A random sample of 12 families were asked how many kids they had. The data are given below. According to the US Census, families have on average 2.4 kids per family. What is the p-value for the alternative hypothesis that the average number of kids per family is different than the US Census value? 1 1 2 2 2 3 3 3 3 4 4 6
A) p-value = 1.070.
B) p-value = 0.154.
C) p-value = 0.846.
D) p-value = 0.308.

0.308

Step-by-step explanation:

Given the data:

1, 1, 2, 2, 2, 3, 3, 3, 3, 4, 4, 6

We need to obtain the sample mean and sample standard deviation :

Sample mean, xbar = Σx/n

Where n = sample size (1+1+2+2+2+3+3+3+3+4+4+6) / 12 = 2.833

Sample standard deviation, s

Using the formula :

s = √Σ((x - xbar)² / n-1)

s = 1.40

Hypothesis :

H0 : μ = 2.4

H1 : μ ≠ 2.4

The test statistic :

(xbar - μ) ÷ (s/√(n))

(2.833 - 2.40) / (1.40/sqrt(12))

Test statistic = 1.0713

Using the Pvalue from Test score calculator :

df = n - 1 = 12 - 1 = 11 ; two tailed

Pvalue(1.071, 11)

= 0.3069

= 0.307

step-by-step explanation:

ok so im not 100% sure but you would start by adding 5.1t and 1.9t beacuse they have the same variable. that is 7t.   then add -3.6 + 1.2 which is -2.4

so then keep that as -2.4 + 7t. i apologize if this is incorrect. best luck on your test < 3