An industrial psychologist consulting with Lara music stores knows that the average number of complaints management receives each month throughout the industry is 6, but the standard deviation is unknown. Twenty of the Lara music stores were randomly selected to record complaints for one month; the mean of this sample was 3 and its standard deviation was 1.5. Using the .05 significance level, is the number of complaints received by Lara music stores different from the number of complaints received by music stores in general

Step-by-step explanation:

H0 : μ ≥ 6

H1 : μ < 6

Test statistic :

n = 20

Sample Mean, x = 3

σ = 1.5

Test statistic = (x - μ) ÷ σ/sqrt(n)

Test statistic = (3 - 6) ÷ 1.5/sqrt(20)

Test statistic = - 3 / 0.3354101

Test statistic = −8.944271

Since, sample size is < 30 ;

Obtaining the Critical value using the from T;

Tcritical at 0.05, df = 19 = - 1.729

Tstatistic < Tcritical

−8.944271 < - 1.729

Hence, we reject the Null:

We can conclude that there is significant evidence that the number of complaints received by Lara music store is different from number of complaints received by music stores in general.

the distance between point a and point b is 6 units

definitely a. the others have almost all data points on one side of the line.

step-by-step explanation: