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Sagot :
Answer: 8.323 (approximate)
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Explanation:
s = 32 = sample standard deviation
n = 40 = sample size
Despite not knowing the population standard deviation (sigma), we can still use the Z distribution because n > 30. When n is this large, the student T distribution is approximately the same (more or less) compared to the standard Z distribution. The Z distribution is nicer to work with.
At 90% confidence, the z critical value is roughly z = 1.645. This can be found using either a Z table or a calculator.
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We have these values:
- z = 1.645 (approximate)
- s = 32
- n = 40
Plug these values into the margin of error formula.
[tex]E = z*\frac{s}{\sqrt{n}}\\\\E \approx 1.645*\frac{32}{\sqrt{40}}\\\\E \approx 8.32311480156318\\\\E \approx 8.323\\\\[/tex]
The margin of error is roughly 8.323
I rounded to 3 decimal places because z = 1.645 is rounded to 3 decimal places. If your teacher wants some other level of precision, then be sure to follow those instructions.
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