Suppose the preliteracy scores of three-year-old students in the United States a
ID: 3055050 • Letter: S
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 (*), sample standard deviation (S), and standard error (SE) of the students' scores. Round your answers to four decimal places S- SE 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: upper limit: Which is the correct interpretation of the confidence interval? 0 There is a 95% chance that the population mean is between 91.269 points and 100.631 points. O Shelia is certain that the true population mean is between 91.269 points and 100.631 points. O Shelia is 95% confident that the true population mean is between 91.269 points and 100.631 points. O Shelia is 95% confident that the true population mean is between 91.566 points and 100.334 points. 0 There is a 95% chance that the true population mean is between 91.566 points and 100.334 points.Explanation / Answer
x bar = 95.95 , s = 10.0025 , SE = 10.0025/sqrt(20) = 2.2366
t critical avlue = 2.093
margin of error = t * SE
= 2.0930 * 2.2366 = 4.681
lower limit = xbar - margin of error
= 95.95 - 4.681
= 91.269
lower limit = xbar + margin of error
= 95.95 + 4.681
= 100.631
Shelia is 95% confident that the true population mean is between 91.269 and 100.631 points
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