Math assignment based on Minitab

Math assignment based on Minitab

Question 1

Normal Distribution

Descriptive Statistics: C1, xbar2, xbar15, xbar30, xbar5

Variable Mean SE Mean StDev Minimum Q1 Median Q3

C1 48.799 0.965 15.261 6.536 37.910 48.797 58.201

xbar2 49.590 0.658 10.400 20.018 41.594 48.846 56.906

xbar15 49.889 0.238 3.755 39.611 47.421 50.227 52.449

xbar30 49.750 0.176 2.780 42.582 47.778 49.581 51.673

xbar5 49.820 0.406 6.414 31.614 45.553 50.422 53.546

Histograms



Normal Probability Plots



Uniform Distribution

Descriptive Statistics: C1, xbar2, xbar15, xbar30, xbar5

Variable N Mean SE Mean StDev Minimum Q1 Median Q3

C1 250 45.55 5.47 86.4 -99.67 -24.54 52.96 118.30

xbar2 250 50.90 3.82 60.47 -89.41 11.03 50.73 93.94

xbar15 250 48.31 1.44 22.75 -6.44 31.50 48.26 62.76

xbar30 250 48.95 1.03 16.31 8.41 38.13 48.44 59.12

xbar5 250 48.66 2.40 37.98 -54.24 23.27 51.05 73.06

Histograms



Probability Plots



Exponential Distribution

Descriptive Statistics: C1, xbar2, xbar15, xbar30, xbar5

Variable Mean SE Mean StDev Minimum Q1 Median Q3 Maximum

C1 10.707 0.664 10.503 0.006 3.202 6.798 15.488 52.717

xbar2 10.714 0.493 7.792 0.701 4.718 9.392 14.216 50.359

xbar15 10.204 0.168 2.651 4.830 8.344 10.02 11.933 18.943

xbar30 10.132 0.124 1.961 5.492 8.772 9.947 11.396 17.132

xbar5 10.175 0.281 4.439 1.356 6.983 9.669 12.525 27.485

Histograms



Probability Plots



Question 2

Using the Hypothesis:

H0 = There is no significant difference in the means of Quiz 1 and Quiz 10

H1 = There is a significant difference in the means of Quiz 1 and Quiz 10

Using the paired t - test statistics we get the following results from the Minitab:

99% CI for mean difference: (-20.94, 18.74)

T-Test of mean difference = 0 (vs. not = 0)

T-Value = -0.18 P-Value = 0.861

The major assumption that has been made in this hypothesis is that the means of the quizzes have not been taken equal. The level of significance used in the hypothesis testing is 0.01 i.e. our hypothesis will be rejected if it is even less than 99% in the acceptance region. The t - value of the paired t - statistics is found to be -0.18 which is less the tabulated values and also the level of significance i.e. p - value is greater than 0.01 which shows that we cannot reject our null hypothesis i.e. there is no significant difference in the means of Quiz 1 and Quiz 10.

Question 3

Finding the unknowns through statistical formulas:

R - square = SSM / SST = 10490 / 19996

R - square = 0.524

Correlation coefficient (r) = = 0.724

Finding the slope coefficient = r Sx/Sy = 0.724(3.140/14.36) = 0.158

Now, Y = + X

29.57 = + 0.158 (4.357)

= 28.87

The model can be written as:

Hours Spent on Social Media = 28.87 + 0.158 (Preparedness)

The coefficient of the regression equation shows that preparedness explains 15.8% of the data i.e. if a person is fully prepared for the exams i.e. 0 level preparedness then he will be giving 28.87 hours to social media. Similarly if the person is not prepared for exams i.e. 10 level preparedness than he will be giving 30.45 hours. The equation shows that there is a very dependency of preparedness and spending hours in social ...