You want to rent an unfurnishedone-bedroomapartment for next semester. The mean monthlyrent for a random
sample of 10 apartments advertised in the localnewspaper is $580.
Assume that the standard deviation is$90. Find a 95% confidence
interval for the mean monthly rentfor unfurnished one bedroom
apartments available for rent in thiscommunity.

Answers

Answer 1
Answer:

Answer: (\$524.22,\ \$635.78)

Step-by-step explanation:

Confidence interval for population mean is given by :-

\overline{x}\pm z^* (\sigma)/(√(n))

, where \overline{x} = Sample mean

z* =  critical z-value.

\sigma = Population standard deviation.

n= Sample size.

Let x be the denotes the monthly rent for unfurnished one bedroom  apartments available for rent in this community.

As per given , we have

n= 10

\overline{x}=\$580

\sigma=\$90

Critical value for 95% confidence : z* = 1.96

So the 95% confidence  interval becomes,

580\pm (1.96) (90)/(√(10))

=580\pm (1.96) (90)/(3.162278)

=580\pm (1.96)(28.46)

=580\pm 55.78

=(580-55.78,\ 580+55.78)=(524.22,\ 635.78)

Hence, a 95% confidence  interval for the mean monthly rent for unfurnished one bedroom  apartments available for rent in this community= (\$524.22,\ \$635.78)


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Answers

Answer:

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Answer:

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Step-by-step explanation:

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Answers

Answer:

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Step-by-step explanation:

hope this helps :)

Answer:

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Step-by-step explanation:

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Answers

Answer:

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Step-by-step explanation:

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Answers

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Final answer:

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Explanation:

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Answers

Answer:

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Step-by-step explanation:

Steps:

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