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Answer:
Yes, this provide enough evidence to show a difference in the proportion of households that own a desktop.
Step-by-step explanation:
We are given that National data indicates that 35% of households own a desktop computer.
In a random sample of 570 households, 40% owned a desktop computer.
<em><u>Let p = population proportion of households who own a desktop computer</u></em>
SO, Null Hypothesis, : p = 25% {means that 35% of households own a desktop computer}
Alternate Hypothesis, : p
25% {means that % of households who own a desktop computer is different from 35%}
The test statistics that will be used here is <u>One-sample z proportion</u> <u>statistics</u>;
T.S. = ~ N(0,1)
where, = sample proportion of 570 households who owned a desktop computer = 40%
n = sample of households = 570
So, <u><em>test statistics</em></u> =
= 2.437
<em>Since, in the question we are not given with the level of significance at which to test out hypothesis so we assume it to be 5%. Now at 5% significance level, the z table gives critical values of -1.96 and 1.96 for two-tailed test. Since our test statistics doesn't lies within the range of critical values of z so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.</em>
Therefore, we conclude that % of households who own a desktop computer is different from 35%.
Answer:
[a,b] divides n
Step-by-step explanation:
Let us denote the least common multiple of a and b [a,b]=m.
We want to prove that m divides n, where n is a multiple of a and b.
We suppose m does not divide n, then by the Division Theorem, there exists q and r integers such that:
(1) ... n=mq+r, where 0<r<m
As n is a multiple of a and b, there exists s and t integers such that:
sa=n and tb=n
Same thing happens to m as it is the least common multiple, there exists u and v such that:
ua=m and vb=m
So (1) has the following form:
n=mq+r ⇒ sa=uaq+r ⇒sa-uaq=r⇒(s-uq)a=r and
n=mq+r ⇒ tb=vbq+r ⇒ tb-vbq=r⇒ (t-vq)b=r
So r is a multiple of a and b, but r<m which is a contradiction as, m is the least common multiple of a and b. So this concludes the proof.
So this means that is and integer.
As m= vb, then is an integer, lets say
=v; and as m=ua, then
=u.
So v=
=a, so
divides a; on the other hand,
u=
=b, so
divides b. From this we can conclude that
is a common divisor of a and b.