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kirza4 [7]
3 years ago
10

Select the correct answer.

Mathematics
2 answers:
kaheart [24]3 years ago
8 0

Answer:

Correct Answer (A)

We can be 95% certain that the sample mean will fall within <u>$44.87 </u>and <u>$46.93</u>

Step-by-step explanation:

Hope this helped :)

#Platolivesmatter

tigry1 [53]3 years ago
6 0

Answer:

We can be 95% certain that the sample mean will fall within $44.87 and $46.93

Step-by-step explanation:

Empirical Rule:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

Central Limit Theorem:

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Population:

Mean $45.90, standard deviation $10.31.

Sample:

By the Central limit theorem, mean $45.90, standard deviation s = \frac{10.31}{\sqrt{400}} = \frac{10.31}{20} = 0.5155

Which interval can the branch manager be 95% certain that the sample mean will fall within?

By the Empirical Rule, within 2 standard deviations of the mean. So

45.90 - 2*0.5155 = $44.87

45.90 + 2*0.5155 = $46.93

We can be 95% certain that the sample mean will fall within $44.87 and $46.93

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z=\frac{0.881 -0.8}{\sqrt{\frac{0.8(1-0.8)}{110}}}=2.124  

Null hypothesis:p\geq 0.8  

Alternative hypothesis:p < 0.8  

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1) Data given and notation  

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\hat p=\frac{97}{110}=0.882 estimated proportion of students who completed the modules before class

p_o=0.8 is the value that we want to test

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z would represent the statistic (variable of interest)

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Alternative hypothesis:p < 0.8  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

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3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.881 -0.8}{\sqrt{\frac{0.8(1-0.8)}{110}}}=2.124  

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