Astudy shoed 53% of college applications were submitted online. Assume this result is based on sa simple random sample of 1000 applications with 530 submitted online. Use 0.01 sig.level to test claim that among all student applications the percentage submitted online is equal to 50%.
Null hypothesis:
Ho: p = .50 -->meaning: population proportion is equal to .50
Alternative hypothesis:
Ha: p does not equal .50 -->meaning: population proportion does not equal .50
Using a formula for a binomial proportion one-sample z-test with your data included, we have:
z = (.53 - .50)/√[(.50)(.50)/1000]
Finish the calculation.
Use a z-table to find the critical values for a two-tailed test at .01 level of significance. Compare the test statistic calculated above to the critical values from the table. If the test statistic exceeds either critical value, reject the null and conclude p does not equal .50. If the test statistic does not exceed either critical value from the table, do not reject the null.
I hope this will help get you started.
To test the claim that the percentage of college applications submitted online is equal to 50%, we will conduct a hypothesis test using the given data.
Step 1: Determine the null and alternative hypotheses
The null hypothesis (H0) states that the percentage of college applications submitted online is equal to 50%..
H0: p = 0.50
The alternative hypothesis (Ha) states that the percentage of college applications submitted online is not equal to 50%.
Ha: p ≠ 0.50 (two-tailed test)
Step 2: Determine the significance level
The significance level (also known as alpha, denoted by α) is given as 0.01 in this case. This means we want to be 99% confident in our conclusions.
Step 3: Calculate the test statistic
We need to calculate the test statistic, which depends on the sample proportion and the assumed population proportion under the null hypothesis. The test statistic for proportion is calculated using the formula:
z = (p̂ - p) / √(p(1-p) / n)
where p̂ is the sample proportion, p is the assumed population proportion under H0, and n is the sample size.
In this case, p̂ (sample proportion) is 530/1000 = 0.53, p (assumed population proportion under H0) is 0.50, and n (sample size) is 1000.
Calculating the test statistic:
z = (0.53 - 0.50) / √(0.50(1-0.50) / 1000)
z = 0.03 / √(0.25 / 1000)
z ≈ 0.03 / 0.01581
z ≈ 1.90
Step 4: Determine the critical value(s)
Since this is a two-tailed test, we need to find the critical value(s) for the given significance level of 0.01. The critical value can be found using a z-table or a statistical software. In this case, the critical value is approximately ±2.58.
Step 5: Compare the test statistic with the critical value(s)
If the test statistic falls outside the range defined by the critical value(s), we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
In this case, the test statistic (z = 1.90) does not fall outside the range defined by the critical values (-2.58 to 2.58). Therefore, we fail to reject the null hypothesis.
Step 6: Draw conclusions
Based on the test, there is not enough evidence to support the claim that the percentage of college applications submitted online is different from 50% at a significance level of 0.01. However, it is important to note that failing to reject the null hypothesis does not prove the alternative hypothesis to be true.