Question

The demand for the video games provided by Mid-Tech Video Games Inc. has exploded in the last several years. Hence, the owner needs to hire several new technical people to keep up with the demand. Mid-Tech gives each applicant a special test that Dr. McGraw, the designer of the test, believes is closely related to the ability to create video games. For the general population, the mean on this test is 100. Below are the scores on this first test for the applicants. 95 105 120 81 90 115 99 100 130 10 The owner is interested in the overall quality of the job applicants based on this test. Compute the mean and the median scores for the 10 applicants. What would you report to the owner

Answer 1

Answer:

Mean: 94.5.

Median: 99.5

Standard deviation: 33.1

We can tell the owner that the applicants don't have a score significantly below from 100.

Step-by-step explanation:

First, we analize the sample and calculate the statistics (mean, median and standard deviation).

Mean of the sample:

$M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M=\dfrac{1}{10}(95+105+120+81+90+115+99+100+130+10)\\\\\\M=\dfrac{945}{10}\\\\\\M=94.5$

The median, as the sample size is an even number, can be calculated as the average between the fifth and sixth value, sort by value:

$\text{Median}=\dfrac{99+100}{2}=99.5$

The standard deviation is:

$s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{9}((95-94.5)^2+(105-94.5)^2+(120-94.5)^2+. . . +(10-94.5)^2)}\\\\\\s=\sqrt{\dfrac{9834.5}{9}}\\\\\\s=\sqrt{1092.7}=33.1$

To tell if this sample has a value significantly lower than the expected score of 100, we should make a hypothesis test.

The claim is that the mean score is significantly lower than 100.

Then, the null and alternative hypothesis are:

$H_0: \mu=100\\\\H_a:\mu< 100$

The significance level is 0.05.

The sample has a size n=10.

The sample mean is M=94.5.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=33.1.

The estimated standard error of the mean is computed using the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{33.1}{\sqrt{10}}=10.467$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{94.5-100}{10.467}=\dfrac{-5.5}{10.467}=-0.53$

The degrees of freedom for this sample size are:

$df=n-1=10-1=9$

This test is a left-tailed test, with 9 degrees of freedom and t=-0.53, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=P(t<-0.53)=0.306$

As the P-value (0.306) is bigger than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the mean score is significantly lower than 100.