In a survey, 27 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $31 and standard deviation of $3. Construct a confidence interval at a 99% confidence level. Give your answers to one decimal place.

CI (29.4, 32.6)

Step-by-step explanation:

Given

n = 27 (Size)

X = 31 (Mean)

sd = 3 (Standard deviation)

Required to construct 99% Confidence Interval CI

X±z*б/√n

What we do have is the z value and to find it we need degrees of freedom and the significance level

DF = 27-1 = 26

SL = 100%-99% = 1%

The z value in a two tailed test with (26,1%) = 2.779

Back to formula

Lower limit = 31 - 2.779 * 3/√27 = 29.4

Upper limit = 32.6

Answer: (29.4, 32.6)

Step-by-step explanation:

From the question, we know that

Sample size (n) = 27

Sample mean (x) = $31

Sample standard deviation (s) = $3

We are to construct a 99% confidence interval interval for average amount spent on gift.

The formulae for constructing a 99% confidence interval for population mean is given as

u = x + tα/2 * s/√n ... For upper limit

u = x - tα/2 * s/√n ... For lower limit

tα/2 is the critical value for a 2 tailed test. This value of gotten from a t distribution table by checking the degree of freedom (27 - 1 = 26) against the level of significance (100% - 99% = 1%).

Hence tα/2 = 2.779

Let us substitute our parameters and solve

For lower limit

u = 31 - 2.779 * 3/√27

u = 31 - 2.779 (0.5773)

u = 31 - 1.6045

u = 29.4

For upper limit

u = 31 + 2.779 * 3/√27

u = 31 + 2.779 (0.5773)

u = 31 + 1.6045

u = 32.6

Hence the 99% confidence interval for population mean is given as 29.4, 32.6