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LUCKY_DIMON [66]
3 years ago
9

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.
Mathematics
1 answer:
Lana71 [14]3 years ago
7 0

Answer:

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

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Compute the mean of this sample as follows:

\bar x_{N}=\frac{x_{1}+1+x_{2}+1+x_{3}+1+...+x_{n}+1}{n}\\=\frac{(x_{1}+x_{2}+x_{3}+...+x_{n})}{n}+\frac{(1+1+1+...n\ times)}{n}\\=\bar x+1\\=4+1\\=5

The new mean is 5.

Compute the standard deviation of this sample as follows:

s_{N}=\frac{1}{n-1}\sum (x_{i}-\bar x)^{2}\\=\frac{1}{n-1}\sum ((x_{i}+1)-(\bar x+1))^{2}\\=\frac{1}{n-1}\sum (x_{i}+1-\bar x-1)^{2}\\=\frac{1}{n-1}\sum (x_{i}-\bar x)^{2}\\=s

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Answer:

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Step-by-step explanation:

we know that

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sum=\frac{n}{2}[2a1+(n-1)d]

where

a1 is the first term

n is the number of terms (number of friends)

d is the common difference in the arithmetic sequence

In this problem we have

sum=275\ stickers

a1=5\ stickers

d=5 ----> the common difference

substitute in the formula and solve for n

275=\frac{n}{2}[2(5)+(n-1)(5)]

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see the attached figure

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