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Ferulic acid is a compound that may play a role in disease resistance in corn. A botanist measured the con- \(-\) centration of soluble ferulic acid in corn seedlings grown in the dark or in a light/dark photoperiod. The results (nmol acid per gm tissue) were as shown in the table. \({ }^{41}\) $$ \begin{array}{|ccc|} \hline & \text { Dark } & \text { Photoperiod } \\ \hline n & 4 & 4 \\ \bar{y} & 92 & 115 \\ s & 13 & 13 \\ \hline \end{array} $$ (a) Construct a \(95 \%\) confidence interval for the difference in ferulic acid concentration under the two lighting conditions. (Assume that the two populations from which the data came are normally distributed.) [Note: Formula (6.7.1) yields 6 degrees of freedom for these data. (b) Repeat part (a) for a \(90 \%\) level of confidence.

Short Answer

Expert verified
The 95% confidence interval for the difference in ferulic acid concentrations between dark and photoperiod-grown corn seedlings is approximately \([-45.497, -0.503]\). The 90% confidence interval is approximately \([-40.865, -5.135]\).

Step by step solution

01

State the Formula for the Confidence Interval

For estimating the confidence interval for the difference between two means with the assumption of the populations being normally distributed, we use the following formula: \[ CI = (\bar{x}_1 - \bar{x}_2) \pm t_{\alpha/2, u} \cdot \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} \]where \(\bar{x}_1\) and \(\bar{x}_2\) are the sample means, \(s_1\) and \(s_2\) are the sample standard deviations, n_1 and n_2 are the sample sizes, \(t_{\alpha/2, u}\) is the t-value for \(\alpha/2\) significance level with \(u\) degrees of freedom.
02

Calculate the Standard Error of the Difference

The standard error (SE) of the difference between the two means is given by the following formula: \[ SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} \]Substitute the given values: \[ SE = \sqrt{\frac{13^2}{4} + \frac{13^2}{4}} = \sqrt{\frac{169}{4} + \frac{169}{4}} = \sqrt{\frac{338}{4}} = \sqrt{84.5} = 9.1924 \]
03

Determine the t-Value for 95% Confidence Level

For a 95% confidence interval with 6 degrees of freedom, we use a t-distribution table or calculator to find the t-value for \(\alpha/2 = 0.025\). The t-value \(t_{0.025, 6}\) is approximately 2.447.
04

Calculate the 95% Confidence Interval for the Difference

Now we can apply the values to the confidence interval formula: \[ CI = (92 - 115) \pm 2.447 \cdot 9.1924 \]\[ CI = (-23) \pm 22.497 \]This results in the following confidence interval: \[ CI = [-45.497, -0.503] \]
05

Determine the t-Value for 90% Confidence Level

For a 90% confidence interval with 6 degrees of freedom, the t-value for \(\alpha/2 = 0.05\) is approximately 1.943.
06

Calculate the 90% Confidence Interval for the Difference

Using the 90% confidence level t-value: \[ CI = (92 - 115) \pm 1.943 \cdot 9.1924 \]\[ CI = (-23) \pm 17.865 \]This results in the following confidence interval: \[ CI = [-40.865, -5.135] \]

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Key Concepts

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

Difference Between Two Means
Understanding the difference between two means is essential in statistics, especially when comparing the results of two groups. In our exercise, we're interested in the difference in ferulic acid concentrations in corn seedlings grown under dark conditions compared to those grown in light/dark photoperiod. The mean difference is calculated simply by subtracting one sample mean from the other, as demonstrated by \( \bar{x}_1 - \bar{x}_2 \).

This calculation tells us if there's a significant variation between the two groups. However, knowing just the mean difference isn't enough; we need to determine whether this difference is statistically significant, which is where the confidence interval comes into play. It's a way to quantify the uncertainty around the mean difference, helping us understand the range wherein the true mean difference between the two populations lies with a certain level of confidence.
Standard Error
When we talk about the standard error (SE), we refer to the precision of our sample mean estimate. It's a measure of how far we can expect the sample mean to be from the true population mean. The SE is especially important when comparing the means from two different groups because it takes both the sample variability and the size into account.

In our exercise, the SE for the difference between two means is calculated with the formula \( SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} \), where \(s_1\) and \(s_2\) are the sample standard deviations and \(n_1\) and \(n_2\) are the sample sizes for the dark and photoperiod groups, respectively. A smaller SE suggests a more precise estimate of the mean difference which enhances the reliability of the confidence interval we are trying to estimate.
T-Distribution
The t-distribution plays a pivotal role when we're dealing with small sample sizes, which is the case in our corn seedlings exercise. It's similar to the normal distribution but accounts for the extra uncertainty that comes with smaller samples.

When constructing a confidence interval for the difference between two means, we use a t-distribution to find the critical value known as the t-value. This reflects the level of confidence we want for our interval. For the 95% and 90% confidence intervals in our exercise, we looked up the respective t-values for 6 degrees of freedom. These t-values are crucial in stretching our interval to capture the true mean difference with a specified level of confidence.
Degrees of Freedom
Degrees of freedom (df) are a concept linked to the t-distribution and refer to the number of independent pieces of information in our data set that can be used to estimate a parameter, such as variance. In the formula for finding a confidence interval for the difference between two means, we used 6 degrees of freedom — a number derived from the sample sizes of our two groups.

The df often equals the sample size minus one for each group, but since we're comparing two means here, we use a combined approach to calculate the degrees of freedom, which in this case, yielded 6. This number then guides which t-value we should use from the t-distribution table to construct our confidence intervals. The concept of degrees of freedom is crucial for ensuring the validity of our t-distribution application in confidence interval estimation.

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Most popular questions from this chapter

Over a period of about 9 months, 1,353 women reported the timing of each of their menstrual cycles. For the first cycle reported by each woman, the mean cycle time was 28.86 days, and the standard deviation of the 1,353 times was 4.24 days. \(^{52}\) (a) Construct a \(99 \%\) confidence interval for the population mean cycle time. (b) Because environmental rhythms can influence biological rhythms, one might hypothesize that the population mean menstrual cycle time is 29.5 days, the length of the lunar month. Is the confidence interval of part (a) consistent with this hypothesis?

Compute the standard error of \(\left(\bar{Y}_{1}-\bar{Y}_{2}\right)\) for the following data: $$ \begin{array}{|ccc|} \hline & \text { Sample 1 } & \text { Sample 2 } \\ \hline n & 5 & 7 \\ \bar{y} & 44 & 47 \\ s & 6.5 & 8.4 \\ \hline \end{array} $$

Researchers were interested in the short-term effect that caffeine has on heart rate. They enlisted a group of volunteers and measured each person's resting heart rate. Then they had each subject drink 6 ounces of coffee. Nine of the subjects were given coffee containing caffeine, and 11 were given decaffeinated coffee. After 10 minutes each person's heart rate was measured again. The data in the table show the change in heart rate; a positive number means that heart rate went up, and a negative number means that heart rate went down. \(^{47}\) (a) Use these data to construct a \(90 \%\) confidence interval for the difference in mean effect that caffeinated coffee has on heart rate, in comparison to decaffeinated coffee. [Note: Formula (6.7.1) yields 17.3 degrees of freedom for these data. (b) Using the interval computed in part (a) to justify your answer, is it reasonable to believe that caffeine may not affect heart rates? (c) Using the interval computed in part (a) to justify your answer, is it reasonable to believe that caffeine may affect heart rates? If so, by how much? (d) Are your answers to (b) and (c) contradictory? Explain. $$ \begin{array}{|ccc|} \hline \text { Caffeine } & \text { Decaf } \\ \hline & 28 & 26 \\ & 11 & 1 \\ & -3 & 0 \\ & 14 & -4 \\ & -2 & -4 \\ & -4 & 14 \\ & 18 & 16 \\ & 2 & 8 \\ & 2 & 0 \\ & & 18 \\ & & -10 \\ \hline n & 9 & 11 \\ \bar{y} & 7.3 & 5.9 \\ s & 11.1 & 11.2 \\ \text { SE } & 3.7 & 3.4 \\ \hline \end{array} $$

In a field study of mating behavior in the Mormon cricket (Anabrus simplex), a biologist noted that some females mated successfully while others were rejected by the males before coupling was complete. The question arose whether some aspect of body size might play a role in mating success. The accompanying table summarizes measurements of head width \((\mathrm{mm})\) in the two groups of females. \(^{45}\) (a) Construct a \(95 \%\) confidence interval for the difference in population means. [Note: Formula (6.7.1) yields 35.7 degrees of freedom for these data. (b) Interpret the confidence interval from part (a) in the context of this setting. (c) Using your interval computed in (a) to support your answer, is there strong evidence that the population mean head width is indeed larger for successful maters than unsuccessful maters? $$ \begin{array}{|lcc|} \hline & \text { Successful } & \text { Unsuccessful } \\ \hline n & 22 & 17 \\ \bar{y} & 8.498 & 8.440 \\ s & 0.283 & 0.262 \\ \hline \end{array} $$

A group of 101 patients with end-stage renal disease were given the drug epoetin. \({ }^{19}\) The mean hemoglobin level of the patients was \(10.3(\mathrm{~g} / \mathrm{dl})\), with an \(\mathrm{SD}\) of 0.9 . Construct a \(95 \%\) confidence interval for the population mean.

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