Computers in some vehicles calculate various quantities related to performance.

Question

Computers in some vehicles calculate various quantities related to performance. One of these is the fuel efficiency, or gas mileage, usually expressed as miles per gallon (mpg). For one vehicle equipped in this way, the mpg were recorded each time the gas tank was filled, and the computer was then reset. Here are the mpg values for a random sample of 20 of these records: 37.8 45.7 38.3 48.9 44.2 40.3 48.4 43.0 44.5 47.1 49.4 44.0 45.2 42.6 35.2 46.0 51.7 38.4 44.0 47.8

(b) Find the mean. (Round your answer to two decimal places.)

Find the standard deviation. (Round your answer to four decimal places.)

Find the standard error. (Round your answer to four decimal places.)

Find the margin of error for 95% confidence. (Round your answer to four decimal places.)

(c) Report the 95% confidence interval for , the mean mpg for this vehicle based on these data. (Round your answers to four decimal places.)

Since previous studies have reported that elite athletes are often deficient in their nutritional intake (for example, total calories, carbohydrates, protein), a group of researchers decided to evaluate Canadian high-performance athletes. A total of n = 324 athletes from eight Canadian sports centers participated in the study. One reported finding was that the average caloric intake among the n = 201 women was 2403.7 kilocalories per day (kcal/d). The recommended amount is 2811.5 kcal/d.

For one part of the study, n = 114 male athletes from eight Canadian sports centers were surveyed. Their average caloric intake was 3077.0 kilocalories per day (kcal/d) with a standard deviation of 986.0. The recommended amount is 3422.4. Is there evidence that Canadian high-performance male athletes are deficient in their caloric intake?

(b) Carry out the test. (Round your answer for t to three decimal places.) t =

Give the degrees of freedom.
(c) Construct a 95% confidence interval for the daily average deficiency in caloric intake. (Round your answers to one decimal place.)

Explanation / Answer

We shall analyse this using the open source statistical package R , The complete R snippet along with the results is as follows

###

> data <- c(37.8, 45.7, 38.3, 48.9, 44.2, 40.3, 48.4, 43.0, 44.5, 47.1, 49.4, 44.0, 45.2, 42.6, 35.2, 46.0, 51.7 ,38.4 ,44.0, 47.8)
>
> # mean of the data
> mean(data)
[1] 44.125
>
> # standard deviation of the data
>
> sd(data)
[1] 4.356227
>
> # standard error of the data is sd/sqrt(n), n is the sample size
>
> length(data)
[1] 20
>
> sd(data)/sqrt(length(data))
[1] 0.9740819
>
> # margin of error is given as z*SE/sqrt(n) or Z*standard error
> #now z from the z tables for 95% CI is 1.96
> 1.96*sd(data)/sqrt(length(data))
[1] 1.909201
>
> # the 95% confidence interval is
> # mean +- 1.96 * SE
>
> mean(data) + 1.96*sd(data)/sqrt(length(data))
[1] 46.0342
> mean(data) - 1.96*sd(data)/sqrt(length(data))
[1] 42.2158

###

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