The director of admissions at a large university says that 15% of high school juniors to whom she sends university literature eventually apply for admission. In a sample 0f 300 persons to whom materials were sent, 30 students applied for admission. In a two-tail test at the 0.05 level of significance, should we reject director’s claim?
Null hypothesis:
Ho: p = .15 -->meaning: population proportion is equal to .15
Alternative hypothesis:
Ha: p does not equal .15 -->meaning: population proportion is not equal to .15 (this is a two-tailed test because the alternative hypothesis doesn't specify a specific direction)
Using a formula for a binomial proportion one-sample z-test with your data included, we have:
z = .10 - .15 -->test value (30/300 = .10) minus population value (.15)
divided by
√[(.15)(.85)/300] -->note: .85 is 1 - .15
Finish the calculation (hint: the z-test statistic will exceed the negative critical cutoff value using a z-table to reject the null hypothesis).
I hope this will help.
Thank you very kindly. Helped immensely!
To determine whether we should reject the director's claim, we can perform a hypothesis test.
Step 1: State the null hypothesis (H0) and the alternative hypothesis (Ha):
- Null hypothesis (H0): The proportion of high school juniors who apply for admission is equal to 15%.
- Alternative hypothesis (Ha): The proportion of high school juniors who apply for admission is not equal to 15%.
Step 2: Set the level of significance (α):
In this case, the level of significance is given as 0.05 (or 5%).
Step 3: Find the test statistic:
We can use the formula for a test statistic for proportions. The formula is:
z = (p̂ - p) / √(p(1-p)/n)
Where:
- p̂ is the sample proportion (number of students who applied divided by the sample size)
- p is the hypothesized proportion (15%)
- n is the sample size
The sample proportion (p̂) = 30/300 = 0.1
Substituting the values into the formula:
z = (0.1 - 0.15) / √(0.15*(1-0.15)/300)
z ≈ -1.7859
Step 4: Find the critical value:
Since this is a two-tail test, we need to find the critical values for rejection in both tails.
At the 0.05 level of significance (α = 0.05), the critical values for a two-tail test are ±1.96.
Step 5: Make a decision:
If the test statistic falls within the critical values, we fail to reject the null hypothesis. Otherwise, if the test statistic falls outside the critical values, we reject the null hypothesis.
In this case, the test statistic (z = -1.7859) falls outside the critical values (-1.96, 1.96). Therefore, we reject the null hypothesis.
Conclusion:
Based on the given sample, there is sufficient evidence to reject the director's claim that 15% of high school juniors who receive university literature eventually apply for admission.
To determine whether we should reject the director's claim, we need to conduct a hypothesis test with the given information.
Let's set up our hypotheses:
- Null Hypothesis (H₀): The proportion of high school juniors who apply for admission is 15% (or 0.15).
- Alternative Hypothesis (H₁): The proportion of high school juniors who apply for admission is not equal to 15% (or 0.15).
Next, we will perform the hypothesis test using a two-tail test at the 0.05 level of significance.
To do this, we'll calculate the test statistic and compare it to the critical value or p-value.
Here are the steps to calculate the test statistic:
1. Calculate the sample proportion:
sample_proportion = number of students who applied / total sample size
= 30 / 300
= 0.10
2. Calculate the standard error:
standard_error = sqrt[(p₀ * (1 - p₀)) / n]
= sqrt[(0.15 * (1 - 0.15)) / 300]
≈ 0.0217
3. Calculate the test statistic (z-score):
test_statistic = (sample_proportion - hypothesized_proportion) / standard_error
= (0.10 - 0.15) / 0.0217
≈ -2.30
4. Determine the critical value or p-value:
Since we are conducting a two-tail test at the 0.05 level of significance, we need to split the significance level in half (0.025 for each tail).
Using a Z-table or calculator, we find that the critical z-values for a two-tail test at the 0.025 level of significance are approximately -1.96 and 1.96.
Now, let's interpret the results:
The calculated test statistic (-2.30) falls in the rejection region beyond the critical values of -1.96 and 1.96.
Since the test statistic falls in the rejection region, we can reject the null hypothesis (H₀) and conclude that there is evidence to suggest that the proportion of high school juniors who apply for admission is not equal to 15%.
In other words, based on the sample data, it appears that the director's claim of 15% may be inaccurate.