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).

80,82,83,85,86,91,91,92,92,93,95,97,99,100,100,103,107,108,111,112

a) Calculate the sample mean, sample standard deviation, and standard error (SE) of the students' scores. Round your answers to four decimal places.
b) Determine the t-critical value (t) and margin of error (m) for a 95% confidence interval. Round your answers to three decimal places.
c) What are the lower and upper limits of a 95% confidence interval? Round your answers to three decimal places.
d) Which is the correct interpretation of the confidence interval:

Shelia is 95% confident that the true population mean is between 91.129 points and 99.571 points.

There is a 95% chance that the true population mean is between 91.129 points and 99.571 points.

Shelia is 95% confident that the true population mean is between 90.842 points and 99.858 points.

There is a 95% chance that the population mean is between 90.842 points and 99.858 points.

Shelia is certain that the true population mean is between 90.842 points and 99.858 points.

Answer 1

For the given sample details the mean is calculated as:

a) Mean = (80 + 82 + 83 + 85 + 86 + 91 + 91 + 92 + 92 + 93 + 95 + 97 + 99 + 100 + 100 + 103 + 107 + 108 + 111 + 112)/20
= 1907/20
Mean = 95.35

and the sample standard deviation as:

Standard Deviation σ =s = √(1/20 - 1) x ((80 - 95.35)² + (82 - 95.35)² + (83 - 95.35)² + (85 - 95.35)² + (86 - 95.35)² + (91 - 95.35)² + (91 - 95.35)² + (92 - 95.35)² + (92 - 95.35)² + (93 - 95.35)² + (95 - 95.35)² + (97 - 95.35)² + (99 - 95.35)² + (100 - 95.35)² + (100 - 95.35)² + (103 - 95.35)² + (107 - 95.35)² + (108 - 95.35)² + (111 - 95.35)² + (112 - 95.35)²)
= √(1/19) x ((-15.35)² + (-13.35)² + (-12.35)² + (-10.35)² + (-9.35)² + (-4.35)² + (-4.35)² + (-3.35)² + (-3.35)² + (-2.35)² + (-0.34999999999999)² + (1.65)² + (3.65)² + (4.65)² + (4.65)² + (7.65)² + (11.65)² + (12.65)² + (15.65)² + (16.65)²)
= √(0.0526) x ((235.6225) + (178.2225) + (152.5225) + (107.1225) + (87.4225) + (18.9225) + (18.9225) + (11.2225) + (11.2225) + (5.5225) + (0.1225) + (2.7225) + (13.3225) + (21.6225) + (21.6225) + (58.5225) + (135.7225) + (160.0225) + (244.9225) + (277.2225))
= √(0.0526) x (1762.55)
= √(92.71013)
= 9.6315

and the standard error is caluclated as:

b) The critical value tc is calculted uisng the given confience level and degree of freedom = n-1=20-1=19.

the formula used is =T.INV.2T(0.05,19) which results in tc = 2.093.

The margin of error is calculated as:

ME = 4.508

c) The confidence interval is calculated as:

d) The correct interpretation of the confidence interval is :

Shelia is 95% confident that the true population mean is between 90.842 points and 99.858 points.