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3 years ago

A survey of 225 students showed the mean number of hours spent studying per week was 20.6 and the standard deviations was 2.7

Mathematics

2 answers:

Step2247 [10]3 years ago

6 0

Answer:

.3 is the answer.

Step-by-step explanation:

LekaFEV [45]3 years ago

5 0

Answer:

The margin of error is approximately 0.3

Step-by-step explanation:

The following information has been provided;

The sample size, n =225 students

The sample mean number of hours spent studying per week = 20.6

The standard deviation = 2.7

The question requires us to determine the margin of error that would be associated with a 90% confidence level. In constructing confidence intervals of the population mean, the margin of error is defined as;

The product of the associated z-score and the standard error of the sample mean. The standard error of the sample mean is calculated as;

$\frac{sigma}{\sqrt{n} }$

where sigma is the standard deviation and n the sample size. The z-score associated with a 90% confidence level, from the given table, is 1.645.

The margin of error is thus;

$1.645*\frac{2.7}{\sqrt{225}}=0.2961$

Therefore, the margin of error is approximately 0.3

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Answer:

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Step-by-step explanation:

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4 years ago

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3 years ago

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Dmitry_Shevchenko [17]

Answer:

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Step-by-step explanation:

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3 Possible Outcomes:

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