in a sample of 81 u.s. adults aged 18 - 24 who celebrate halloween, the mean amount spent on a costume was…

in a sample of 81 u.s. adults aged 18 - 24 who celebrate halloween, the mean amount spent on a costume was $37.7 with a standard deviation of $14. round answers to at least 4 decimal places.\na) what parameter are we estimating? select an answerv\nb) we can work this problem because: select an answerv\nc) what is the point estimate of that parameter? select an answerv\nd) what is the standard deviation of the sampling distribution for the sample means?\ne) find the margin of error for a 90% confidence interval for the true average amount spent on a costume.\nf) we are 90% confident that the true average amount spent on a halloween costume for u.s. adults aged 18 - 24 is between $ and $. do not round between steps, and round answer to 4 decimal places.\nquestion help: message instructor

in a sample of 81 u.s. adults aged 18 - 24 who celebrate halloween, the mean amount spent on a costume was $37.7 with a standard deviation of $14. round answers to at least 4 decimal places.\na) what parameter are we estimating? select an answerv\nb) we can work this problem because: select an answerv\nc) what is the point estimate of that parameter? select an answerv\nd) what is the standard deviation of the sampling distribution for the sample means?\ne) find the margin of error for a 90% confidence interval for the true average amount spent on a costume.\nf) we are 90% confident that the true average amount spent on a halloween costume for u.s. adults aged 18 - 24 is between $ and $. do not round between steps, and round answer to 4 decimal places.\nquestion help: message instructor

Answer

Explanation:

Step1: Parameter estimation

We are estimating the population mean amount spent on a costume by U.S. adults aged 18 - 24 who celebrate Halloween.

Step2: Workability of the problem

We can work this problem because the sample size (n = 81\geq30) (by the Central Limit Theorem, for a sample of size (n\geq30), the sampling distribution of the sample mean (\bar{X}) is approximately normal).

Step3: Point estimate

The point estimate of the population mean (\mu) is the sample mean (\bar{x}). So, the point estimate is (\bar{x}=37.7)

Step4: Standard deviation of the sampling distribution

The formula for the standard deviation of the sampling distribution of the sample mean (also known as the standard error) is (\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}). Given (\sigma = 14) and (n = 81), we have (\sigma_{\bar{x}}=\frac{14}{\sqrt{81}}=\frac{14}{9}\approx1.5556)

Step5: Margin of error for 90% confidence interval

For a 90% confidence interval, the critical value (z_{\alpha/2}) (where (\alpha=1 - 0.90=0.10) and (\alpha/2=0.05)) from the standard normal distribution is (z_{0.05}\approx1.645). The margin of error (E = z_{\alpha/2}\times\sigma_{\bar{x}}). Substituting (z_{\alpha/2}=1.645) and (\sigma_{\bar{x}}=\frac{14}{9}), we get (E=1.645\times\frac{14}{9}\approx2.5639)

Step6: Confidence interval

The confidence interval for the population mean (\mu) is (\bar{x}-E<\mu<\bar{x} + E). Substituting (\bar{x}=37.7) and (E = 2.5639), we have (37.7-2.5639=35.1361) and (37.7 + 2.5639=40.2639)

Answer:

a) Population mean b) Sample size (n\geq30) (Central Limit Theorem) c) (37.7) d) (\frac{14}{9}\approx1.5556) e) (1.645\times\frac{14}{9}\approx2.5639) f) (35.1361) and (40.2639)