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.

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 with min (n1-1,n2-1)df

The pooled variance estimate Sp equals:

Sp²=2.7456

Sp=1.65699

Degrees of freedom=n1+n2-2=32

Under null hypothesis:

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

we have three similar triangles, because each has a right angle and shares an angle.   let's write the angles in order: opposite to short leg, long leg, hypotenuse.

cab similar to bad similar to cbd

or as ratios,

ca: ab: cb = ba: ad: bd = cb: bd: cd

we also know

ac = ad + cd

(ad+cd): ab: cb = ba: ad: bd = cb: bd: cd

(ad+cd)/cb=cb/cd

we have cb=6, ad=5 and seek x=cd.

we reject the negative root and conclude x=4

answer: 4

320ft

step-by-step explanation:

120 divided by 3 equals 40

40 times 4=160

160 times 2=320