Question

On April 15th, 1912, the R.M.S. Titanic sank into the North Atlantic Ocean, resulting in the loss of more than 1,500 passengers and crew. Below is a table which summarizes the total number of passengers and the number of survivors for the crew, first-, second-, and third-class. Find the proportion of survivors by type of service (crew, first, second, third). Test the hypothesis at the 95% confidence level that the proportion of first-class survivors is the same as .50. What is a type I error? What is a type II error? For a test at the 95% confidence level, what is alpha? What is the power of a statistical test? What do you calculate for the test-statistic? What is the critical statistic? What is the conclusion of your hypothesis test?

Class	Passengers	Survived
Crew	885	212
First	325	202
Second	285	118
Third	706	178
Total	2201	710

Answer 1

Proportion of survivors:

Crew : 212/885 = 0.2395 = 23.95%

First : 202/325 = 0.6215 = 62.15%

Second : 118/285 = 0.4140 = 41.40%

Third : 178/706 = 0.2521 = 25.21%

Hypothesis test for the proportion of first class survivors:

Observed sample proportion, p = 0.6215

Hypothesised proportion, P = 0.50

Sample size, n = 325

Significance level (type I error), = 1 - 95% = 0.05

Type I error is the rejection of a true null hypothesis. The significance level or the probability of type I error() of 0.05 implies that it is acceptable to have a 5% probability of incorrectly rejecting the null hypothesis.

Type II error() is failing to reject a false null hypothesis.

Power of a a statistical test = 1 - = The probability of rejecting the null hypothesis when it is false.

Null Hypothesis(H0):

The proportion of first class survivors is same as 0.50. P=0.50

Alternative Hypothesis(H1):

The proportion of first class survivors is significantly different from 0.50. P ≠ 0.50.

Standard deviation of the sampling distribution is:

= = = 0.027

Test statistic:

Z = (p - P)/ = (0.6215 - 0.50)/0.027 = 4.5

p-value for Z = 4.5 is less than 0.00001

Thus, p-value < of 0.05

Critical statistic:

Z_crit (critical value of Z) at = 0.05 for a two-tailed test is 1.96 (from cumulative Z-table, corresponding to 0.975 probability or area because, 1- (/2) = 1 - 0.025 = 0.975)

Conclusion:

We reject the null hypothesis at 0.05 significance level because the test statistic of 4.5 is greater than the critical value of Z of 1.96

(OR)

Since, p-value is less than the significance level of 0.05, we reject the null hypothesis.

Thus, the proportion of first class survivors is significantly different from the hypothesised proportion of 0.50. Therefore, P ≠ 0.50