Answer:

1. (a) A Wilcoxon signed-rank test may not be appropriate because the salary data are highly skewed and contain an extreme outlier, so the distribution of salaries is unlikely to be symmetric.

2. (b) There are 4 salaries above \$32500 and 11 below. For a two-sided sign test, the two-sided p-value is \[ 2P(X \le 4)=2(0.0592)=0.1184. \] Since \[ 0.1184 > 0.10, \] there is insufficient evidence to reject the company's claim that the median salary is \$32500.

Solution:

1. (a) The Wilcoxon signed-rank test uses the sizes of the positive and negative differences from the hypothesised median, and it is suitable when the underlying distribution is reasonably symmetric. Here, the salaries include a very large value, \$125000, compared with the rest of the sample. This suggests the salary distribution is skewed and contains an outlier, so the symmetry assumption is doubtful.

2. (b) Let \(m\) be the population median salary.

   The hypotheses are

   \[ H_0: m = 32500, \qquad H_1: m \ne 32500. \]

   Compare each salary with \$32500. The salaries above \$32500 are \$125000, \$42500, \$52500 and \$33000, so there are 4 salaries above \$32500. The remaining 11 salaries are below \$32500, and there are no ties.

   Under \(H_0\), each salary is equally likely to be above or below the median, so if \(X\) is the number of salaries above \$32500, then

   \[ X \sim \mathrm{B}(15,0.5). \]

   The observed value is \(X=4\). Since the test is two-sided, use the probability of being at least as extreme in either tail:

   \[ P(X \le 4) = \sum_{r=0}^{4} \binom{15}{r}(0.5)^{15} = 0.0592. \]

   So the two-sided p-value is

   \[ 2 \times 0.0592 = 0.1184. \]

   Since

   \[ 0.1184 > 0.10, \]

   the result is not significant at the 10% level. Therefore, there is insufficient evidence to reject the company's claim that the median salary is \$32500.

Answer: Let the difference be practical mark minus written mark.

Use the sign test with hypotheses:

\(H_0\): the median difference is \(0\) (so practical and written marks are equally likely to be higher).

\(H_1\): the median difference is greater than \(0\) (practical marks tend to be higher).

From the data, there are \(10\) positive signs out of \(15\).

Under \(H_0\), \(X \sim \text{Bin}(15,0.5)\), where \(X\) is the number of positive signs.

\(p\text{-value}=P(X\ge 10)=0.151\).

Since \(0.151 \gt 0.10\), we do not reject \(H_0\).

There is insufficient evidence at the \(10\%\) level to suggest that, on average, the practical examination marks are higher than the written examination marks.

For a sign test, compare each student's practical mark with their written mark and record the sign of

\(\text{practical} - \text{written}\).

We test:

• \(H_0\): the median of \(\text{practical} - \text{written}\) is \(0\)

• \(H_1\): the median of \(\text{practical} - \text{written}\) is greater than \(0\)

The question asks whether practical marks are higher on average, so this is a one-tailed test in the positive direction.

From the sample of \(15\) students, there are \(10\) positive differences, so the test statistic is \(X=10\).

If \(H_0\) is true, a positive sign and a negative sign are equally likely, so

\(X \sim \text{Bin}(15,0.5)\).

We need the probability of getting \(10\) or more positive signs:

\(P(X\ge 10)=1-P(X\le 9)\).

Using binomial probabilities,

\(P(X\ge 10)=0.151\) approximately.

This is the p-value for the test.

Now compare with the significance level \(0.10\):

\(0.151 \gt 0.10\).

So the result is not significant, and we do not reject \(H_0\).

Conclusion: there is not enough evidence at the \(10\%\) significance level to suggest that the practical examination marks are higher than the written examination marks.