Mathematics, 01.07.2021 16:10 amichellecam2

College national study finds that students buy coffee from a coffee shop on average 12 times a week, I believe it may be different for UML students. I collect data from a sample of 36 UML students and find that they buy coffee on average 8 times a week, with a standard deviation of 6 days. What is the T value for this data, and can you reject the null?

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Answer from: jessicavasquez081

The t-value for this data is -4.

The p-value of the test is 0.0003 < 0.05, which means that the null hypothesis can be rejected.

Step-by-step explanation:

College national study finds that students buy coffee from a coffee shop on average 12 times a week, I believe it may be different for UML students.

At the null hypothesis, we test if the mean is of 12, that is:

$H_0: \mu = 12$

At the alternative hypothesis, we test if the mean is different of 12, that is:

$H_1: \mu \neq 12$

The test statistic is:

We have the standard deviation for the sample, so the t-distribution is used to solve this question.

$t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, s is the standard deviation and n is the size of the sample.

12 is tested at the null hypothesis:

This means that $\mu = 12$

I collect data from a sample of 36 UML students and find that they buy coffee on average 8 times a week, with a standard deviation of 6 days.

This means that n = 36, X = 8, s = 6

Value of the test statistic:

$t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}$

$t = \frac{8 - 12}{\frac{6}{\sqrt{36}}}$

t = -4

The t-value for this data is -4.

P-value of the test:

Considering a standard significance level of 0.05.

Test if the mean is different from a value, so two-tailed test, with 36 - 1 = 35 df and t = -4. Using a t-distribution calculator, the p-value is of 0.0003.

The p-value of the test is 0.0003 < 0.05, which means that the null hypothesis can be rejected.

Answer from: Quest

mr. flanders’ class sold 47 doughnuts and mr. rodriquez’s class sold 38 cartons of milk

step-by-step explanation:

let x be the number of doughnuts sold by mr. flanders’ class and y be the number of cartons sold by mr. rodriquez’s class.

1. together, the classes sold 85 items, then

x+y=85.

2. both classes earn earned $104.49 for their school. if doughnuts were sold for $1.35 each, then x doughnuts costed $1.35x. if cartons of milk were sold for $1.08 each, then y cartons costed $1.08y. hence,

1.35x+1.08y=104.49.

3. solve the system of two equations:

$\left\{\begin{array}{l}x+y=85\\1.35x+1.08y=104.49\end{array}\right.\rightarrow \left\{\begin{array}{l}x=85-y\\1.35(85-y)+1.08y=104.49\end{array}\right.$

$114.75-1.35y+1.08y=104.49,\\ \\114.75-104.49=1.35y-1.08y,\\ \\10.26=0.27y,\\ \\y=38,\\ \\x=85-38=47.$

Answer from: Quest

a reflection across the x-axis , followed by the by a reflection across the y-axis , and then a scale dilation of 0.5 .

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