Question

An industrial psychologist conducted an experiment in which 40 employees that were identified as "chronically tardy" by their managers were divided into two groups of size 20. Group 1 participated in the new "It's Great to be Awake!" program, while Group 2 had their pay docked. The following data represent the number of minutes that employees in Group 1 were late for work after participating in the program.
Does the probability plot suggest that the sample was obtained from a population that is normally distributed? Provide TWO reasons for your classification.

Answer 1

Answer:

The probability plot of this distribution shows that it is approximately normally distributed..

Check explanation for the reasons.

Step-by-step explanation:

The complete question is attached to this solution provided.

From the cumulative probability plot for this question, we can see that the plot is almost linear with no points outside the band (the fat pencil test).

The cumulative probability plot for a normal distribution isn't normally linear. It's usually fairly S shaped. But, when the probability plot satisfies the fat pencil test, we can conclude that the distribution is approximately linear. This is the first proof that this distribution is approximately normal.

Also, the p-value for the plot was obtained to be 0.541.

For this question, we are trying to check the notmality of the distribution, hence, the null hypothesis would be that the distribution is normal and the alternative hypothesis would be that the distribution isn't normal.

The interpretation of p-valies is that

When the p-value is greater than the significance level, we fail to reject the null hypothesis (normal hypothesis) and but if the p-value is less than the significance level, we reject the null hypothesis (normal hypothesis).

For this distribution,

p-value = 0.541

Significance level = 0.05 (Evident from the plot)

Hence,

p-value > significance level

So, we fail to reject the null or normality hypothesis. Hence, we can conclude that this distribution is approximately normal.

Hope this Helps!!!