STA101 Statistics For Business And The Negative Relationship

Answer:

Data: Years of Experience () and Salary ()

Table 1

 Covariance and It interpretation

According to Hahs-Vaughn& Lomax (2013), the covariance between  is given by the formula;

 To work out the covariance between X and Y, all the items in the above formula will be computed and presented in table:

Table 2

 Therefore,

Interpretation:

The covariance explains the relationship between two variables, say X and Y (Hassett &Stewart 2006). According to Hassett &Stewart (2006), a negative covariance between two variables, indicates a negative relationship, that two are negatively associated.  Therefore, since the covariance between X and Y, -12.71, is negative, Years of Experience () and Salary () are said to be negatively related. This implies that a change in one influence the other in opposite direction.

1. Reason for the Negativity of the covariance between X and Y in a

The covariance between X and Y is negative, since small values of X (year of experience) are associated with   large values of Y (salary) and vice versa. This is an indicator of negative relationship.

1. Coefficient  of  correlation  and its  Interpretation

According to Hassett &Stewart (2006), coefficient of correlation is computed using the formula

The data from table 2 will be borrowed to work out Variance(X) and Variance(Y)

 Therefore,

 Then use the two variances and the covariance obtained in part a to find the coefficient of correlation.

 Hence, the correlation coefficient is -0.9911, which is very close to -1.  This implies X (years of experience) and Y (salary) are highly negatively related.  Just like the covariance, it shows that X and Y are negatively associated.

1. Reason for negative correlation coefficient.

The covariance between X and Y was negative, which influence the sign of correlation level between the two variables.

Question Two

Data:  mean   of an exponential distribution is 3 minute

The probability density function (pdf) of an exponential distribution is;  

 Given this pdf, the mean (T) is given by    .Also the cumulative probability function (Cdf) is given by  

1. The value   of  parameter for the exponential distribution

1. The proportion of customers who will hold for more than 1.5 minutes.

This will be given by the probability of waiting for more than 1.5 minutes. This will be computed by integrating the pdf of exponential distribution between 1.5 and infinity.

1. Waiting time at which 10% of the customer will continue to hold.

This the probability that ,  which can  computed from the Cdf of the  exponential  distribution.  Therefore,

Since,    then,

1. Probability a randomly selected customer is placed on hold for 3 to 6 minutes.

This will be given by  

Therefore,

Question Three

Data: H0:  = 950 hours vs. H1:  950 hours,  

1. Computation of the probability of a Type II error when = 1000 and  = 0.10.

Below hypothesis will be considered;

First, the critical value of sample mean will be computed. According to Goos & Meintrup (2016), critical value of mean is given by

Therefore,

This given by

Thus,

1. Calculate the power of the test when = 1000 and = 0.10.

The power of the test is given by

1. Interpretation of power of the test

, is the probability of correctly rejecting the null hypothesis that.

1. Effects of increasing the sample size.

To explain this, we shall assume that the sample size of the population in question was changed from 25 to 36. Then compute the new

This given by

Thus, ,  which is less than initial . This clear show that the when sample size is increased   probability of a Type II error will decrease. On the other hands, the power of test will increase as it depends on the value of  ,  a smaller  results to a bigger power of test.

Question Four

Data: Production Filling Operations,  

Hypothesis test at 5% significance level

Since the population mean and standard deviation are known, the hypothesis test will be based on -score. According to Goos & Meintrup (2016, p.100), -score is computed by the formula

Hence, -score is 1.6

To decide the significance of the test, critical value of, will be determined from tables at 5% significance level. Since, the above test single tailed (right), the critical   will be 1.645, which is greater than , 1.6.

Interpretation:

Since  computed is less than critical, then null hypothesis will not be rejected (Goos & Meintrup 2016). This implies that the population mean is not greater than 47 at 5% significance level

References

Hahs-Vaughn, D.L. and Lomax, R.G., 2013. An introduction to statistical concepts. Routledge.

Hassett, M.J. and Stewart, D., 2006. Probability for risk management. Actex Publications.

Goos, P. and Meintrup, D., 2016. Statistics with JMP: Hypothesis Tests, ANOVA and Regression. John Wiley & Sons.