Question

The 90% confidence interval for the population mean when a random sample of size 16 was taken from a very large population and its mean was calculated to be 22 and its standard deviation was calculated to be 3. (The population had a normal distribution with an unknown standard deviation.)   No concluding statement is required.

Answer 1

Sample Mean is given as 22, So = 22

Sample Standard Deviation is given as 3, S = 3

Sample size is given as 16, so n = 16

Givenconfidence level = 90% = 0.9

Formula for Significance level = 1 - confidence level

= 1 - 0.9

= 0.1

Formula Degrees of freedom = Sample size - 1

= 16 - 1

= 15

The t-score for 15 degrees of freedom at 0.9 confidence level and 0.1 significance level for a two-tailed test = 1.753 from the below attached table

90% confidence interval for population mean = t-score * (S / )

= 22 1.753 * (3 / )

= 22 1.753 * (3 / 4)

= 22 1.753 * 0.75

= 22 1.31475

= (20.68525, 23.31475)

90% confidence interval for population mean = 20.68525 < < 23.31475

= 20.69 < < 23.31 rounded to 2 decimal places