Please help me to get to these answers, I don\'t know how my professor got these
ID: 3334870 • Letter: P
Question
Please help me to get to these answers, I don't know how my professor got these answers can someone explain thank you.
crosoft Wordx Chapter 32) 2p x D 01) (2)p x %20-%20SAMPLE%20Midterm%20(Answers)( 1 )%20( 1 ).pdf -1 Dental Receptionist PART TIME BILLIN pa Ro Microsoft Wordx LH Chapter 2 ng Huang at Baruc M McGraw-Hill Con Using the aforementioned sample of 50 Baruch College alumni, you divide your dataset into two groups based on SAT score. Alumni who earned a score above 1250 on their SAT are denoted as the "HiSAT" group- and alumni who earned a score of 1250 or less on their SAT are denoted as the "LowSAT" group. You import two columns of data into R via a CSV file. The column entitled "HISAT GPA contains the GPAs for all alumni in the HISAT group. The column entitled "LowSAT GPA contains the GPAs for all alumni in the LowSAT group. HiSAT GPA 3.65 3.20 3.80 3.95 2.63 -2.95 3.67 3.33 3.85 2.30 3.79 3.75 4.00 3.55 3.50 3.18 3.45 3.32 4.00 2.25 3.85 PA LowSAT GPA 2.52 3.12 2.91 3.22 2.95 3.25 2.65 2.84 3.15 3.35 2.25 3.10 2.10 2.37 2.75 3.35 2.65 2.49 2.90 2.20 3.25 3.92 3.38 3.20 2.19 2.85 3.25 3.75 4.00Explanation / Answer
14) In this part, we have asked the most accurate regarding the unpooled variance differences in means t-test provided.
Null hypothesis is "No significant difference between specified populations".
So the correct option is c) Your null hypothesis is that the average GPA for the HiSAT group = the average GPA for the LowSAT group.
16) Here p value is 7.403e-05
That is p-value = 0.00007403
0.01% means 0.0001000
P-value < 0.0001000 so we reject null hypothesis.
So correct option is d) You reject the null hypothesis at the 0.01% significance level.
17) Sample size for HiSAT = n1 = 29
Sample size for LowSAT = n2 = 21
Degrees of freedom = n1 + n2 - 2 = 29 + 21 - 2 = 48
Test statistic t = 1.475
At the 1% level of significance and degrees of freedoms are 48 the critical value = 2.704 ( using t table)
So critical value > test statistic we accept null hypothesis.
So the correct option is d) Given a t statistic of 1.475 you would have accepted the null hypothesis at the 1% significance level.