A family is relocating from St. Louis, Missouri, to California. Due to an increa
ID: 3242470 • Letter: A
Question
A family is relocating from St. Louis, Missouri, to California. Due to an increasing inventory of houses in St. Louis, it is taking longer than before to sell a house. The wife is concerned and wants to know when it is optimal to put their house on the market. Her realtor friend informs them that the last 24 houses that sold in their neighborhood took an average time of 135 days to sell. The realtor also tells them that based on her prior experience, the population standard deviation is 88 days. Use Table 1.
What assumption regarding the population is necessary for making an interval estimate for the population mean?
Construct the 90% confidence interval for the mean sale time for all homes in the neighborhood. (Round intermediate calculations to 4 decimal places. Round "z-value" to 3 decimal places and final answer to 2 decimal places.)
A family is relocating from St. Louis, Missouri, to California. Due to an increasing inventory of houses in St. Louis, it is taking longer than before to sell a house. The wife is concerned and wants to know when it is optimal to put their house on the market. Her realtor friend informs them that the last 24 houses that sold in their neighborhood took an average time of 135 days to sell. The realtor also tells them that based on her prior experience, the population standard deviation is 88 days. Use Table 1.
Explanation / Answer
Solution:-
a). What assumption regarding the population is necessary for making an interval estimate for the population mean?
Ans :- Assume that the population has a normal distribution.
b). Construct the 90% confidence interval for the mean sale time for all homes in the neighborhood. (Round intermediate calculations to 4 decimal places. Round "z-value" to 3 decimal places and final answer to 2 decimal places.)
Ans:- CI for Mean 104.283 < µ < 165.717
Calculation,
If (n<30),
CI = x ± t/2 × (/n)
Where, x = Mean
= Standard Deviation
= 1 - (Confidence Level/100)
t/2 = t-table value
CI = Confidence Interval
CI = x ± t/2 × (/n)
= 135 + t0.05 * (88/24)
90% CI is [104.283, 165.717]
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