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Dvinal [7]
3 years ago
13

A math teacher tells her students that eating a healthy breakfast on a test day will help their brain function and perform well

on their test. During finals week, she randomly samples 46 students and asks them at the door what they ate for breakfast. She categorizes 26 students into Group 1 as those who ate a healthy breakfast that morning and 20 students into Group 2 as those who did not. After grading the final, she finds that 50% of the students in Group 1 earned an 80% or higher on the test, and 40% of the students in Group 2 earned an 80% or higher. Can it be concluded that eating a healthy breakfast improves test scores? Use a 0.05 level of significance.
Mathematics
1 answer:
kogti [31]3 years ago
5 0

Answer:

Step-by-step explanation:

Hello!

To test the claim that eating a healthy breakfast improves the performance of students on their test a math teacher randomly asked 46 students what did they have for breakfast before they took the final exam and classified them as:

<u>Group 1</u>: Ate healthy breakfast

X₁: Number of students that ate a healthy breakfast before the exam and earned 80% or higher.

n₁= 26

<u>Group 2: </u>Did not eat healthy breakfast

X₂: Number of students that did not eat a healthy breakfast before the exam and earned 80% or higher.

n₂= 20

After the test she counted the number of students that got 80% or more in the test for each group obtaining the following sample proportions:

p'₁= 0.50

p'₂= 0.40

The parameters of study are the population proportions, if the claim is true then p₁ > p₂

And you can determine the hypotheses as

H₀: p₁ ≤ p₂

H₁: p₁ > p₂

α: 0.05

Z= \frac{(p'_1-p'_2)-(p_1-p_2)}{\sqrt{p'(1-p')[\frac{1}{n_1} +\frac{1}{n_2}] } } }≈N(0;1)

pooled sample proportion: p'= \frac{x_1+x_2}{n_1+n_2} =\frac{13+8}{46} = 0.46

Z_{H_0}= \frac{(0.5-0.4)-0}{\sqrt{0.46(1-0.46)[\frac{1}{26} +\frac{1}{20}] } } }= 0.67

p-value: 0.2514

The decision rule is:

If p-value ≤ α, reject the null hypothesis.

If p-value > α, do not reject the null hypothesis.

The p-value: 0.2514 is greater than the significance level 0.05, the test is not significant.

At a 5% significance level you can conclude that the population proportion of math students that obtained at least 80% in the test and had a healthy breakfast is equal or less than the population proportion of math students that obtained at least 80% in the test and didn't have a healthy breakfast.

So having a healthy breakfast doesn't seem to improve the grades of students.

I hope this helps!

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