Question #98710
Question #3 You want to rent an unfurnished one-bedroom apartment in Boston next year. A 95% confidence interval for the true mean monthly rent of all one-bedroom apartments in Boston is (1263.64, 1536.36). What is the margin of error for this interval?
Question #4 A student curious about the average number of chocolate chips in a commercial brand of cookie estimated the standard deviation to be 8. If he wants to be 99% confident in his results, how manychocolate chip cookies will he need to sample to estimate the mean to within 2?
1
Expert's answer
2019-11-15T09:59:05-0500

1.


margin of error=12×(length of the interval)margin\ of\ error={1 \over 2}\times (length\ of\ the\ interval)

margin of error=12×(1536.361263.64)margin\ of\ error={1 \over 2}\times(1536.36-1263.64)

margin of error=136.36margin\ of\ error=136.36

2.Based on the information provided, the significance level is α=0.01\alpha=0.01 (99%99\% confidence interval), and the critical value for a two-tailed test is zc=2.58z_c=2.58


margin of error=zc×σnmargin\ of\ error=z_c\times{\sigma \over\sqrt{n}}

Given that σ=8,margin of error2.\sigma=8, margin\ of\ error\leq2. Then


2.58×8n22.58\times{8 \over\sqrt{n}}\leq2

n(2.58×82)2n\geq({2.58\times8 \over 2})^2

n106.5n\geq106.5

Since nn is integer, then n107.n\geq107.



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