Chapter 9.3, Problem 74E

a)

To determine

To verify whether there are any outliers in given data.

Explanation of Solution

Given:The Minitab output of data of broker’s returns for a random sample of 36 weeks and a significance test to compare the returns with market performance are shown below

  

Concept used:The boundaries for outliers are shown below

  $\begin{array}{l}Lowerboundary=Q_1-1.5*IQR\\ Upperboundary=Q_3+1.5*IQR\end{array}$

Any values which does not lie between these boundaries are considered as outliers in a data distribution.

Calculation:From the given information

  $\begin{array}{l}{Q}_1=-3.418\\ Q_3=1.543\\ IQR=Q_3-Q_1=1.543-(-3.418)=4.961\end{array}$

We get Inter Quartile Range as 4.961, using this we calculate the boundaries as shown below

  $\begin{array}{l}Lowerboundary=Q_1-1.5*IQR\\ =-3.418-1.5*4.961=-10.8595\\ Upperboundary=Q_3+1.5*IQR\\ =1.543+1.5*4.961=8.9845\end{array}$

Interpretation:

Here the minimum value is -10.270 and maximum value is 7.340. Hence all values do lie between the boundaries of outliers. Hence there are no outliers in the data.

b)

To determine

To interpret the p value from Minitab output.

Explanation of Solution

From the Minitab output the p value is 0.003.

If the average percentage of return is 0.95% per month, then the probability of getting a random sample of 36 weeks with an average returns -1.441 or less is approximately 0.003. It also concludes that the average returns for 36 weeks is less than 0.95% per month.

c)

To determine

To verify if the data provide convincing evidence to support the lawyer’s case.

Explanation of Solution

Concept used:The conditions to perform a one sample t testfor µ.The sample has to be randomly selected.

The population should follow a normal distribution.

The sample should be less than 10% of the population.

Calculation:

To test the claim that the average returns per month µ is less than its true value 0.95%.

The null and alternate hypothesis are shown below

  $\begin{array}{l}{H}_0:\mu =0.95\\ H_a:\mu <0.95\end{array}$

The sample of 36 weeks when stock market did well are selected randomly. The population is of large size, so we can approximately assume it to be a normal distribution. There will be more than 360 weeks where stock market did well, so 10% condition is also satisfied.

A one sample t test is run and we get test statistic of $t=-2.98$ and the corresponding p value as 0.003. As the p value is less than the level of significance (0.05), we have sufficient evidence at 5% level of significance to reject the null hypothesis.

Conclusion:At 5% level of significance there is evidence to conclude that the returns are less than 0.95% per month.