3 years ago

HELP PLZZZ ASAP BRAINLYIST IS AVAILABLE FOR WHOEVER ANSWERS

Mathematics

1 answer:

raketka [301]3 years ago

4 0

1. 15% decay
2. 50% growth
3. 15% growth
4. 5% growth
5. 50% decay

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In a survey of 300 T.V. viewers, 40% said they watch network news programs. Find the margin of error for this survey if we want

bixtya [17]

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Step-by-step explanation:

The formula for calculating the margin of error is expressed as;

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Substituting this values into the formula will give;

$M.E = 1.96*\sqrt{\frac{0.4(1-0.4)}{300} }\\ \\M.E = 1.96*\sqrt{\frac{0.4(0.6)}{300} }\\\\\\M.E = 1.96*\sqrt{\frac{0.24}{300} }\\\\\\M.E = 1.96*\sqrt{0.0008}\\\\\\M.E = 1.96*0.02828\\\\M.E \approx 0.05543$

Hence, the margin of error for this survey if we want 95% confidence in our estimate of the percent of T.V. viewers who watch network news programs is approximately 0.05543

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