Question

MONTE CARLO HYPOTHESIS TEST HELP?!

We are given the following averages in categories (How much time you spend a week on HW, on FB, READING, and LISTENING TO MUSIC:

These were the averages for each category:

HW = 12.0276 /wk

FB = 385.403 /wk

BOOKS = 42.4919 /wk

MUSIC = 18.8011 /wk

-- (Basically my hypothesis are testing whether or not my class does more or less when compared to graduate students)

I NEED TO USE MY AVERAGES IN A MONTE CARLO SIMULATION TO PERFORM A HYPOTHESIS TEST ON THE GRADUATE STUDENT SAMPLES FOR EACH CATEGORY (grad data below).

AND

I ALSO NEED TO USE THE SAMPLE MEANS TO FIND A P-VALUE!!! (I have no idea how to do that either).

Please help!

<The numbers below the red line are the average for each of the categories for grad students>

005 5 001055005 880 5 322 - 22 1-1 2 0050000000000005 5 07 0403550554517 2 1 21 1 50064007 01800000 5 2 01 8 520 414 06003 66243 45000005 70505500 5 333 24143 131 2

Answer 1

In the given problem we need to test the hypothesis whether or not my class does more or less when compared to graduate students for each of the 4 categories, HW, FB, BOOKS, and MUSIC

Given sample size, n =16

n - 1 = 16 - 1 = 15

We use t-test for single mean under the null hypothesis H0

0 7t

$S^{2} = \frac{\sum_{i=1}^{n}(x_{i} - \overline{x})^{2}}{n-1}$

t-critical value at 5% level of significance and 15 degrees of freedom, t(0.05, 15) = 2.131

For HW:

To test

Null Hypothesis:

Но : μ0 12.0276

versus

Alternate Hypothesis:

H1 : μο 12.0276

From the given data

S :-tr-T) 2 = 2849 189.93 rl 15

$\overline{x} = 22.25$

The test statistic is calculated as

... (1)

E=22.25 112.027. 104454.2 t- man = 3.4454 = 2.9670 189.93 16

From (1) since the calculated > t-tabulated (t(0.05,15)= 2.131) we reject null hypothesis H0, and infer that the average of the class under the category HW is not equal to the average of HW for graduate students

리 = 2.9670

For FB:

To test

Null Hypothesis:

Но : 140 385.403

versus

Alternate Hypothesis:

H, : μ0 385.403

From the given data

s2 _ Σί_I(zi-T)2-20090004. T9 1339333.6523 rl 15

T-298.8125

The test statistic is calculated as

... (2)

T_μ0 298.8125-385.403 VE V LİSELER!--289.32 40 =-0.2993 1339333.6523 16

From (2) since the calculated < t-tabulated (t(0.05,15)= 2.131) we do not reject null hypothesis H0, and infer that the average of the class under the category FB is equal to the average of FB for graduate students

t0.2993

For BOOKS:

To test

Null Hypothesis:

H0 : μ,-42.4919

versus

Alternate Hypothesis:

H1 : μοメ42.4919

From the given data

Ση -1(zi-2) 2-1213450 R _ 808966667 rl 15

T= 147.5

The test statistic is calculated as

... (3)

-rVE- ー-140 147.5-42.4919 105.0081 80896.6667 16 424919-105008-1470 1.4768 71.1058

From (3) since the calculated < t-tabulated (t(0.05,15)= 2.131) we do not reject null hypothesis H0, and infer that the average of the class under the category BOOKS is equal to the average of BOOKS for graduate students

리 1.4768

For MUSIC:

To test

Null Hypothesis:

Но : ,10 18.8011

versus

Alternate Hypothesis:

H, : t10メ18.8011

From the given data

$S^{2} = \frac{\sum_{i=1}^{n}(x_{i} - \overline{x})^{2}}{n-1} = \frac{1603.75}{15} = 106.9167$

$\overline{x} = 17.625$

The test statistic is calculated as

$t = \frac{\overline{x} - \mu_{0}}{\sqrt{\frac{S^{2}}{{n}}}} = \frac{17.625 - 18.8011}{\sqrt{\frac{106.9167}{16}}} = \frac{-1.1761}{2.5850} = -0.4550$ ... (4)

From (4) since the calculated < t-tabulated (t(0.05,15)= 2.131) we do not reject null hypothesis H0, and infer that the average of the class under the category MUSIC is equal to the average of MUSIC for graduate students

0.4550