Answered: Thanks to an initiative to recruit top…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Question

Thanks to an initiative to recruit top students, an administrator at a college claims that this year's entering class must have a greater mean IQ score than that of entering classes from previous years. The administrator tests a random sample of 15 of this year's entering students and finds that their mean IQ score is 118, with a standard deviation of 14. The college records indicate that the mean IQ score for entering students from previous years is 114. Is there enough evidence to conclude, at the 0.05 level of significance, that the population mean IQ score, μ, of this year's class is greater than that of previous years? To answer, assume that the IQ scores of this year's entering class are approximately normally distributed. Perform a one-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal places.

A. Find the value of the test statistic and round to 3 or more decimal places. (I have posted a picture of an example problem and the equation to use, with the correct answer as every expert I have asked thus far has gotten this problem wrong.)

B. Find the critical value. (Round to three or more decimal places.)

The Journal de Botanique reported that the mean height of Begonias grown while being treated with a
particular nutrient is 36 centimeters. To check whether this is still accurate, heights are measured for a
random sample of 14 Begonias grown while being treated with the nutrient. The sample mean and sample
standard deviation of those height measurements are 32 centimeters and 11 centimeters, respectively.
Assume that the heights of treated Begonias are approximately normally distributed. Based on the sample,
can it be concluded that the population mean height of treated begonias, µ, is different from that reported in
the journal? Use the 0.10 level of significance. Perform a two-tailed test. Then complete the parts below. Carry
your intermediate computations to three or more decimal places.
MATL
BU
Ch

(c) Finding the value of the test statistic
Since we're assuming the null is true, we use μ = 36. We also have that x = 32, s = 11, and n = 14. So we get the following.
32-36
11
✓√14
t
x-μ
S
√n
≈ 1.361

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

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