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mr_godi [17]
2 years ago
15

11. An experiment is set up to study the weights of 45 frogs after they are given a hormone treatment. The population mean is 22

.70 grams with a standard deviation of 2.96 grams.
What is the standard error of the sample mean? Round your answer to the nearest hundredth.

The standard error of the sample mean is approximately (? ).
Mathematics
1 answer:
sveta [45]2 years ago
5 0

Answer: The standard error of the sample mean = 0.44

Step-by-step explanation:

Standard error of the sample mean = \dfrac{\sigma}{\sqrt{n}}<em> , where </em>\sigma = population standard deviation , n= sample size.

Given : n= 45 , \sigma= 2.96 grams

The e standard error of the sample mean =\dfrac{2.96}{\sqrt{45}}

=\dfrac{2.96}{6.7082039325}\\\\\approx\ 0.44

Hence, the standard error of the sample mean = 0.44

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The solution to the product of 8 and three-seventh is 2 2/7. According to Laura, she made mistake in step 2 by adding 8 and 3 instead of multiplying

Multiplication of fractions and integers

Fractions are written as a ratio of two integers. For instance a/b is a fraction where a and b are integers.

Given the following equation

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This is expressed mathematically as;

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Step 2: Group the numerator

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The solution to the product of 8 and three-seventh is 2 2/7. According to Laura, she made mistake in step 2 by adding 8 and 3 instead of multiplying

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As a store manger you would like to find out the average time it takes to unload the truck which delivers the merchandise for yo
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Answer:

The upper boundary of the 95% confidence interval for the average unload time is 264.97 minutes

Step-by-step explanation:

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The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

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Now, we have to find a value of T, which is found looking at the t table, with 34 degrees of freedom(y-axis) and a confidence level of 1 - \frac{1 - 0.9}{2} = 0.975([tex]t_{975}). So we have T = 2.0322

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The upper end of the interval is the sample mean added to M. So it is 204 + 60.97 = 264.97

The upper boundary of the 95% confidence interval for the average unload time is 264.97 minutes

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