A newspaper headline includes the following statement...

A newspaper headline includes the following statement: 'average range for electric vehicles now 220 miles'. In this context, range refers to the distance an electric vehicle can travel before the battery needs to be recharged. A consumer group decides to check the newspaper's statement by testing 23 of the most popular electric cars on the market. The cars were each fully charged and then driven around a track in the same conditions and at the same speed until the charge was fully depleted. The number of completed miles travelled when the battery ran out was recorded. The results are summarised in Table 2. Table 2 Summary statistics for the range (in miles) of electric cars tested Sample size Mean Standard deviation 23 211 30 (a) What is the most appropriate version of the t-test if you want to test the null hypothesis that the underlying population mean distance an electric vehicle can travel before recharging is what the newspaper headline claims and the alternative hypothesis is that the underlying population mean distance an electric vehicle can travel before recharging is not what the newspaper headline claims? Justify your choice. (b) What are the number of degrees of freedom and the 5% critical value for the most appropriate t-test applied to these data? (c) Without using Minitab, calculate a 95% confidence interval for the population mean distance an electric vehicle can travel before the battery needs to be recharged. Your answer should be rounded to the nearest whole number. (d) Based on your calculated 95% confidence interval, is it plausible that the underlying population mean distance an electric vehicle can travel before the battery needs to be recharged is 220 miles as claimed by the newspaper headline? Justify your answer.

Transcript

00:01 So in the given question, we are told that an electric car manufacturer is advertising the car range of 500 kilometers and due to the suspicion of overreporting of a reach in marketing, an independent test is conducted for six randomly selected cars from the manufacturer, right? and the mileage that these cars gave are as follows it was 465 460 490 480 480 470 and 480 and 485 so these were the the mileages, the mileages in kilometers for six randomly selected cars.

01:05 Six randomly selected cars.

01:13 Now what we are asked in the question is first we are asked to find the sample average and the standard deviation.

01:22 So, the sample average can simply be calculated by taking the sum of the observations, sum of the observations and dividing it with the number of observations, right? the number of observations.

01:43 And this would be equal to, when we take this, this one.

01:49 Be equal to let's write this is the mileage is 465 460 490 480 470 and 486 okay so we will take the sum of 465 plus 460 plus 470 plus 490 plus 490 plus 490 plus 480 plus 486 divided by 6 now when we add 465 plus 460 plus 490 plus 480 plus 480 plus 486 and divided by 6 what we get as the sampled average is 475 .17 so this would be the mileage in kilometers which would be which we can take as the sample average right now next we are told to find the standard deviation right and the standard deviation can be found using a formula which is the summation of i from 1 to n x i squared minus 1 by n times the summation of i equal to 1 to n x i whole squared and this whole thing is divided by n minus 1 so this is the formula for the standard deviation squared right so now what we would do is x i is x1 x2 x3 x4 x5 and x6 which are the observations in order that is 465 460 460 490 480 480 4 70 for 70 for 70 for 70 for 70 and 480 for 70 and 480 and we have already calculated the mean of this data and now what we should do is we should take the square of this data right that is sigma x i square sigma x i square minus one by n times sigma x i whole square as what we should take.

04:51 So when we evaluate this, what we would get as s -square is a value of 140.

05:01 And from this we can write the value of the standard deviation as the root of 140, which is 11 .832.

05:14 So, this would be the required standard deviation.

05:20 Now let's move on to the second part of the question, right? so the second part of the question, what we are asked is we have to create a 95 percentage two -sided confidence interval based on the t observer.

05:37 So the t -observer is simply the t -statistic value and what? we do is we are given to find a confidence interval of how much a 95 percentage right yeah a 95 percentage confidence interval and we are told to find we are told to use the t observer which means we will use the formula of the confidence interval to be c -i as equal to x bar x bar plus or minus the critical value of the t observer times the standard deviation divided by the square root of sample size so this would be the formula that we use now in order to find the critical value of t we need two things one is alpha which is the level of significance so we calculate level of significance by just referring to what percentage confidence interval that we have to take.

06:49 So if it is 95 percentage, the alpha is 0 .05.

06:55 And the next parameter is the degree of freedom, which is simply n minus 1.

07:01 So over here, n minus 1 is what, 6 minus 1, right? which is equal to 5.

07:10 So at these parameters, we would find.

07:14 That t is actually equal to 2 .571 so the critical value of c at degree of freedom 5 and a level of significance of 0 .05 is 2 .571 so now we can write x bar is already found as 475 point what 475 .177 right 17 plus or minus 2 .571 times the standard deviation is 11 .832 and it is divided by the square root of 6.

07:57 Now when we find this confidence interval it would be from it would be from 462 .583 which is 475 minus this value.

08:14 And the upper limit would be for 75 plus the same value which is 487 .417.

08:25 So this is the required confidence in a 95 percentage confidence interval based on the t observer.

08:34 Right.

08:35 And now in the question, we are asked to use a one -sided t test to investigate.

08:44 Whether there is a reason to believe that the electric car manufacturer has overreported the range of the car that is the mileage of the car right so what we would do in this case is we are told to actually verify or verify certain verify a claim that is made by the car manufacturer.

09:11 So we would take a null hypothesis that the mileage is indeed the mean mileage is indeed 500 kilometers, which is the claim made by the manufacturer and we would take an alternate hypothesis that would say that no, it is less than 500, right? since we are told to take a one -sided test, we can write either we can write that the mean is not 500 kilometers...

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