Answer to Question #151377 in Statistics and Probability for Yashasvi

Question

The following are the amounts of time, in minutes, that it took a random sample of 16 technicians to perform a certain task: 18.1, 20.3, 18.3, 15.6, 22.5, 16.8, 17.6, 16.9, 18.2, 17.0, 19.3, 16.5, 19.5, 18.6, 20.0 and 18.8. Assuming that this sample came from a symmetrical continuous population, use the signed-rank test at the 0.05 level of significance to test the null hypothesis that the mean of this population is 19.4 minutes against the alternative hypothesis that it is not 19.4 minutes. (Z0.025 = 1.96)

Answer 1

Null hypothesis H₀: μ = 19.4

Alternative hypotheses H₁: μ ≠ 19.4

Use the test statistic X, the observed number of plus signs.

Replacing each value exceeding 19.4 with a plus sign, each value less than

19.4 with minus sign, we get

– + – – + – – – – – – – + – + – – – – –

n = 20, x = 4 and n/2 = 10

Here x < n/2

Therefore, from table the computed P-Value is

P = 2P(X ≤ 4 when θ = $\frac{1}{2}$ )

= 2 (0.0059)

= 0.0118

Since the P value, 0.0118, is less than 0.05, the null hypothesis must be rejected, and we conclude that the mean of the population is not 19.4 minutes.