Question

Suppose the preliteracy scores of three-year-old students in the United States are normally distributed. Shelia, a preschool teacher, wants to estimate the mean score on preliteracy tests for the population of three-year-olds. She draws a simple random sample of 20 students from her class of three-year-olds and records their preliteracy scores (in points). 74, 79, 83, 85, 88, 90, 94, 95, 95, 97,99, 99, 100, 103, 105, 105, 106, 107, 107, 108 Click to download the data in your preferred format. CrunchIt! CSV Excel JMP Mac Text Minitab PC Text R SPSS TI Calc Calculate the sample mean (X), sample standard deviation (s), and standard error (SE) of the students' scores. Round your answers to four decimal places. X = Determine the t-critical value (t) and margin of error (m) for a 95% confidence interval. Round your answers to three decimal places. What are the lower and upper limits of a 95% confidence interval? Round your answers to three decimal places. lower limit: lower limit: upper limit: upper limit:

Answer 1

	Values ( X )	Σ ( Xi- X̅ )2
	74	481.8025
	79	287.3025
	83	167.7025
	85	119.9025
	88	63.2025
	90	35.4025
	94	3.8025
	95	0.9025
	95	0.9025
	97	1.1025
	99	9.3025
	99	9.3025
	100	16.4025
	103	49.7025
	105	81.9025
	105	81.9025
	106	101.0025
	107	122.1025
	107	122.1025
	108	145.2025
Total	1919	1900.95

Mean X̅ = Σ Xi / n
X̅ = 1919 / 20 = 95.95

Sample Standard deviation S_X = √ ( (Xi - X̅ )² / n - 1 )
S_X = √ ( 1900.95 / 20 -1 ) = 10.0025

Standard Error = S/√(n) = 2.2366

Critical value  t(α/2, n-1) = t(0.05 /2, 20- 1 ) = 2.093

Margin of Error = t(α/2, n-1) S/√(n) = 4.681

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.05 /2, 20- 1 ) = 2.093
95.95 ± t(0.05/2, 20 -1) * 10.0025/√(20)
Lower Limit = 95.95 - t(0.05/2, 20 -1) 10.0025/√(20)
Lower Limit = 91.269
Upper Limit = 95.95 + t(0.05/2, 20 -1) 10.0025/√(20)
Upper Limit = 100.631
95% Confidence interval is ( 91.269 , 100.631 )

Shelia is 95% confident that the true population mean is between 91.269 points and 100.631 points.