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 25 houses that sold in their neighborhood took an average time of 85 days to sell. The realtor also tells them that based on her prior experience, the population standard deviation is 12 days.

a. What assumption regarding the population is necessary for making an interval estimate for the population mean?○ Assume that the population has a normal distribution.○ Assume that the central limit theorem applies. b. Construct the 99% confidence interval for the…

The amount of weight a high school student carries in his or her bookbag follows a normal distribution with a standard deviation of σ = 1.9 pounds. Suppose a random sample of 23 bookbags produced a mean of 12.2 pounds. Construct a 95% confidence interval to estimate the mean bookbag weight for all high school students.

Answer:Given: Sample Mean $ (x\bar)=12.2 $ Population Standard Deviation $ (\sigma)=1.9 $ Sample Size $ (n) = 23 $ Confidence Interval Level $ (CI) = 95\% $ Solution: The level of significance $ (\alpha):$ $ \alpha = 1 – 0.95…

The amount of weight a high school student carries in his or her bookbag follows a normal distribution with a standard deviation of σ = 2 pounds. Suppose a random sample of 17 bookbags produced a mean of 13.8 pounds. Construct a 95% confidence interval to estimate the mean bookbag weight for all high school students.

Answer:Given: Sample Mean $ (x\bar)=13.8 $ Population Standard Deviation $ (\sigma)=2 $ Sample Size $ (n) = 17 $ Confidence Interval Level $ (CI) = 95\% $ Solution: The level of significance $ (\alpha):$ $ \alpha = 1 – 0.95…

The amount of weight a high school student carries in his or her bookbag follows a normal distribution with a standard deviation of σ = 2.7 pounds. Suppose a random sample of 21 bookbags produced a mean of 16 pounds. Construct a 95% confidence interval to estimate the mean bookbag weight for all high school students.

Answer:Given: Sample Mean $ (x\bar)=16 $ Population Standard Deviation $ (\sigma)=2.7 $ Sample Size $ (n) = 21 $ Confidence Interval Level $ (CI) = 95\% $ Solution: The level of significance $ (\alpha):$ $ \alpha = 1 – 0.95…

The amount of weight a high school student carries in his or her bookbag follows a normal distribution with a standard deviation of σ = 3 pounds. Suppose a random sample of 15 bookbags produced a mean of 14.2 pounds. Construct a 95% confidence interval to estimate the mean bookbag weight for all high school students.

Answer:Given: Sample Mean $ (x\bar)=14.2 $ Population Standard Deviation $ (\sigma)=3 $ Sample Size $ (n) = 15 $ Confidence Interval Level $ (CI) = 95\% $ Solution: The level of significance $ (\alpha):$ $ \alpha = 1 – 0.95…

The amount of weight a high school student carries in his or her bookbag follows a normal distribution with a standard deviation of σ = 1.6 pounds. Suppose a random sample of 20 bookbags produced a mean of 12.5 pounds. Construct a 95% confidence interval to estimate the mean bookbag weight for all high school students.

Answer:Given: Sample Mean $ (x\bar)=12.5 $ Population Standard Deviation $ (\sigma)=1.6 $ Sample Size $ (n) = 20 $ Confidence Interval Level $ (CI) = 95\% $ Solution: The level of significance $ (\alpha):$ $ \alpha = 1 – 0.95…

The amount of weight a high school student carries in his or her bookbag follows a normal distribution with a standard deviation of σ = 2.2 pounds. Suppose a random sample of 24 bookbags produced a mean of 13.5 pounds. Construct a 95% confidence interval to estimate the mean bookbag weight for all high school students.

Answer:Given: Sample Mean $ (x\bar)=13.5 $ Population Standard Deviation $ (\sigma)=2.2 $ Sample Size $ (n) = 24 $ Confidence Interval Level $ (CI) = 95\% $ Solution: The level of significance $ (\alpha):$ $ \alpha = 1 – 0.95…

In a survey, 10 people were asked how much they spent on their child’s last birthday gift. The results were roughly bell-shaped with a mean of 48 and standard deviation of 13. Construct a confidence interval at a 99% confidence level.

Answer : Given : The sample mean ${(x\bar)} = 48$ The sample standard deviation $(s) = 13$ The sample size $(n) = 10$ Confidence level = 99% Solution : For a 99% confidence level, $$\alpha = 1 – 0.99 =…

Construct the confidence interval for the population mean μ. c=0.94, xˉ=7.8, σ=0.9, and n=55. A 94% confidence interval for μ is ___ , ___. (Round to two decimal places as needed.)

Answer:Given:Sample mean $(\bar{x}) = 7.8$ Population standard deviation $(\sigma) = 0.9$ Sample size $(n) = 55$ Confidence interval level $= 94\%$ Solution:The significance level $(\alpha):$$\therefore (\alpha) = 1 – 0.94 = 0.06$ The critical value $(z_c):$The critical value at 0.06…

Construct the confidence interval for the population mean μ. c=0.96, xˉ=9.3, σ=0.6, and n=45. A 96% confidence interval for μ is ___ , ___. (Round to two decimal places as needed.)

Answer:Given:Sample mean $(\bar{x}) = 9.3$ Population standard deviation $(\sigma) = 0.6$ Sample size $(n) = 45$ Confidence interval level $= 96\%$ Solution:The significance level $(\alpha):$$\therefore (\alpha) = 1 – 0.96 = 0.04$ The critical value $(z_c):$The critical value at 0.04…