# Calculate for the standard error

The instructor of a Chemistry Laboratory Class, which is composed of 10 students, asked each individual to record the time it took for a certain chemical to change its color after the third trial. The data (in minutes) were as follows: 5.18, 9.02, 13.35, 10.91, 8.57, 9.24, 10.16, 8.01, 9.33, and 11.28. What is the margin of error for the estimate at 95% confidence?

$mean = 9.505\$ mins

$\sigma = 2.0613\$ mins

$E = 1.96 *\frac{2.0613}{\sqrt{10}}$

$E = 1.2776$

I'm pretty sure I solved it correctly but my answer isn't in the choices on my worksheet so I want to know if I actually answered it correctly.

2.84 minutes, 2.46 minutes, 2.87 minutes, 2.69 minutes, 2.64 minutes, 2.33 minutes, 2.61 minutes, 2.71 minutes, 2.39 minutes, 2.30 minutes, 2.50 minutes, 2.92 minutes, 2.41 minutes, 2.57 minutes, 2.79 minutes

• I don't know your course material but I'd say it was more common to use (sample size $-$ 1) in estimating the SD and t-based confidence intervals for the mean with a sample of 10. – Nick Cox Mar 11 '18 at 11:17
• using $n-1$ in estimating $\sigma$ gives me 2.1728, together with a t-tail value of 1.833 givies me $E = 1.2594$ ... this is still not in the choices tho... – Richard Marin Mar 11 '18 at 11:25
• Can you tell us the possible choices? – Sven Hohenstein Mar 11 '18 at 11:31
• Here are all the possible choices : 2.84 minutes, 2.46 minutes, 2.87 minutes, 2.69 minutes, 2.64 minutes, 2.33 minutes, 2.61 minutes, 2.71 minutes, 2.39 minutes, 2.30 minutes, 2.50 minutes, 2.92 minutes, 2.41 minutes, 2.57 minutes, 2.79 minutes... and yes there are 15 choices.. – Richard Marin Mar 11 '18 at 11:36
• The question title says standard error and body text requires computation of margin of error . which one is your target ? – Subhash C. Davar Mar 11 '18 at 16:15

Here is a way to derive a value of about 2.69 minutes.

Let $\{x_1, ..., x_{N}\}$ denote the $N = 10$ observed values. The mean is indeed

$$\bar x = \frac{\sum_{i=1}^{N} x_i}{N} = 9.505\,.$$

We don't know the variance of the population. The sample standard deviation is

$$s = \sqrt{\frac{\sum_{i=1}^{N} (x_i - \bar x)^2}{N-1}} \approx 2.172777\,.$$

Since the standard deviation of the population is unknown, we should use the $t$-distribution for calculation the confidence interval. However, you have to use the normal distribution to obtain a value of the margin of error that is present in the choices.

We compute the margin of error for a 95% confidence interval ($\alpha = .05$),

$$E = \Phi^{-1}\left(1 - \frac{\alpha}{2}\right) \frac{s}{\sqrt{N}} \approx 1.346677\,,$$

where $\Phi$ denotes the cumulative distribution function of the standard normal distribution.

As you can see, margin of error $E$ is considerably lower than the values of the choices. I suppose, you have to find the full width of the confidence interval instead:

$$2 E \approx 2.69\,.$$

R code for these calculations:

x <- c(5.18, 9.02, 13.35, 10.91, 8.57, 9.24, 10.16, 8.01, 9.33, 11.28)
N <- length(x)
mu <- mean(x)
s <- sd(x)
alpha <- 0.05
E <- qnorm(1 - alpha / 2) * s / sqrt(N)
2 * E
#  2.693353

• "for a α = 0.95 confidence intervall " does not make a sense. It should be eithe stated as α =.05 or 95% confidence interval. – Subhash C. Davar Mar 11 '18 at 16:27
• As you can see, error of margin E is considerably lower than the values of the choices. I suppose, you have to find the full width of the confidence interval instead: 2E≈2.69. This part is incorrect. – Subhash C. Davar Mar 11 '18 at 16:42
• @subhashc.davar Do you know the correct approach? – Sven Hohenstein Mar 11 '18 at 19:28
• i said - This part is incorrect. – Subhash C. Davar Mar 12 '18 at 10:49