Chapter 11.2, Problem 10P

(a)

To determine

Find the level of significance.

State the null and alternative hypothesis.

Answer to Problem 10P

The level of significance is 0.01.

Explanation of Solution

Calculation:

From the given information the value of $\alpha$ is 0.01, and to test the claim that there is no difference in the distributions of time to complete a doctorate for the two fields.

Hence, the level of significance is 0.01.

The null and alternative hypothesis is,

Null hypothesis:

$H_0:$ The distributions of average time to earn a doctorate from humanities field and social sciences field are same.

Alternative hypothesis:

$H_1:$ The distributions of average time to earn a doctorate from humanities field and social sciences field are different.

(b)

To determine

Identify the sampling distribution to be used.

Mention the conditions to use the distribution.

Find the value of the sample test statistic.

Answer to Problem 10P

The sampling distribution to be used is normal distribution.

The value of the sample test statistic is 1.38.

Explanation of Solution

Calculation:

Test statistic:

The z value for the sample test statistic R is,

$z=\frac{R-{\mu }_R}{{\sigma }_R}$

In the formula R is the sum of ranks from the sample of size $n_1$ (smaller sample),

$\begin{array}{l}{\mu }_R=\frac{n_1\left(n_1+n_2+1\right)}{2}\\ {\sigma }_R=\sqrt{\frac{n_1{n}_2\left(n_1+n_2+1\right)}{12}}\end{array}$

$n_1$ be the sample size of the smaller sample and $n_2$ be the sample size of the larger sample

The standard normal distribution is used as the sampling distribution for the rank sum test.

The necessary conditions to use the rank sum test are,

• $n_1>10$
• $n_2>10$

Procedure for assigned rank to data values:

• First combine both the samples.
• Arrange the data values in ascending order.
• Rank each of the data value in sequential order.

The rank for each of tied data is computed as,

• First assign the sequential position ranks for all the values that are same.
• Take the mean of all the position ranks that are assigned for same data values.
• Assign the mean rank as the rank position for all the tied data.
• The assigned mean rank is used in calculating the test statistic.

The ranks are,

Years	Rank	Field
5.4	1	B
5.8	2	B
5.9	3	B
6.0	4	A
6.2	5	B
6.3	6	B
6.4	7	A
7.0	8	B
7.1	9	A
7.2	10	A
7.5	11.5	A
7.5	11.5	A
7.6	13	B
7.7	14	B
7.8	15	B
7.9	16	B
8.0	17	A
8.3	18.5	A
8.3	18.5	B
8.5	20	B
8.7	21	A
8.9	22	A
9.2	23	A

The smaller sample is 11 which corresponds the field A, and field B sample size is 12. That is, $n_1=11,n_2=12$.

The value of R is,

$\begin{array}{c}R=4+7+9+10+11.5+11.5+17+18.5+21+22+23\\ =154.5\end{array}$

The value of R is 154.5.

The mean of R is,

$\begin{array}{c}{\mu }_R=\frac{11\left(11+12+1\right)}{2}\\ =\frac{11\left(24\right)}{2}\\ =\frac{264}{2}\\ =132\end{array}$

The standard deviation of R is,

$\begin{array}{c}{\sigma }_R=\sqrt{\frac{\left(11\times 12\right)\left(11+12+1\right)}{12}}\\ =\sqrt{\frac{132\left(24\right)}{12}}\\ =\sqrt{\frac{3,168}{12}}\\ =\sqrt{264}\\ =16.25\end{array}$

Test statistic:

Substitute R as 154.5, ${\mu }_R$ as 132 and ${\sigma }_R$ as 16.25 in the test statistic formula

$\begin{array}{c}z=\frac{154.5-132}{16.25}\\ =\frac{22.5}{16.25}\\ =1.38\end{array}$

Hence, the z value is 1.38.

(c)

To determine

Find the P-value of the sample test statistic.

Answer to Problem 10P

The P-value is 0.1676.

Explanation of Solution

Calculation:

Step by step procedure to obtain P-value using MINITAB software is given below:

• Choose Graph > Probability Distribution Plot choose View Probability > OK.
• From Distribution, choose ‘Normal’ distribution.
• Click the Shaded Area tab.
• Choose X Value and Both Tail, for the region of the curve to shade.
• Enter the X value as 1.38.
• Click OK.

Output using MINITAB software is given below:

From Minitab output, the P-value is 0.0838 which is one sided value.

The two-tailed P-value is,

$\begin{array}{c}P\text{-value}=2\times 0.0838\\ =0.1676\end{array}$

Hence, the P-value is 0.1676.

(d)

To determine

Mention the conclusion of the test.

Answer to Problem 10P

The null hypothesis is failed to be rejected.

Explanation of Solution

Calculation:

From part (c), the P-value is 0.1676.

Rejection rule:

• If the P-value is less than or equal to $\alpha$, then reject the null hypothesis and the test is statistically significant. That is, $P\text{-value}\le \alpha$.

Conclusion:

The P-value is 0.1676 and the level of significance is 0.01.

The P-value is greater than the level of significance.

That is, $0.1676\left(=P\text{-value}\right)>0.01\left(=\alpha \right)$.

By the rejection rule, the null hypothesis is failed to be rejected.

Hence, the data is not statistically significant at level 0.01.

(e)

To determine

Interpret the conclusion in the context of the application.

Explanation of Solution

Calculation:

From part (d), the null hypothesis is failed to be rejected. This shows that, there is no sufficient evidence that the distributions of average time to earn a doctorate from humanities field and social sciences field are different at level of significance 0.01.