This project was not the most in-depth study, but we did get some results, and made an infographic that we posted at a later date on the site. You’re welcome to see if this is helpful to you in any way.

I don’t personally know much about the electric car industry in the Middle East myself, but I don’t think they have much incentive over there to switch to an electric car, considering how much of their wealth and industry is invested in petroleum and gasoline. At this point, they may view gasoline as a world necessity, and as the price goes up, they will try and milk it for all that it is worth. They will obviously have to invest in future technologies soon, to prevent becoming obsolete when electric cars become a standard, but right now electric cars are probably still a good five, ten years out from even being a quarter of the cars on the road.

-Tyler

P.S. Here’s the link to the infographic.

http://derekbruff.org/blogs/math216/?p=775

I happen to be writing a thesis on electric cars and i found your article really interesting. Its true that its hard to compare electric cars with petrol/gasoline cars. I was hoping you would have any sort of information/ stats or figures regarding electric cars compared to petrol cars. Any sort of help would be really appreciated. Please keep posting such interesting reads.Also I would love to have your opinion on the growth of the electric car industry in the MIddle East such as Dubai where the city has the means to build the industry but do you think that only when oil prices rise will it looks at such alternatives as the electric car?

Thanks and regards,

Gautham Venkatesan

(High School Senior/Editor of Scholl Newsletter) ]]>

We’ve recently improved our sampling techniques.

– We use a Agilent 34110A which has a testing accuracy of around .0002% to as high as .01% towards the date when it’s due for recalibration. A typical multimeter will have an accuracy of around 1-2%.

– Batches are tracked with a batch number so you can see on our website what samples were recorded for the kit that you purchased.

You can see all this on our website: http://www.joeknowselectronics.com/?page_id=956

Regards,

Founder/CEO

Joe B.

1) normal probability plot with theoretical distributions.

2) calculate 99% of the data is between +-3 sigma, 95% between +-2 sigma, and 66% between +-1sigma.

B)

First of all, The number of students are way small compare to the puploation. Also, It depedns on many other factors (living place, income, …etc). For that, we can not build an idea from this method. Saying thatby finding P-value, we can determine some ideas:

H0: mean_starbucks = mean_panera

Ha: mean_starbucks > mean_panera

SE = 0.31

Then, we got that the P-value is smaller than 0.05. For that, The H0 is true. Than, students have the same perferance for both brands.

C)

Thanks, I am sorry for spelling mistake because I am kind in rush. Sorry ðŸ™‚

2) Ho: the mean at starbucks is = to the mean at panera

Ha: the mean at starbucks is > the mean at panera

t= 2.21, so p <.05 ……. Students tend to prefer Starbucks coffee over Panera's

3) None

]]>2. Yes this is sufficient evidence as .5 std is not at great as .7 (point est) margin of victory. Use a confidence interval.

3. How do t-tests and p-tests relate? are they derived in a similar method?

]]>2. SE=sqrt(.5^2/10), T = (.7-0)/SE = 4.42719., degrees of freedom = 9, so we have a p value between 0 and 0.010 which is very small. Therefore, we can reject the null hypothesis.

3. If we’re only given one standard deviation and sample population as we are in this problem, how should we use the formula for Standard Error given in the book that requires two of each? Should we just say SE=sqrt(s^2/n)? ]]>

2. I donâ€™t think the sample size is big enough to make a conclusion about all students at Vanderbilt. However, if you assume that itâ€™s a representative sample, you can perform a t-test and be within the 95% confidence level that Vanderbilt students prefer Starbucks coffee to Panera.

3. How are we going to recognize that you are asking us to do a t-test versus some other statistical analysis method on the final? ]]>

2. NO it is not.4

3. The reading was clear. ]]>

2.)

ms = mean of starbucks

mp = mean of paneraHo: mean_star – mean_pan = 0

Ha: ms – mp > 0

SE = sqrt(2s^2/n) = sqrt(2(.5)^2/10) = .316

t = (.7-0)/.316 = 2.21

df = 9

p < .05

We can conclude that students prefer starbucks coffee to Panera coffee

3)Im not clear on how the df is really found.

]]>2) No we would need the standard deviations and test statistics for Starbucks and Panera, not the standard deviation of the differences.

]]>2. Am now seeing that there is a reading assignment and because I do not have my book on me cannot answer this question.

3. How do we do two sample t tests…

]]>2) No, it is not sufficient evidence

3) No questions about the reading

]]>2.) To test this, we employ the t-distribution with mean 0.7 and standard deviation 0.5. The test statistic is then T = (0.7 – 0)/0.5 = 1.4. With 10 students in our sample, we have the degrees of freedom as df = 9. We find the p-value for this test to be relatively high, therefore we refuse to reject the null hypothesis that there is no difference between the sample means.

3.) Would the t-test work if we were interesting in estimating the sum of two sample means?

]]>2) No since we have a very small degree of freedom.

3) N/A

]]>a) probability plot

b) look at a histogram and see if it is symmetric

2. Suppose you recruit 10 Vanderbilt undergraduates at random to sample the coffee at Starbucks and Panera, rating each store’s coffee on a scale of 1 to 5. The average of the differences in the 10 pairs of ratings is 0.7 in favor of Starbucks, with a standard deviation of 0.5. Is this sufficient evidence to conclude that all Vanderbilt students prefer Starbucks coffee to Panera coffee? (Assume that the underlying populations here are normal.)

X1=Starbucks X2=Panera

Ho: X1-X2=0

Ha: X1-X2>0

p-value=P(x1-x2>.7|X1-X2=0)=P(t>(0.7-0)/SE)

SE=2*.5/sqrt(10)=1/sqrt(10)

d.o.f=9

P(t>2.214)=.027

if Vanderbilt students had no preference, we would only get such a large skew 3% of the time. That is pretty convincing evidence that Vanderbilt students prefer Starbucks to Panera coffee. ]]>

One method may be plotting the data in a quartile plot to see if there are numerous outliers. Another may be to plot the data using a histogram to check for any large skew to one side.

2.

SE = sq(2*s^2/n) = 0.224

t = (.7 – 0) / 0.224 = 3.125

df = 10 -1 = 9

using table C2 we can see our p-value is significantly less than 0.05, and so we can conclude students prefer starbucks to panera coffee.

3.

Why do we take the smaller of the df values? How does the computer calculate them? ]]>

2) H0: Starbucks = Panera; Ha: Starbucks> Panera

SE = sqrt(2s^2/10) = sqrt(2(.5^)2/10) = .316

t = (.7-0)/.316 = 2.21 so our t value is .0136 which means that we can reject the null hypothesis and conclude that we do have sufficient evidence to say that students prefer Starbucks to Panera.

3) Is a sample t test better to use than a sample p test? ]]>

2) replacing SEx1-x2 with the standard deviation of the difference and df=9, we get T=.7/(.5/10)^.5 = 3.13 which gives us a p-value between .01 and .005 which is significant evidence.

3) none ]]>

2. Constructing a t test: null: mean starbucks=mean panera. Alternate: mean starbucks > mean panera. T= pt estimate-null /SE . T= (.7-0)/ 0.316 = 2.21. df=9. p < 0.05, so reject the null hypothesis. There is evidence that Vanderbilt students prefer starbucks over panera coffee.

3. How do you create confidence intervals for t tests in comparison to when using z scores? ]]>

2. Ho = meanPanera >= meanStarbucks

Ha = meanStarbucks 0.012 (p-value)

Since our p-value is less than .05, we can safely reject our null hypothesis and say that there is strong evidence that students prefer Starbucks to Panera coffee (as they should).

3. I really don’t feel like I computed the standard error in question 2 correctly… is there a different way to do this?

]]>2) Ho: Starbucksmean = Paneramean

Ha: Starbucksmean > Paneramean

SE = sqrt(2s^2/n) = sqrt(2(.5)^2/10) = .316

t = (.7-0)/.316 = 2.21 using df = 9 ====> p < .05

This is strong evidence that students prefer Starbucks to Panera!

3) The book makes an adjustment look necessary as a product of having two samples – is this the case?

]]>2. Yes (T = 3.162)

3.None ]]>

2) Ho: Mean Starbucks = Mean Panera

Ha: Mean Starbucks > Mean Panera

SE = sqrt(2s^2/n) = sqrt(2(.5)^2/10) = .316

t = (.7-0)/.316 = 2.21

df = 9 –> p strong evidence.

3) Why are they called degrees of freedom?

]]>2. t=.7/(.5*sqrt(1/5)) = 3.13 which p-value for 2-tailed is 0.0058 so this is strong evidence students prefer Starbucks.

3. What’s the R function for t-table lookup?

]]>b. look at a normal distribution chart and see if it fits the normal curve

2. Ho: Coffees are the same (panera = starbucks)

Ha: Starbucks is preferred (starbucks > panera)

DOF = 9

SE = sqroot(s^2/n + s^2/n)=sqroot(.5^2/10 + .5^2/10)= 0.223606798

t = (point estimate – null) / SE = (.7 – 0) / .224 = 3.13 w/ DOF = 9

p < .005

Therefore, we will reject our Ho and accept our Ha. Startbucks coffee is prefered.

3. When its a single tail to we use the same s and n in the SE equation?

]]>