Question

How Big Are Gribbles? Gribbles are small, pale white, marine worms that bore through wood. While sometimes considered a pest since they can wreck wooden docks and plers, they are now being studied to determine whether the entryme they secrete will allow us to turn inedible wood and plant waste into biofuel. A sample of 50 gribbles finds an average length of 3.1 mm with a standard deviation of 0.72. Give a best estimate for the length of gribbles, a margin of error for this estimate (with 95% confidence), and a 95% confidence interval Round your answer for the point estimate to one decimal place, and your answers for the margin of error and the confidence interval to two decimal places point estimate- margin of error- The 95% confidence intervalisi

Answer 1

point estimate = 3.1

margin of error=

18 Your input: find the margin of error for the sample size n = 50, standard deviations = confidence level 95.0% using t-distribution. op, and First, find the critical value: t = 2.00957523712924 Next, find the standard error of the mean: SE= S 92 n 125

Finally, the margin of error is ME=t.SE= 2.00957523712924. (92 125 ~0.204621735956926

margin of error is 0.20

The 95% confidence Interval

The provided sample mean is X = 3.1 and the sample standard deviation is s = 0.72. The size of the sample is n = 50 and the required confidence level is 95%. The number of degrees of freedom are df = 50 – 1 = 49, and the significance level is a = 0.05. Based on the provided information, the critical t-value for a = 0.05 and df = 49 degrees of freedom is te=2.01. The 95% confidence for the population mean u is computed using the following expression CI= (* to monitoramento) Therefore, based on the information provided, the 95% confidence for the population mean u is C1 = (3.1 – 2.0160.12, 31 + 2.012012) V50 50 = (3.1 – 0.205, 3.1 +0.205) = (2.895, 3.305)

The 95% confidence Interval (2.90,3.30)

note here tc that is t-critical value is calculate using statistical table