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irina1246 [14]
3 years ago
6

You measure 33 backpacks' weights, and find they have a mean weight of 36 ounces. Assume the population standard deviation is 12

.1 ounces. Based on this, what is the maximal margin of error associated with a 90% confidence interval for the true population mean backpack weight.
Mathematics
1 answer:
saul85 [17]3 years ago
8 0

Answer:

E=3.4363

Step-by-step explanation:

#Given a significance level of \alpha=1-CI=1-0.90=0.1 and

\bar x=36, \ \sigma=12.1 \ n=41, the critical value is,z_\alpha_/_2=z_\alpha_/_0_._0_5=1.645

#the margin of error can then be calculated as:

E=z_\alpha_/_2\times\frac{\sigma}{\sqrt n}=1.645\times \frac{12.1}{\sqrt 33}\\\approx 3.4363

Hence the margin of error is 3.4363

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