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A small amount of the trace element selenium, \(50-200\) micrograms ( \(\mu \mathrm{g}\) ) per day, is considered essential to good health. Suppose that random samples of \(n_{1}=n_{2}=30\) adults were selected from two regions of the United States and that a day's intake of selenium was recorded for each person. The mean and standard deviation of the selenium daily intakes for the 30 adults from region 1 were \(\bar{x}_{1}=\) 167.1 and \(s_{1}=24.3 \mu \mathrm{g},\) respectively. The corresponding statistics for the 30 adults from region 2 were \(\bar{x}_{2}=\) 140.9 and \(s_{2}=17.6 \mu \mathrm{g} .\) Find a \(95 \%\) confidence interval for the difference in the mean selenium intakes for the two regions. Interpret this interval.

Step by step solution

01

Identify the given information

For Region 1, we have: Sample size: \(n_1 = 30\) Mean daily intake: \(\bar{x}_1 = 167.1 \mu g\) Standard deviation: \(s_1 = 24.3 \mu g\) For Region 2, we have: Sample size: \(n_2 = 30\) Mean daily intake: \(\bar{x}_2 = 140.9 \mu g\) Standard deviation: \(s_2 = 17.6 \mu g\)

02

Calculate the pooled standard deviation

The pooled standard deviation is given by the formula: \(sp = \sqrt{\frac{(n_1 - 1) s_1^2 + (n_2 - 1) s_2^2}{n_1 + n_2 - 2}}\) Plug in the given values: \(sp = \sqrt{\frac{(30 - 1)(24.3)^2 + (30 - 1)(17.6)^2}{30 + 30 - 2}}\) Calculate the pooled standard deviation: \(sp \approx 21.15 \mu g\)

03

Calculate the standard error of the difference

The standard error of the difference is calculated using the formula: \(SE_d = sp \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}\) Plug in the values: \(SE_d = 21.15 \sqrt{\frac{1}{30} + \frac{1}{30}}\) Calculate the standard error of the difference: \(SE_d \approx 5.49 \mu g\)

04

Determine the critical value for a 95% confidence interval

We are looking for a 95% confidence interval, which implies that we want a \(Z\)-score that corresponds to the critical value of \(0.025\) in each tail. From the standard normal table, we find that critical value is: \(Z_{0.025} = 1.96\)

05

Calculate the margin of error

We calculate the margin of error by multiplying the critical value by the standard error of the difference: \(ME = Z_{0.025} * SE_d = 1.96*5.49\) Compute the margin of error: \(ME \approx 10.76 \mu g\)

06

Calculate the confidence interval

The confidence interval for the difference in mean selenium intakes is given by: \((\bar{x}_1 - \bar{x}_2) \pm ME\) Plug in the values: \((167.1 - 140.9) \pm 10.76\) Compute the confidence interval: \(26.2 \pm 10.76\) Therefore, the confidence interval is approximately: \((15.44, 36.96) \mu g\)

07

Interpret the interval

We can interpret this confidence interval as follows: We are 95% confident that the true difference in the mean selenium intakes between the two regions lies between 15.44 and 36.96 micrograms. This suggests that people from Region 1 consume more selenium on average than those from Region 2.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Selenium Intake

Selenium is a trace mineral that is crucial for various bodily functions, including antioxidant protection and immune system support. It is essential to consume an adequate amount of selenium daily to maintain optimal health. The recommended intake is between 50 and 200 micrograms (\(\mu g\)) per day.
Selenium can be found in foods such as nuts, seafood, and meats. Its intake varies in different geographical regions, which may lead to differences in average selenium consumption.

• Factors influencing selenium intake include soil composition, dietary habits, and availability of selenium-rich foods.
• Deficiency in selenium can lead to health issues like weakened immunity and thyroid problems, while excessive intake might cause toxicity.

This study aims to investigate the average selenium intake among adults in two US regions, helping to assess whether residents of different areas have similar dietary selenium levels.

Pooled Standard Deviation

To compare the variability of selenium intake between two different regions, the pooled standard deviation (\(s_p\)) is used. It gives a combined measure of spread for two independent samples. Calculating \(s_p\) involves the individual standard deviations and sample sizes from both groups.
Pooled standard deviation is given by the formula:\[s_p = \sqrt{\frac{(n_1 - 1) s_1^2 + (n_2 - 1) s_2^2}{n_1 + n_2 - 2}}\]

• In our exercise, the pooled standard deviation gives a clearer picture of the overall variability in selenium intake across the two regions.
• A larger \(s_p\) indicates greater variability in intake, while a smaller value suggests more consistency.

Pooled variance ensures that differences in sample sizes do not skew the overall finding, making comparisons more reliable.

Standard Error of the Difference

The standard error of the difference (\(SE_d\)) is a vital metric in assessing the mean difference of selenium intake between the two regions. It provides a measure of how much the observed difference in sample means (\[\bar{x}_1 - \bar{x}_2\]) is likely to vary by chance.
Calculated using the pooled standard deviation:\[SE_d = s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}\]

• \(SE_d\) helps to determine how much uncertainty there is around the mean difference.
• A smaller \(SE_d\) indicates a more precise estimate of the true difference in population means.

Understanding \(SE_d\) is crucial because it forms the basis for constructing confidence intervals and conducting hypothesis tests, revealing the strength of the difference observed.

Margin of Error

The margin of error (ME) provides a range that expresses the uncertainty in estimating the true difference in selenium intakes between the regions. It accounts for sampling variability and uses the standard error of the difference to quantify this uncertainty. The wider the margin, the less precise the estimate is.
Calculated as:\[ME = Z_{0.025} \times SE_d\]

• The critical value \(Z_{0.025}\) is derived from the Z-distribution for a 95% confidence interval, indicating the extremes of the normal distribution we are willing to accept.
• By knowing the margin of error, researchers can understand the confidence level of their estimate.

This knowledge aids in creating a confidence interval, which suggests that 95% of the time, the actual mean difference should fall within the bounds established by adding and subtracting the margin of error from the observed difference.