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Nimfa-mama [501]
3 years ago
7

If two samples A and B had the same mean and standard deviation, but sample A had a larger sample size, which sample would have

the wider 95% confidence interval? Sample A as it comes first Sample B as its sample is more dispersed Sample A as it has the larger sample Sample B as it has the smaller sample
Mathematics
1 answer:
Advocard [28]3 years ago
7 0
<h3>Answer: Sample B as it has the smaller sample (choice #4)</h3>

===========================================================

Explanation:

Recall that the margin of error (MOE) is defined as

MOE = z*s/sqrt(n)

The sample size n is located in the denominator, meaning that as n gets bigger, the MOE gets smaller. The same happens in reverse: as n gets smaller, the MOE gets bigger.

Put another way, a small sample size means we have more error because small samples mean they are less representative of the population at large. The bigger a sample is, the better estimate we will have of the parameter.

We are told that "sample A had a larger sample size" indicating that sample A has a more narrow confidence interval.

Therefore, sample B would have a wider confidence interval.

This is true regardless of what the confidence level is set at.

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Answer:

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Step-by-step explanation:

We have to find the quotient of the following division, \frac{15p^{-4}q^{-6} }{- 20p^{-12} q^{-3}}.

Now, \frac{15p^{-4}q^{-6} }{- 20p^{-12} q^{-3}}

= - \frac{3}{4} p^{[- 4 - (- 12)]} q^{[-6 - (- 3)]} {Since all the terms in the expression are in product form, so we can treat them separately}

{Since we know the property of exponent as \frac{a^{b} }{a^{c} } = a^{(b - c)}}

= - \frac{3}{4} p^{8} q^{-3}

= - \frac{3}{4} \times  \frac{p^{8} }{q^{3} } (Answer)

{Since we know, a^{-b} = \frac{1}{a^{b} }}

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Answer:

\large\boxed{-2,\ -\dfrac{3}{4},\ -0.45,\ 3\%,\ 0.36}

Step-by-step explanation:

We convert some numbers to decimal fractions:

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other method

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