A randomly selected 14 people were asked how long they slept at night. The mean time was 5.5 hours, and the standard deviation was 0.77 hour. Find the 80% confidence interval of the mean time. Assume the variable is normally distributed.

a. The $t_{\alpha/2}$ at 80% confidence level is equal to:b. Find the best point estimate of the population mean. Detailed Answer with Explanation: Calculating an 80% Confidence Interval for Average Sleep Time Sleep is a vital part of daily life,…

A randomly selected 14 people were asked how long they slept at night. The mean time was 5.5 hours, and the standard deviation was 0.77 hour. Find the 80% confidence interval of the mean time. Assume the variable is normally distributed.

a. The $t_{\alpha/2}$ at 80% confidence level is equal to:b. Find the best point estimate of the population mean. Answer: Given Data: Solution: a) The $ t_{\frac{\alpha}{2}} $ at the $ 80% $ confidence level: The significance level is calculated…

Find the critical values for a 90% confidence interval using the chi-square distribution with 13 degrees of freedom. Round the answers to three decimal places.

Finding Critical Values for a 90% Confidence Interval Using the Chi-Square Distribution In statistics, the Chi-Square distribution is frequently used to estimate population variances and test hypotheses involving categorical data. One of its important applications is in constructing confidence intervals…

The Chartered Financial Analyst (CFA) designation is fast becoming a requirement for serious investment professionals. Although it requires a successful completion of three levels of grueling exams, the designation often results in a promising career with a lucrative salary. A student of finance is curious about the average salary of a CFA charterholder. He takes a random sample of 49 recent charterholders and computes a mean salary of $100,000 $with a standard deviation of $42,000.$ Use this sample information to determine the 90% confidence interval for the average salary of a CFA charterholder.

Answer: we’ll calculate the 90% confidence interval for the average salary of CFA charterholders based on sample data. Problem Statement We are given the following sample data from a survey of recent CFA charterholders: The task is to calculate the…

The Chartered Financial Analyst (CFA) designation is fast becoming a requirement for serious investment professionals. Although it requires a successful completion of three levels of grueling exams, the designation often results in a promising career with a lucrative salary. A student of finance is curious about the average salary of a CFA charterholder. He takes a random sample of 49 recent charterholders and computes a mean salary of $100,000 $with a standard deviation of $42,000.$ Use this sample information to determine the 90% confidence interval for the average salary of a CFA charterholder.

Answer:Given Data: Solution: The significance level: $ \alpha = 1 – 0.90 = 0.10 $ The degrees of freedom: $ df = n – 1 = 49 – 1 = 48 $ The critical value: $ t_c = t_{\frac{\alpha}{2}, df}…

A random sample of 100 credit sales in a department store showed an average sale of $130.00.$ From past data, it is known that the standard deviation of the population is $35.00.$

(a) Determine the standard error (in dollars) of the mean.(b) With a 0.95 probability, determine the margin of error (in dollars).(c) What is the 95% confidence interval of the population mean (in dollars)? Answer : Given information : Sample mean,…

In a random sample of 10 residents of the state of Florida, the mean waste recycled per person per day was 2.8 pounds with a standard deviation of 0.64 pounds. Determine the 95% confidence interval for the mean waste recycled per person per day for the population of Florida. Assume the population is approximately normal.

Answer: Given: Sample mean $(x\bar)=2.8$ Sample standard deviation $(s)=0.64$ Sample size $(n)=10$ Confidence interval level $=95%$ Solution: → The significance level $\alpha = 1 – 0.95 = 0.05$ → The degree of freedom $\text{df} = n – 1 = 10…

Karen wants to advertise how many chocolate chips are in each Big Chip cookie at her bakery. She randomly selects a sample of 77 cookies and finds that the number of chocolate chips per cookie in the sample has a mean of 13.4 and a standard deviation of 3.5. What is the 99% confidence interval for the number of chocolate chips per cookie for Big Chip cookies? Enter your answers accurate to one decimal place (because the sample statistics are reported accurate to one decimal place).

Answer: Given Information from Question: The Sample Mean $(x\bar)=13.4$ The Sample Standard Deviation $(s)=3.5$ The Sample Size $(n)=77$ The Confidence Interval level $=99%$ The critical value at significance level = 1-0.99 = 0.01, and df = n – 1 =…