A new medical company claims that 70% of the workers at ABSA are satisfied with their new medical scheme. A random sample of 150 workers showed that 100 are satisfied with their new medical scheme. Test the claim at a 10% significant level.

Answer 1

To test the claim, we need to set up the null and alternative hypotheses.

Let p be the true proportion of workers at ABSA who are satisfied with their new medical scheme.

Null hypothesis: p = 0.70
Alternative hypothesis: p ≠ 0.70 (two-tailed test)

Next, we can calculate the test statistic. The test statistic for testing proportions is the z-score, which can be computed as:

z = (p̂ - p) / √[(p * (1-p)) / n]

where p̂ is the sample proportion, p is the hypothesized proportion, and n is the sample size.

In this case, p̂ = 100/150 = 0.6667, p = 0.70, and n = 150.

Plugging in these values, we get:

z = (0.6667 - 0.70) / √[(0.70 * (1-0.70)) / 150]
= -0.0333 / √[(0.70 * 0.30) / 150]
= -0.0333 / 0.0322
= -1.0311

Now, we need to compare this z-score to the critical value(s) to determine if we can reject the null hypothesis.

At a 10% level of significance, we perform a two-tailed test, so we divide the significance level by 2 to get 0.05 for each tail.

The critical z-score for a 0.05 significance level is approximately ±1.96.

Since -1.0311 is not less than -1.96 or greater than 1.96, we do not have enough evidence to reject the null hypothesis.

Therefore, we fail to reject the claim that 70% of the workers at ABSA are satisfied with their new medical scheme at a 10% significant level.