A group of 20 women took the njcbspt and 17 passed the algebra portion. A group of 35 men took the test and 28 passed. Does this support the claim that men's success rates are lower than women's ? Sigma=.05
To determine if there is enough evidence to support the claim that men's success rates are lower than women's, we can perform a hypothesis test using the given information.
First, let's define the null and alternative hypotheses:
Null Hypothesis (H₀): Men's success rates are not lower than women's.
Alternative Hypothesis (H₁): Men's success rates are lower than women's.
Next, we need to choose an appropriate test statistic. In this case, since we are comparing proportions, we can use the z-test for proportions.
The test statistic for comparing two proportions is calculated as:
z = (p₁ - p₂) / √[(p̂ * (1 - p̂) / n₁) + (p̂ * (1 - p̂) / n₂)]
Where:
- p₁ and p₂ are the sample proportions (p₁ represents women's success rate, p₂ represents men's success rate),
- p̂ is the pooled proportion, calculated as (x₁ + x₂) / (n₁ + n₂), where x₁ and x₂ are the number of successes and n₁ and n₂ are the sample sizes for women and men, respectively.
Let's calculate the sample proportions, the pooled proportion, and the test statistic using the given information:
Sample proportion of women: p₁ = 17 / 20 = 0.85
Sample proportion of men: p₂ = 28 / 35 = 0.8
Pooled proportion: p̂ = (17 + 28) / (20 + 35) = 0.819
Now we can calculate the test statistic:
z = (0.85 - 0.8) / √[(0.819 * (1 - 0.819) / 20) + (0.819 * (1 - 0.819) / 35)]
Using a standard normal distribution table or calculator, we can find the critical z-value for a significance level of 0.05. Assuming a two-tailed test, the critical z-value is ±1.96.
If the calculated test statistic falls outside the range of ±1.96, we can reject the null hypothesis in favor of the alternative hypothesis.
By calculating the test statistic and comparing it to the critical value, we can determine if there is enough evidence to support the claim that men's success rates are lower than women's.