Mathematics, 15.01.2020 21:31 traceyschmidtk3562

The upcoming championship high school football game is a big deal in your little town. the problem is, it is being played in the next biggest town, which is two hours away! to get as many people as you can to attend the game, you decide to come up with a ride-sharing app, but you want to be sure it will be used before you put all the time in to creating it. you determine that if more than three students share a ride, on average, you will create the app. you conduct simple random sampling of 20 students in a school with a population of 300 students to determine how many students are in each ride-share (carpool) on the way to school every day to get a good idea of who would use the app. the following data are collected: 6 5 5 5 3 2 3 6 2 25 4 3 3 4 2 5 3 4 5construct a 95% confidence interval for the mean number of students who share a ride to school, and interpret the results. part a: state the parameter and check the conditions. part b: construct the confidence interval. be sure to show all your work, including the degrees of freedom, critical value, sample statistics, and an explanation of your process. part c: interpret the meaning of the confidence interval. part d: use your findings to explain whether you should develop the ride-share app for the football game.

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a) The parameter of interest on this case is $\mu$ who represent the average number of students in order to create an app.

b) $3.85 - 2.09 \frac{1.348}{\sqrt{20}}=3.22$

$3.85 + 2.09 \frac{1.348}{\sqrt{20}}=4.48$

The 95% confidence interval is given by (3.22;4.48)

c) For this case we have 95% of confidence that the true mean for the average number of students in order to create an app is between 3,22 and 4.48.

d) We have the following criteria in order to decide: "You determine that if more than three students share a ride, on average, you will create the app"

So then since the lower limti for the confidence interval is higher than 3 we can conclude that at 5% of significance we have more than 3 students share a ride so then makes sense create the app.

Step-by-step explanation:

Part a

The parameter of interest on this case is $\mu$ who represent the average number of students in order to create an app.

Part b

Data: 6 5 5 5 3 2 3 6 2 2 5 4 3 3 4 2 5 3 4 5

n=20 represent the sample size

$\bar X$ represent the sample mean

represent the sample standard deviation

m represent the margin of error

Confidence =95% or 0.95

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Calculate the mean and standard deviation for the sample

On this case we need to find the sample standard deviation with the following formula:

$s=sqrt{\frac{\sum_{i=1}^20 (x_i -\bar x)^2}{n-1}}$

And in order to find the sample mean we just need to use this formula:

$\bar x =\frac{\sum_{i=1}^{20} x_i}{n}$

The sample mean obtained on this case is $\bar x= 3.85$ and the deviation s=1.348

Calculate the critical value tc

In order to find the critical value is important to mention that we don't know about the population standard deviation, so on this case we need to use the t distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . The degrees of freedom are given by:

df=n-1=20-1=19

We can find the critical values in excel using the following formulas:

"=T.INV(0.025,19)" for $t_{\alpha/2}=-2.09$

"=T.INV(1-0.025,19)" for $t_{1-\alpha/2}=2.09$

The critical value $tc=\pm 2.09$

Calculate the confidence interval

The interval for the mean is given by this formula:

$\bar X \pm t_{c} \frac{s}{\sqrt{n}}$

And calculating the limits we got:

$3.85 - 2.09 \frac{1.348}{\sqrt{20}}=3.22$

$3.85 + 2.09 \frac{1.348}{\sqrt{20}}=4.48$

The 95% confidence interval is given by (3.22;4.48)

Part c

For this case we have 95% of confidence that the true mean for the average number of students in order to create an app is between 3.22 and 4.48.

Part d

We have the following criteria in order to decide: "You determine that if more than three students share a ride, on average, you will create the app"

So then since the lower limti for the confidence interval is higher than 3 we can conclude that at 5% of significance we have more than 3 students share a ride so then makes sense create the app.

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