95% CI for the population mean when population s.d. is unknown

Question 1

What is the formula for a 95% confidence interval if the population mean is unknown?

Answer 1

95% CI if pop. mean is unknown=

sample mean - (c x e.s.e.)
sample mean + (c x e.s.e.)

Question 2

How do you find ‘c’ when you want to calculate the 95% CI if the pop. mean is unknown?

sample mean - (c x e.s.e.)
sample mean + (c x e.s.e.)

Answer 2

How to find c:

1) Find out v = N-1

2) Refer to the parameter v / t-score table and find your v = N-1

3) Find the 5% two-tailed (or 2.5% one-tailed) t-score on the table

4) The number under the 5% two-tailed (or 2.5% one-tailed) is the c value

Question 3

Sample: {1, 6, 8, 7, 10 , 12}
(N = 6)
[m = 7.33, s = 3.78, e.s.e. = 1.54]

What is the 95% CI for µ?

Answer 3

Find c=
v = N-1
v = 6 - 1
v = 5
5% two-tailed of v = 5 is 2.571

95% CI =
sample mean - (c x e.s.e.)
7.33 - (c x 1.54)
7.33 - (2.571 x 1.54) = 3.37

sample mean + (c x e.s.e.)
7.33 + (c x 1.54)
7.33 + (2.571 x 1.54) = 11.30

Answer = (3.37, 11.30)

Question 4

Aliens visiting from the planet Ziltoidia 10 are keen to establish an estimate for the population mean of their height while on earth. The population s.d. is unknown (since 100,000 Ziltoidians have travelled to earth and there isn’t enough time to measure the height of everyone in the population).

You take a sample of 25 Ziltoidians and find that the sample has a mean height of 23.2cm and s.d. of 2.2cm.

What is the value c that isolates 95% of the area under the appropriate t-distribution that enables you to subsequently work out the 95% CI for the population mean?

a.	c = 2.060

b.	c = 1.960

c.	c = 2.064

d.	c = 1.711

Answer 4

Find c=
v = N-1
v = 25 - 1
v = 24
5% two-tailed of v = 24 is 2.064

Answer = c. 2.064

Question 5

Aliens visiting from the planet Ziltoidia 10 are keen to establish an estimate for the population mean of their height while on earth. The population s.d. is unknown (since 100,000 Ziltoidians have travelled to earth and there isnt enough time to measure the height of everyone in the population).

You take a sample of 25 Ziltoidians and find that the sample has mean height of 23.2cm and s.d. of 2.2cm.

Using your answer to Q1a., what is the 95% CI for the population mean centred on the sample mean?

a.	(22.2936, 24.1064)

b.	(22.2918, 24.1082)

c.	(18.6592, 27.7408)

d.	(22.3376, 24.0624)

Answer 5

Find c=
v = N-1
v = 25 - 1
v = 24
5% two-tailed of v = 24 is 2.064

e.s.e. = sample s.d. / SQRT sample size
e.s.e. = 2.2 / SQRT 25
e.s.e. = 0.44

95% CI =
sample mean - (c x e.s.e.)
23.2 - (c x 0.44)
23.2 - (2.064 x 0.44) = 22.29184

sample mean + (c x e.s.e.)
23.2 + (c x 0.44)
23.2 + (2.064 x 0.44) = 24.1081

Answer = b. (22.2918, 24.1082)

Question 6

You are interested in anxiety levels in UK undergraduates and collect data on a well-established anxiety questionnaire (which measures anxiety on a 20 point scale: 1 = lowest anxiety; 20 = highest anxiety) from a sample of 10 of course mates. The anxiety scores obtained are:

{13, 12, 16, 6, 7, 14, 6, 9, 10, 11}

Using the critical values table for the t-distribution and working to 4 decimal places, calculate the 95% confidence interval for the population mean anxiety centred on the sample mean.

Hint 1: the sample s.d. is 3.4383

Hint 2: you will need to calculate the estimated standard error (e.s.e.)

a.	(7.9775, 12.8225)

b.	(2.6226, 18.1774)

c.	(7.9405, 12.8595)

d.	(8.2689, 12.5311)

Answer 6

Find c=
v = N-1
v = 10 - 1
v = 9
5% two-tailed of v = 9 is 2.262

e.s.e. = sample s.d. / SQRT sample size
e.s.e. = 3.4383 / SQRT 10
e.s.e. = 1.087

sample mean = 13, 12, 16, 6, 7, 14, 6, 9, 10, 11
sample mean = 10.4

95% CI =
sample mean - (c x e.s.e.)
10.4 - (c x 1.087)
10.4 - (2.262 x 1.087) = 7.9405

sample mean + (c x e.s.e.)
10.4 + (c x 1.087)
10.4 + (2.262 x 1.087) = 12.8595

Answer = c. (7.9405, 12.8595)