Question

A chemist made six independent measurements of the sublimation point of carbon dioxide (the temperature at which it changes to dry ice). She obtained a sample mean of 196.64 K with a standard deviation of 0.66 K.

1. Use the Student’s t distribution to find a 95% confidence interval for the sublimation point of carbon dioxide.
2. Use the Student's t distribution to find a 98% confidence interval for the sublimation point of carbon dioxide.
3. If the six measurements had been 196.35, 196.32, 196.4 198.02, 196.36, 196.39, would the confidence intervals above be valid? Explain.

Answer 1

(195.733, 197.547)

Explanation:

Given that:

Mean, m = 196.64 K

Standard deviation, s = 0.66 K

Sample size, n = 6

Confidence interval :

Mean ± margin of error

Margin of error =Tcritical * s/sqrt(n)

Tcritical at α = 98% ; df = 6 - 1 = 5

Tcritical = 3.3648

Margin of Error = 3.3648 * 0.66/sqrt(6) = 0.9066

Confidence interval = 196.64 ± 0.9066

Lower boundary = 196.64 - 0.9066 = 195.733

Upper boundary = 196.64 + 0.9066 = 197.547

(195.733, 197.547)

Given the measurement :

X = 196.35, 196.32, 196.4 198.02, 196.36, 196.39

Mean = ΣX / n = 1179.84 / 6 = 196.64

Yes, confidence interval will be valid as theean obtained is within the calculated interval