Question

33 46 48 21 30 32 10 31 20 29 A survey of 25 randomly selected customers found the ages shown (in years). The mean is 31.76 years and the standard deviation is 10.37 years. a) Construct a 98 % confidence interval for the mean age of all customers, assuming that the assumptions and conditions for the confidence 20 38 40 35 40 interval have been met. 22 37 10 41 26 b) How large is the margin of error? c) How would the confidence interval change if you had assumed that the standard deviation was known to be 11.0 years? 29 34 36 38 48

Answer 1

Given : Sample mean ($\bar{x}$) = 31.76 , Sample sd ( s ) = 10.37 , sample size ( n ) = 25

a) Population standard deviation σ is unknown and we are given sample standard deviation therefore we use T interval.

Confidence interval is given by,

Lower bound = $\bar{x}$  – E

Upper bound = $\bar{x}$ + E

Margin of error E =     

t * Vn

; t is critical value follows t distribution with degrees of freedom (d.f ) = n - 1

We have n = 25 and confidence level (c) = 98% or 0.98

d.f = n-1 = 25 -1 = 24

α = 1 – c = 1- 0.98 = 0.02

Therefore t_(0.02,24) = 2.492 ----- ( from t distribution table , for two tail )

Margin of error E = 5.1684

Lower bound = 31.76 - 5.1684 = 26.59

Upper bound = 31.76 + 5.1684 = 36.93

98% confidence interval for mean age for all customer is ( 26.59,36.93 )

b) Margin of error E = 5.1684

c) If standard deviation is known  σ = 11

Since standard deviation is known we have to use Z interval.

Margin of error E =

/n

z is critical value follows standard normal distribution, we can find its value using z score table .

α = 1 – c = 1- 0.98 = 0.02

α = 0.02/ 2 = 0.01

Therefore z_(0.01) = 2.33 ----- ( from z score table )

E =

2.33 11 V25

E = 5.126

Lower bound = 31.76 - 5.126 = 26.63

Upper bound = 31.76 + 5.126 = 36.89

98% confidence interval for mean age for all customer is ( 26.63,36.89 )

So width of confidence interval is decreases ,as standard deviation increases.