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3 years ago

100 POINTS!! I WILL REPORT ANSWERS THAT DO NOT HELP!!!

1 answer:

7 0

A reasonable person might come to the conclusion that monetary policy should be set by rules that limit the discretion of policymakers - for the reason that policymakers cannot be trusted to do what they say - because monetary policy better combats economic slowdowns and reduces the cost of investment. The input of policymakers should be limited because of effect that monetary policy has on an important tell of how policymakers are doing - unemployment rate. When more money is pumped into the economy through monetary policies, the unemployment rate drops, so it is easy to see how tempting it is for policymakers to foucs too much on monetary policies or twist them to benefit themselves and their campaign the most.

To improve policymakers' performance, I would set limits to how money could be added to an economy at any time, or try to spread out the time over which the money is added. This would put a limit on policymakers that would potentially reduce the temptation to overdo monetary policy.

I hope this helps. I'm not sure how long you wanted it.

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company manufactured six television sets on a given day, and these TV sets were inspected for being good or defective. The resul

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Sampling distribution involves the proportions of a data element in a given sample.

The proportion of Good TV set is 0.67
The number of ways of selecting 5 from 6 TV sets is 6
The number of ways of selecting 4 from 6 TV sets is 15

Given

$n = 6$

Sample Space = Good, Good, Defective, Defective, Good, Good

(a) Proportion that are good

From the sample space, we have:

$Good = 4$

So, the proportion (p) that are good are:

$p = \frac{Good}{n}$

$p = \frac{4}{6}$

$p = 0.67$

(b) Ways to select 5 samples (without replacement)

This is calculated using:

$^nC_r = \frac{n!}{(n - r)!r!}$

Where

$r = 5$

So, we have:

$^6C_5 = \frac{6!}{(6 - 5)!5!}$

$^6C_5 = \frac{6!}{1!5!}$

$^6C_5 = \frac{6 \times 5!}{1 \times 5!}$

$^6C_5 = \frac{6}{1}$

$^6C_5 = 6$

Hence, there are 6 ways

(c) All possible sample space of 4

First, we calculate the number of ways to select 4.

This is calculated using:

$^nC_r = \frac{n!}{(n - r)!r!}$

Where

$r = 4$

So, we have:

$^6C_4 = \frac{6!}{(6 - 4)!4!}$

$^6C_4 = \frac{6!}{2!4!}$

$^6C_4 = \frac{6 \times 5 \times 4}{2 \times 1 \times 4!}$

$^6C_4 = \frac{30}{2}$

$^6C_4 = 15$

So, the table is as follows:

$\left[\begin{array}{ccc}TV&Good&Proportion\\1,2,3,4&2&0.5&2,3,4,5&2&0.5&3,4,5,6&2&0.5\\4,5,6,1&3&0.75&5,6,1,2&4&1&6,1,2,3&3&0.75\\1,2,3,5&3&0.75&3,5,6,2&3&0.75&1,3,4,5&2&0.5\\1,3,4,6&2&0.5&1,4,5,2&3&0.75&2,4,6,1&3&0.75\\2,4,6,3&2&0.5&2,4,6,5&3&0.75&3,5,6,1&3&0.75\end{array}\right]$

The proportion column is calculated by dividing the number of Good TVs by the total selected (4) i.e.

$p = \frac{Good}{n}$

(d) The sampling distribution

In (a), we have:

$p = 0.67$ --- proportion of Good TV

The sampling error is calculated as follows:

$SE_n = |p - p_n|$

So, we have:

$\left[\begin{array}{ccc}TV&Good&SE\\1,2,3,4&2&0.17&2,3,4,5&2&0.17&3,4,5,6&2&0.17\\4,5,6,1&3&0.08&5,6,1,2&4&0.33&6,1,2,3&3&0.08\\1,2,3,5&3&0.08&3,5,6,2&3&0.08&1,3,4,5&2&0.17\\1,3,4,6&2&0.17&1,4,5,2&3&0.08&2,4,6,1&3&0.08\\2,4,6,3&2&0.17&2,4,6,5&3&0.08&3,5,6,1&3&0.08\end{array}\right]$

Read more about sampling distributions at:

brainly.com/question/10554762

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3 years ago