Family Size. You selected a random sample of n = 31 families in your neighborhood and found the mean family size for the sample equal to 3.1, the standard deviation for the sample is 1.42? What is the 90% confidence interval for the estimate?

Step to step explanation:

Confidence interval for mean, when population standard deviation is unknown:

, where = sample mean

n= sample size

s= sample standard deviation

= Critical t-value for n-1 degrees of freedom

We assume the family size is normal distributed.

Given, n= 31 , , s= 1.42 ,

Critical t value for and degree of 30 freedom

= 1.697  [By t-table]

The required confidence interval:

Hence,  the 90% confidence interval for the estimate = (2.67, 3.53)

just do 5.2 + 2 ( 7.2) and 1/3 x 3 (.9) and youv'e got your equation. 0.9( 9/10) + 7.2 ^2 = 16.4 hope i could !

step-by-step explanation:

d. vertex

step-by-step explanation: