Video game players and divided attention tasks. Human Factors (May 2014) published the results of a study designed to determine whether video game players are better than non–video game players at crossing the street when presented with distractions. Participants (college students) entered a street-crossing simulator. The simulator was designed to have cars traveling at various high rates of speed in both directions. During the crossing, the students also performed a memory task as a distraction. The researchers found that students who are video game players took an average of 5.1 seconds to cross the street, with a standard deviation of .8 second. Assume that the time, x, to cross the street for the population of video game players has , Now consider a sample of 30 students and let x represent the sample mean time (in seconds) to cross the street in the simulator.

a. Find $P\left(\overline{x}>5.5\right)$

b. The 30 students in the sample are all non–video game players. What inference can you make about and/or for the population of non–video game players? Explain.

Step by step solution

01

Given information

A sample of 30 students is selected with mean 5.1 and the standard deviation 0.8.

02

Calculating the probability

a.

Let X be the time to cross the street in the simulator.

From the given problem $\mu =5.1,\sigma =0.8$ and sample size is n=30.

According to Central limit theorem, if the sample size is large, then the sampling distribution of the sample mean$\left(\overline{x}\right)$ becomes approximately normal.

Then

$\begin{array}{rcl}P\left(\overline{x}>5.5\right)& =& P\left(\frac{\overline{x}-\mu }{\sigma /\sqrt{n}}>\frac{5.5-5.1}{0.8/\sqrt{30}}\right)\\ & =& P\left(Z>\frac{0.4}{0.1461}\right)\\ & =& P\left(Z>2.74\right)\\ & & \end{array}$

$$\begin{gathered} =1-P\left(Z⩽2.74\right) \\ =1-\left\{0.5+P\left(0<Z<2.74\right)\right\} \\ =1-0.5-0.4969 \\ =0.0031 \end{gathered}$$

Thus, $P\left(\overline{x}>5.5\right)=0.0031$

03

Interpretation

b.

From part a., the probability that sample mean is greater than 5.5 is 0.0031. Moreover, the probability for the video game players is very low than the non-video game players. Thus, the population mean $\left(\mu \right)$for the non-video game players is greater than 5.1 and the population standard deviation$\left(\sigma \right)$ for the non-video game players is greater than 0.8.

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