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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.

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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,

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

The pooled variance estimate Sp equals:

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

Sp²=2.7456

Sp=1.65699

Degrees of freedom=n1+n2-2=32

Under null hypothesis:

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

[tex]tcal=0.6989[/tex]

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