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soldier1979 [14.2K]
3 years ago
14

The mean number of children among a sample of 15 low-income households is 2.8. The mean number of children among a sample of 19

high-income households is 2.4. The standard deviations for low- and high-income households are found to be 1.6 and 1.7, respectively. Test the hypothesis of no difference against the alternative that high-income households have fewer children. Use a = 0.05 and a pooled estimate of the variance.
Mathematics
1 answer:
gtnhenbr [62]3 years ago
6 0

Answer:

Step-by-step explanation:

Let's put the data as below:

n1=15    x1=2.8      s1=1.6       and s1²=2.56

n2=19   x2=2.4     s2=1.7        and s2²=2.89

alpha= 0.05

To test the hypothesis:

H0= There is no sufficient evidence that low income household have fewer children

H1=There is sufficient evidence that low income household have fewer children

Assume that population variances are equal.

the t-static for two samples,

t=\frac{x1-x2}{Sp\sqrt{\frac{1}{n1}-\frac{1}{n2} \\} } } ~t with min (n1-1,n2-1)df

The pooled variance estimate Sp equals:

Sp^{2}=\frac{(n1-1)s1^{2}+(n2-1)s2^{2}  }{n1+n2-2}

Sp²=2.7456

Sp=1.65699

Degrees of freedom=n1+n2-2=32

Under null hypothesis:

tcal=\frac{|2.8 - 2.4|}{1.6599\sqrt{\frac{1}{15}+\frac{1}{19}  } }

tcal=0.6989

The critical value ttab=2.0369 for alpha=0.05

So we reject our null hypothesis H0

So there is sufficient evidence that low income households have fewer children than high income households

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Graph attached.

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Answer:

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m∠JLN = 42°

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Step-by-step explanation:

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We know that ∠GHD = 2x° and ∠AGH (x+15)°.

You correctly calculated that x=15°.

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So m∠AGE = 160°

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