In a survey of 1005 adult Americans, 46.6% indicated that they were somewhat interested or very interested in having web access
1 answer:
Answer:
No, the marketing manager was not correct in his claim.
Step-by-step explanation:
We are given that in a survey of 1005 adult Americans, 46.6% indicated that they were somewhat interested or very interested in having web access in their cars.
Suppose that the marketing manager of a car manufacturer claims that the 46.6% is based only on a sample and that 46.6% is close to half, so there is no reason to believe that the proportion of all adult Americans who want car web access is less than 0.50.
<em>Let p = population proportion of all adult Americans who want car web access</em>
SO, Null Hypothesis, : p
50% {means that the proportion of all adult Americans who want car web access is more than or equal to 0.50}
Alternate Hypothesis, : p < 50% {means that the proportion of all adult Americans who want car web access is less than 0.50}
The test statistics that will be used here is<u> One-sample z proportion statistics</u>;
T.S. = ~ N(0,1)
where, = sample proportion of Americans who indicated that they were somewhat interested or very interested in having web access in their cars = 46.6%
n = sample of Americans = 1005
So, <u><em>test statistics</em></u> =
= -2.161
<em>Since in the question we are not given the level of significance so we assume it to be 5%. Now at 5% significance level, the z table gives critical value of -1.6449 for left-tailed test. Since our test statistics is less than the critical value of z so we sufficient evidence to reject our null hypothesis as it will fall in the rejection region.</em>
Therefore, we conclude that the proportion of all adult Americans who want car web access is less than 0.50 which means the marketing manager was not correct in his claim.
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Answer:
a) An alternating series is a whose terms are alternately positive and negative
b) An alternating series where bn = |an|, converges if
for all n, and
0
c) The error involved in using the partial sum sn as an approximation to the total sum s is the remainder Rn = s − sn and the size of the error is bn + 1
Step-by-step explanation:
<em>Part a</em>
An Alternating series is an infinite series given on these three possible general forms given by:
For all
The initial counter can be n=0 or n =1. Based on the pattern of the series the signs of the general terms alternately positive and negative.
<em>Part b</em>
An alternating series where bn = |an| converges if
for all n and
=0
Is necessary that limit when n tends to infinity for the nth term of bn converges to 0, because this is one of two conditions in order to an alternate series converges, the two conditions are given by the following theorem:
<em>Theorem (Alternating series test)</em>
If a sequence of positive terms {bn} is monotonically decreasing and
<em><em>, then the alternating series
converges if:</em></em>
<em>i) </em>
<em>ii) </em>
then <em> converges</em>
<em>Proof</em>
For this proof we just need to consider the sum for a subsequence of even partial sums. We will see that the subsequence is monotonically increasing. And by the monotonic sequence theorem the limit for this subsquence when we approach to infinity is a defined term, let's say, s. So then the we have a bound and then
for all n, and that implies that the series converges to a value, s.
And this complete the proof.
<em>Part c</em>
An important term is the partial sum of a series and that is defined as the sum of the first n terms in the series
By definition the Remainder of a Series is The difference between the nth partial sum and the sum of a series, on this form:
Rn = s - sn
Where represent the partial sum for the series and s the total for the sum.
Is important to notice that the size of the error is at most by the following theorem:
<em>Theorem (Alternating series sum estimation)</em>
<em>If is the sum of an alternating series that satisfies</em>
<em>i) </em>
<em>ii) </em>
Then then
<em>Proof</em>
In the proof of the alternating series test, and we analyze the subsequence, s we will notice that are monotonically decreasing. So then based on this the sequence of partial sums sn oscillates around s so that the sum s always lies between any two consecutive partial sums sn and sn+1.
And this complete the proof.