Question

(a) Test at the 0.05 level of significance whether there is sufficient evidence to conclude that the average salary of articling lawyers in Toronto exceeds $75,000. Use the critical value approach and show manually how the t-statistic is calculated.

(b) Now show how you would calculate the p-value for the result in part (a).

(c) Finally calculate manually a 95% 1-sided confidence interval for the average salary of an articling lawyer in Toronto.

(d) Explain why the p-value and confidence interval do or do not support your conclusion in part (a). (e) Show a boxplot of the data and explain whether the assumptions of the test and confidence interval are reasonable.

salary
72508
77802
79906
87425
69996
66279
79308
81199
99659
78188
83387
87125
81557
69685
55962
78814
89389
82446
82694
76371
77474
72797
76889
70710
71557
77960
95545
75568
59922
89528
65670
78530
90170

Answer 1

Solution:-

Data 78853.34 1665.404 78188 99659 Mean Standard Error Median Mode Standard Deviation 9852.664 Sample Variance97074980 Kurtosis Skewness Range Minimum Maximum Sum Count 0.391163 0.062919 43697 55962 99659 2759867 35

a)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: μ < 75,000
Alternative hypothesis: μ > 75,000

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample t-test.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = s / sqrt(n)

S.E = 1665.404
DF = n - 1

D.F = 34
t = (x - μ) / SE

t = - 2.31

t_critical = - 1.691

where s is the standard deviation of the sample, x is the sample mean, μ is the hypothesized population mean, and n is the sample size.

Interpret results. Since the t-value (- 2.31) is less than the tcritical (-1.691), we have to reject the null hypothesis.

b)

Thus the P-value in this analysis is 0.0135.

Interpret results. Since the P-value (0.0135) is less than the significance level (0.05), we have to reject the null hypothesis.

c) 95% 1-sided confidence interval for the average salary of an articling lawyer in Toronto is C.I = 82237.78.

$C.I = \bar{x}\pm t_{\alpha /2}\times \frac{S.D}{\sqrt{n}}$

C.I = 78853.34 + 2.0322 × 1665.404

C.I = 78853.34 + 3384.43

C.I = 82237.78

d) The p-value and confidence interval do support our conclusion in part (a).

.