<p>Can somebody who's good at Stat please be a kind soul and help me with this stat problem? Oh by the way, it might even help you review for the AP test! but honestly can somebody please help me with it? I'll appreciate it A LOT and i'll love you forever :)</p>
<p>Are you more likely to have a motor vehicle collision when using a cell phone? A study of 699 drivers who were using a cell phone when they were involved in a collision examined this question. These drivers made 26,798 cell phone calls during a 14-month study period. Each of the 699 collisions was classified in various ways. Here are the counts for each day of the week:</p>
<p>Day Sun Mon Tues Wed Thurs Fri Sat Total
Number 20 133 126 159 136 113 12 699</p>
<p>We have a total of 699 accidents involving drivers who were using a cell phone at the time of their accident. Are the accidents equally likely to occur on any day of the week? Follow the inference toolbox. (Include the expected counts.)</p>
<p>Thanks so much again! I really do appreciate it a looott! If theres anything you want to ask me, please ask.</p>
<p>Ho: all distributitions are the same
Ha: at least one of them is significantly different</p>
<p>Conditions: Expected are all > 5, SRS (I may be forgetting one)</p>
<p>Calculations
Expected for each day is 699/7. Oberserved are the numbers.
Do:
Observed - Expected, square the difference, then divide by Expected.
((20-699/7)^2 / (699/7) ) for sunday
Do this for each day, add the numbers up. That’s your chi-squared value. Use a chart or calc, using df=n-2 = 5 and find the P value.</p>
<p>Using alpha = 0.05, if p-value is larger than 0.05, you fail to reject Ho. If p-value is less than 0.05, you reject Ho in favor of Ha. (Then talk about distributions, one is significantly diff,etc).</p>
<p>Sorry, typed that in a rush, tell me if you don’t get parts of it.</p>
<p>Since this is a chi-square goodness of fit test, degrees of freedom is n -1 not n -2</p>
<p>I think I got a chi-square test value of about 208.85, but that may be wrong since my calculator sort of weirded out. If that is the right value, then the p-value is 2.479e-42, so you can reject the null hypothesis. </p>
<p>If you think about it, it makes sense since the expected count for each day is roughly 100 and the observed counts for sunday and saturday are much smaller</p>
<p>Yes I meant n to be number of categories. The poster above me used n - 2 for degrees of freedom which is incorrect in a goodness of fit test. n - 2 is used for degrees of freedom on regression inference</p>
<p>If you have anything lower than an 84, then you have to do chi-square goodness of fit test manually with lists. There is no special test for it in the calculator. If you are using the test that uses matrices, then that’s the wrong one</p>
<p>Matrices are used for chi-square homogeneity of population or ch-square test for independence/association</p>
<p>The null hypothesis is that the distribution of accidents is uniform over the week (each day has the same amount)</p>
<p>The alternative hypothesis is that the distribution of accidents is not uniform, so the days have a different proportion of accidents</p>
<p>Finding expected counts: According to the null hypothesis, the distribution of accidents throughout the week is uniform, so each day has the same number of expected accidents. To find the expected counts, you take the total number of accidents (699) and divide by the number of days (7).</p>
<p>If Ho is rejected, as it was in my calculations (which might be incorrect since I did it in a hurry), then you have significant evidence that an accident is not equally likely to occur on each day of the week i.e. the probability of an accident is different for each day</p>
<p>k thanks. sry but can you help me with 2 more of these questions quickly? i’m really sorry but I’m like freaking out b/c this is due tomorrow and I don’t know what i’m doing…well now i think i kinda do. after you do them i’m gonna check my answers with yours to confirm my answers you’re the best! </p>
<ol>
<li><p>In a certain town, there are about one million eligible voters. A simple random sample of 10,000 eligible voters was chosen to study the relationship between sex and participation in the last election. THe results are summarized in the following 2 x 2 read two by two) contingency table:</p>
<p>We want to check whether being a man or a woman (columns) is independent of having voted in the last election (rows). In other words is “sex and voting independent”? Follow the inference toolbox. Show the expected counts and show the work to find the expected count for women who voted.</p>
<p>Okay so in this case, you are going to want to use a chi-square test of independence/association. This is the one that you use the matrices for.</p>
<p>Your null hypothesis is that there is no association between gender and voting participation
Your alternative hypothesis is that there is an association between gender and voting participation</p>
<p>Conditions: SRS/expected counts are greater than 5</p>
<p>To find the expected counts in a two way table, you find the marginal distributions by adding the values of each column and row. Once you have the marginal distributions, you calculate the expected count for each cell by multiplying the sum of that cell’s row and the cell’s column and then dividing by the total count. For example, for expected number of men who voted you would do (2792+1486)(2792+3592)/(2792+3591+1486+2131) to get 2730.6474. You would do this for each of the cells. (A quicker way to do this is to plug the original data into Matrix A, running the X^2 test, then looking at Matrix B where the calculator will automatically store the expected counts)</p>
<p>To get the X^2 test statistics, you would take the sum of (observed - expected)^2/expected for each cell, or run the X^2 test. You should get 6.660455</p>
<p>The degrees of freedom is equal to (# of rows - 1)(# of columns - 1) = (2-1)(2-1)=1</p>
<p>The p-value that you should get is .0098, so you have significant evidence to reject the null hypothesis.</p>
<p>Therefore, you have significant evidence that there is an association between gender and voting participation</p>
<p>Let me know if you have questions. Hope this helped</p>
<p>honestly you are the best! i think i’m learning more from you than i do from my teacher. short and sweet and to the point lol. hope i’m not bothering you too much but i promise this is the last question (i had 2 be4 and this is the 2nd one).</p>
<p>One of the questions of interest to an author was whether there was an association between the survival after a stroke and level of education. Medical records for a sample of 2333 residents of Vienna, Austria who had suffered a stroke were used to classify each individual according to two variables, survival and level of education. The table below organizes the results.</p>
<pre><code> No basic secondary school technical training/ higher university
education graduation apprenticed secondary graduate
school
degree
</code></pre>
<p>there’s a MINITAB output shown with chi-sq of 12.219 and the expected counts written.</p>
<p>a. I have to write the null and alternate hypotheses.
b. What are degrees of freedom?
c. Using the above data, what is the p-value for the chi-square statistic?
d. Using the above data, what is the best conclusion to the author’s question of association between education level and survival of a stroke?</p>
<p>hope this always helps you study for ur stat exam! thanks so much this whole time!</p>
<p>it should be no basic education, secondary school graduation, technical training/apprentice, higher secondary school degree, and university graduate…</p>
you’re the best! </p>