A group of students titrate a calcium carbonate standard solution repeatedly in six experimental...

Question:

A group of students titrate a calcium carbonate standard solution repeatedly in six experimental trials. Based on the six trials, the group determines that the average concentration of the standard is 75.3 ppm with a standard deviation of 0.6 ppm. Using a t-value of 2.57 for a 95% confidence interval, calculate the confidence interval for which there is a 95% chance of finding the "true value" of the standard solution's concentration based on the data.

Put your answer in the format number lower limit < "true value" < number upper limit.

Confidence Interval:

In this question, we will use the t distribution to calculate and construct the 95% confidence interval of population mean. We are using the t distribution because the sample size is small and the population standard deviation is not known.

Given that,

• Sample size, {eq}n = 6 {/eq}
• Sample mean, {eq}\bar{x} = 75.3 {/eq}
• Sample standard deviation, {eq}s = 0.6 {/eq}

Critical value, {eq}t_{0.05/2} = 2.57 {/eq}

The 95% confidence interval for the population mean is defined as:

{eq}\bar{x} \pm t_{0.05/2}\times \frac{s}{\sqrt{n}} {/eq}

Now,

{eq}75.3\pm 2.57\times \frac{0.6}{\sqrt{6}}\\ 74.67 < \mu < 75.93 {/eq}

A 95% chance of finding the "true value" of the standard solution's concentration = (74.67, 75.93).