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Anni [7]
3 years ago
5

What is the correct way to reach a decision regarding a hypothesis test based on a confidence interval

Mathematics
1 answer:
Softa [21]3 years ago
3 0

<u>Answer</u>:

this is a multiple choice question and the answer to this question has been highlighted in bold letters.

Based on a confidence interval, the best way to get a decision on a hypothesis test is for you to fail to rej<u>e</u><u>c</u><u>t</u><u> </u><u>the </u><u>null</u><u> hypothesis</u><u> (</u><u>that </u><u>is,</u><u> </u><u>you </u><u>will</u><u> </u><u>have</u><u> to</u><u> </u><u>accept</u><u> the</u><u> null</u><u> hypothesis</u><u>)</u><u> </u><u>if </u><u>the </u><u>hypothesized</u><u> </u><u>value</u><u> </u><u>for</u><u> the</u><u> </u><u>parameter</u><u> </u><u>that </u><u>is </u><u>being</u><u> </u><u>tested</u><u> </u><u>is </u><u>within</u><u> the</u><u> </u><u>confidence</u><u> interval</u><u>.</u>

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Answer:

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3 years ago
The population of Henderson City was 3,381,000 in 1994, and is growing at an annual rate 1.8%
liq [111]
<h2>In the year 2000, population will be 3,762,979 approximately. Population will double by the year 2033.</h2>

Step-by-step explanation:

   Given that the population grows every year at the same rate( 1.8% ), we can model the population similar to a compound Interest problem.

   From 1994, every subsequent year the new population is obtained by multiplying the previous years' population by \frac{100+1.8}{100} = \frac{101.8}{100}.

   So, the population in the year t can be given by P(t)=3,381,000\textrm{x}(\frac{101.8}{100})^{(t-1994)}

   Population in the year 2000 = 3,381,000\textrm{x}(\frac{101.8}{100})^{6}=3,762,979.38

Population in year 2000 = 3,762,979

   Let us assume population doubles by year y.

2\textrm{x}(3,381,000)=(3,381,000)\textrm{x}(\frac{101.8}{100})^{(y-1994)}

log_{10}2=(y-1994)log_{10}(\frac{101.8}{100})

y-1994=\frac{log_{10}2}{log_{10}1.018}=38.8537

y≈2033

∴ By 2033, the population doubles.

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3 years ago
What does x=? I know it is vertical
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Answer:no

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