Question

researchers measured the data speeds for a particular smartphone carrier at 50 airports. the highest speed measured was 78.8 mbps. the complete list of 50 data speeds has a mean of x = 18.71 mbps and a standard deviation of s = 19.99 mbps.
a. what is the difference between carriers highest data speed and the mean of all 50 data speeds?
b. how many standard deviations is that the difference found in part (a)?
c. convert the carriers highest data speed to a z score.
d. if we consider data speeds that convert to z scores between - 2 and 2 to be neither significantly low nor significantly high, is the carriers highest data speed significant?

Explanation:

Step1: Calculate the difference in part (a)

The highest data speed is \(x = 78.8\) Mbps and the mean is \(\bar{x}=18.71\) Mbps.
The difference is \(x-\bar{x}=78.8 - 18.71=60.09\) Mbps.

Step2: Calculate the number of standard deviations in part (b)

The standard deviation is \(s = 19.99\) Mbps.
The number of standard deviations is \(\frac{x-\bar{x}}{s}=\frac{60.09}{19.99}\approx3.00\)

Step3: Calculate the z - score in part (c)

The formula for the z - score is \(z=\frac{x-\bar{x}}{s}\)
Substituting \(x = 78.8\), \(\bar{x}=18.71\) and \(s = 19.99\)
\(z=\frac{78.8 - 18.71}{19.99}=\frac{60.09}{19.99}\approx3.00\)

Step4: Check significance in part (d)

Since the z - score \(z = 3.00>2\) (using the rule that \(|z|> 2\) implies significance)

Answer:

a. \(60.09\) Mbps
b. \(3.00\)
c. \(z\approx3.00\)
d. Yes, the carrier's highest data speed is significant.