Question

Suppose a student measuring the boiling temperature of a certain liquid observes the readings (in degrees Celsius) 102.5, 101.7, 103.1, 100.9, 100.5, and 102.2 on 6 samples of the liquid and the measurements follow a normal distribution. (a) What is a point estimate for the mean boiling temperature of this liquid? (b) State the two reasons that we are able to perform a confidence interval test for the mean boiling temperature of the liquid. (c) What is the confidence interval for the mean boiling temperature of this liquid at a 98% confidence level?

Answer 1

Solution:

x	x2
102.5	10506.25
101.7	10342.89
103.1	10629.61
100.9	10180.81
100.5	10100.25
102.2	10444.84
∑x=610.9	∑x2=62204.65

Mean ˉx=∑xn

=102.5+101.7+103.1+100.9+100.5+102.2/6

=610.9/6

=101.8167

Sample Standard deviation S=√∑x2-(∑x)2nn-1

=√62204.65-(610.9)265

=√62204.65-62199.80175

=√4.84835

=√0.9697

=0.9847

Degrees of freedom = df = n - 1 = 6 - 1 = 5

At 98% confidence level the t is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

t /2,df = t0.01,5 =2.797

Margin of error = E = t/2,df * (s /n)

= 2.797 * (0.98 / 6)

= 1.34

Margin of error =1.34

The 98% confidence interval estimate of the population mean is,

- E < < + E

101.8 - 1.34< < 101.8 +1.34

100.45 < < 1.03.14