Bicycle Deaths A researcher wanted to determine whether bicycl...

Question

Bicycle Deaths A researcher wanted to determine whether bicycle deaths were uniformly distributed over the days of the week. She randomly selected 200 deaths that involved a bicycle, recorded the day of the week on which the death occurred, and obtained the following results (the data are based on information obtained from the Insurance Institute for Highway Safety). Is there reason to believe that bicycle fatalities occur with equal frequency with respect to day of the week at the alpha = 0.05 level of significance?

Table showing bicycle death frequencies by day of the week, with counts ranging from 16 to 41 for seven days.

Accepted Answer

Hello everyone, let's look at our next problem. A transportation safety analyst claims that vehicle accidents occur equally across all days of the week. To test this claim, you collect data on 1400 accidents from a recent year and categorize them by the day they occurred. It was also shown in the table below. We have observed values for all 7 days of the week. Come back to that. And then it says at the alpha equals 0.05 significance level, test the analyst's claim that accidents are uniformly distributed across the days of the week. We want to match a distribution. So we're gonna be doing a chi squared test. So let's start by calculating our chi squared value. So, we have our 7 days of the week with observed frequencies or O for a formula. To call that our chi squared formula is going to be the sum of. And in the numerator, we have. Or should we have the say we have the sum of, and then a numerator, observe minus expected. That's squared and then in the denominator we have expected. So we have our O values. Let's calculate our expected values. I'll put them in the boxes underneath our observed values, and this would basically be the probability of something happening on each day of the week, multiplied by. Our total number of accidents, so that will be. 1400 That will be multiplied by 1/7 to reflect the expected value for each of the 7 days of the week. So, all of these expected values are going to be the same since we expect the same probability for each day of the week. So actually, instead of using a table, For this we will just say that expected value for each. is equal to equal to 200. So, what we can write in here then is our value of observe minus expected and that difference squared. Oh Minus E2, and we put that amount in each box. So that gives us on Monday, where our observed value is 180, that gives us, we have an expected value of 200, gives us that O minus C squared value of 400. Tuesday, our observed value was 190, and then our calculation gives us 100. Wednesday, our observed frequency is 195. And our O minus E squared is going to be. 25 Thursday observed by 210. 0 minus e squared is 100. Friday observe 2300 minus E is squared. Is 900. Saturday observed value 205. O minus E squared. Is equal to 25. And on Sunday where our observed value was 190. And our O minus E squared value is 100. So now we'll just add a row down below our table, which will be our observed minus expected. Difference squared, divided by expected to give us the values that we'll need to sum to get our chi squared. So, that's going to give us for Monday, that gives us a value of 2 for Tuesday, 0.5, or Wednesday, 0.125. Thursday 0.5. Friday 4.5. Saturday 0.125. And Sunday, 0.5. So now I'll go ahead and highlight in blue. Those values there. That we'll need to sum up for our chi squared. So we have our formula for chi squared. Let's sum all those values in the bottom of our table. Now, when I wrote these in the table, I didn't add all the correct significant figures. So, I'll write them out with the correct figures here, just for the sake of getting our answer and the correct number of significant figures. So, our squared is 2.00 plus. 0.125. Plus 0.50. Plus 4.50. Plus 0.125. Plus 0.50. So when we add all that up, we get that equals 8. 25, as our chi squared value. So, we now need to compare that to the critical value of chi squared. So, to find that, we need our alpha, which we know equals 0.05, and we need our degrees of freedom. Well, that is the number of categories, or categories of our days of the week, 7 minus 1. So we have 6 degrees of freedom here. So we just look up with six degrees of freedom at that level 0.05, our critical cry squared. Value is equal to 12.592. So looking at our chi squared value of 8.25, This value is less than 12.592. Which is our Critical chi squared value. So, since that chi squared value is less than the critical value, we fail to reject. The null hypothesis. So that means that. So, that means that there is insufficient evidence to say that the accidents are distributed unequally. That's a little complicated there. So again, since our comparative distribution was the accidents distributed so that they occurred equally on all days of the week. So again, our problem had a transportation safety analyst claiming that vehicle accidents occur equally across all days of the week. So we looked at observed values on 1400 accidents, categorized by the day they occurred. And we compared it to the expected value if the accidents occurred equally across every day, you know, equally across the week. We then calculated our chi square value and compared that to the critical chi squared value. Our chi squared value was less than the critical value, and therefore, we fail, failed to reject the null hypothesis, and we have insufficient evidence. And we do not have sufficient evidence to claim that there's an unequal distribution. See you in the next video.

Key Concepts

Chi-Square Goodness-of-Fit Test

Null and Alternative Hypotheses

Significance Level and Decision Rule

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