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a_sh-v [17]
3 years ago
10

A health club annually surveys its members. Last year, 33% of the members said they use the treadmill at least four times a week

. How large a sample should be selected this year to estimate the percentage of members who use the treadmill at least four times a week? The estimate is desired to have a margin of error of 5% with a 95% level of confidence.
Mathematics
1 answer:
timofeeve [1]3 years ago
8 0

Answer:

The large  sample n = 174.24

Step-by-step explanation:

Given 33% of the members said they use the treadmill at least four times a week.

Given the margin of error is 5% = 0.05

we know that the margin of error at 95% 0f level of confidence  = \frac{2S.D}{\sqrt{n} }

Given standard deviation = 33% = 0.33

M .E = \frac{2S.D}{\sqrt{n} }

0.05 = \frac{2X0.33}{\sqrt{n} }

cross multiplication , we get

\sqrt{n}  = \frac{2X0.33}{0.05}

√n = 13.2

squaring on both sides, we get

n = 174.24

<u>Conclusion:-</u>

The large  sample n = 174.24

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<h3>What is trigonometry?</h3>

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Trigonometric Identities are equality statements that hold true for all values of the variables in the equation and that use trigonometry functions. There are numerous distinctive trigonometric identities that relate to a triangle's side length and angle.

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