true population parameter falls within the calculated interval. For example, a 95% confidence level means that if the sampling process is repeated multiple times, 95% of the resulting confidence intervals will contain the true population parameter.

5. How does sample size affect the width of a confidence interval?

As the sample size increases, the width of the confidence interval decreases. This is because a larger sample size provides more information about the population and reduces the margin of error, resulting in a more precise estimate. However, the relationship between sample size and confidence interval width also depends on the variability of the data and the chosen confidence level.

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