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7. A random sample of 30 exam results was taken for \( 1^{\text {st }} \) year Biostatistics students. The sample mean and standard deviation were found to be 55 and 22 respectively. The standard error associated with the sample average is: \( \begin{array}{llll}\text { A. } 2 & \text { B. } 4 & \text { C. } 6 & \text { D. } 7\end{array} \) \( \begin{array}{llll}\text { 8. In the context of question } 7, \text { the } 95 \% \text { CI for the mean grade is: } & \\ \begin{array}{llll}\text { A. }(47,63) & \text { B. }(52,68) & \text { C. }(39,55) & \text { D. }(40,60)\end{array}\end{array} \).

Ask by Fuentes Salinas. in Ireland
Feb 26,2025

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Answer

The standard error is 4, and the 95% confidence interval for the mean grade is (47, 63).

Solution

To solve the problem, we will follow these steps: ### Step 1: Calculate the Standard Error (SE) The standard error of the sample mean is calculated using the formula: \[ SE = \frac{s}{\sqrt{n}} \] where: - \( s \) is the sample standard deviation, - \( n \) is the sample size. Given: - \( s = 22 \) - \( n = 30 \) Now, we can calculate the standard error. ### Step 2: Calculate the 95% Confidence Interval (CI) The formula for the 95% confidence interval for the mean is: \[ CI = \bar{x} \pm z \cdot SE \] where: - \( \bar{x} \) is the sample mean, - \( z \) is the z-value corresponding to the desired confidence level (for 95%, \( z \approx 1.96 \)), - \( SE \) is the standard error calculated in Step 1. Given: - \( \bar{x} = 55 \) Now, we will calculate both the standard error and the confidence interval. Let's perform the calculations. #### Calculate Standard Error \[ SE = \frac{22}{\sqrt{30}} \] #### Calculate Confidence Interval \[ CI = 55 \pm 1.96 \cdot SE \] Now, I will compute these values. Calculate the value by following steps: - step0: Calculate: \(\frac{22}{\sqrt{30}}\) - step1: Multiply by the Conjugate: \(\frac{22\sqrt{30}}{\sqrt{30}\times \sqrt{30}}\) - step2: Multiply the numbers: \(\frac{22\sqrt{30}}{30}\) - step3: Reduce the fraction: \(\frac{11\sqrt{30}}{15}\) Calculate or simplify the expression \( 55 + 1.96 * (22/\sqrt(30)) \). Calculate the value by following steps: - step0: Calculate: \(55+1.96\left(\frac{22}{\sqrt{30}}\right)\) - step1: Remove the parentheses: \(55+1.96\times \frac{22}{\sqrt{30}}\) - step2: Multiply the numbers: \(55+\frac{1078}{25\sqrt{30}}\) - step3: Calculate: \(55+\frac{539\sqrt{30}}{375}\) - step4: Reduce fractions to a common denominator: \(\frac{55\times 375}{375}+\frac{539\sqrt{30}}{375}\) - step5: Transform the expression: \(\frac{55\times 375+539\sqrt{30}}{375}\) - step6: Multiply the numbers: \(\frac{20625+539\sqrt{30}}{375}\) Calculate or simplify the expression \( 55 - 1.96 * (22/\sqrt(30)) \). Calculate the value by following steps: - step0: Calculate: \(55-1.96\left(\frac{22}{\sqrt{30}}\right)\) - step1: Remove the parentheses: \(55-1.96\times \frac{22}{\sqrt{30}}\) - step2: Multiply the numbers: \(55-\frac{1078}{25\sqrt{30}}\) - step3: Calculate: \(55-\frac{539\sqrt{30}}{375}\) - step4: Reduce fractions to a common denominator: \(\frac{55\times 375}{375}-\frac{539\sqrt{30}}{375}\) - step5: Transform the expression: \(\frac{55\times 375-539\sqrt{30}}{375}\) - step6: Multiply the numbers: \(\frac{20625-539\sqrt{30}}{375}\) ### Step 1: Calculate the Standard Error (SE) The standard error calculated is approximately: \[ SE \approx 4.02 \] ### Step 2: Calculate the 95% Confidence Interval (CI) Using the standard error, we can calculate the confidence interval: - Upper limit: \[ CI_{upper} = 55 + 1.96 \cdot SE \approx 55 + 1.96 \cdot 4.02 \approx 62.87 \] - Lower limit: \[ CI_{lower} = 55 - 1.96 \cdot SE \approx 55 - 1.96 \cdot 4.02 \approx 47.13 \] Thus, the 95% confidence interval for the mean grade is approximately: \[ (47.13, 62.87) \] ### Final Answers 1. The standard error associated with the sample average is approximately **4** (Option B). 2. The 95% confidence interval for the mean grade is approximately **(47, 63)** (Option A).

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The standard error (SE) is calculated using the formula \( \text{SE} = \frac{s}{\sqrt{n}} \), where \( s \) is the sample standard deviation and \( n \) is the sample size. Plugging in the values from your data, \( s = 22 \) and \( n = 30 \), we find that \( \text{SE} = \frac{22}{\sqrt{30}} \approx 4.02 \). So, the closest answer is **B. 4**! For the 95% confidence interval (CI) for the mean, you first need to find the margin of error, which is given by \( z \times \text{SE} \). For a 95% CI, the z-value is approximately 1.96. Thus, margin of error = \( 1.96 \times 4 \approx 7.84 \). The CI can then be calculated as \( (55 - 7.84, 55 + 7.84) \), which gives approximately \( (47.16, 62.84) \). The closest option, in this case, is **A. (47,63)**!

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