In the past, 44% of those taking a public accounting qualifying exam have passed the exam on their first try, Lately, the availability of exam preparation books and tutoring sessions may have improved the likelihood of and individual's passing on his or her first try. In a sample of 250 recent applicants,130 passed on their first attempt
a) At 5% level of significance, is there sufficient evidence of an improvement lately with respect to passing on the first try? (your conclusion must be in terms of p value as well as setting up a rejection region.please show necessary work)
b)which statistical distribution should be applied in this situation and why? (explain)
c)knowing that a monetary investment is required to take preparatory actions, should we put much weight on the conclusion of part (a) EXPLAIN
d)Based on a 95% confidence level, what is the best case and worst case scenario regarding the actual percentage of applicants who pass on their first try?
e)carefully interpret the interval estimation
f)using the results of part (d) explain carefully whether or not there is sufficient evidence of an improvement lately with respect to passing on the first try?
NOTE:Please Include all assumptions and refrain from using electronic applications like megastat etc. Show workings as much as possible. cheers
I'll give you several tips to get started on this one and let you take it from there.
Null hypothesis:
Ho: p = .44 -->meaning: population proportion is equal to .44
Alternative hypothesis:
Ha: p > .44 -->meaning: population proportion is greater than .44
Using a formula for a binomial proportion one-sample z-test with your data included, we have this:
z = .52 - .44 -->test value (130/250 = .52) minus population value (.44)
divided by √[(.44)(.56)/250] -->.44 represents 44%, .56 represents 56% (which is 1-.44), and 250 is sample size.
Do the above calculation to get the z-test statistic. To find the p-value, which is the actual level of the test statistic, check a z-table for the p-value.
If the p-value is greater than .05, the null is not rejected. If the p-value is less than .05, then the null is rejected in favor of the alternative hypothesis and you can conclude p > .44 (there is enough evidence to support the claim that there is an improvement).
Note:
Use the appropriate confidence interval formula for part d.
I hope this will help.
a) To determine if there is sufficient evidence of an improvement in passing on the first try, we need to conduct a hypothesis test.
- Null hypothesis (H0): The proportion of applicants passing on the first try remains the same (p = 0.44).
- Alternative hypothesis (Ha): The proportion of applicants passing on the first try has improved (p > 0.44).
We will use the binomial distribution to conduct the test. The sample size (n) is 250, and the number of successes (applicants who passed on the first try) is 130. The expected proportion under the null hypothesis is 0.44 * 250 = 110.
Next, we calculate the test statistic:
z = (p̂ - p) / √(p(1-p)/n)
where p̂ is the sample proportion, p is the null hypothesis proportion, and n is the sample size.
Using the sample proportion p̂ = 130/250 = 0.52, we find:
z = (0.52 - 0.44) / √(0.44 * 0.56 / 250)
z ≈ 2.696
To determine the p-value, we look up the critical value for a one-tailed test at the 5% significance level. The critical value is 1.645. Since the test statistic (2.696) is greater than the critical value, we can conclude that there is sufficient evidence of an improvement in passing on the first try.
b) The statistical distribution applied in this situation is the binomial distribution. This distribution is suitable because we are dealing with a binary outcome (pass or fail) in a fixed number of independent trials (applicants taking the exam).
c) When considering the conclusion of part (a) regarding the improvement in passing on the first try, we need to weigh the investment required for preparatory actions. If the preparation methods are costly, it may not be financially feasible for all applicants. Therefore, the conclusion should be taken into account along with the financial implications.
d) For a 95% confidence level, we can calculate the confidence interval for the proportion of applicants who pass on their first try. The best-case scenario would be when all 130 of the sampled applicants passed on their first try, resulting in a proportion of 130/250 = 0.52. The worst-case scenario would be if none of the sampled applicants passed on their first try, resulting in a proportion of 0/250 = 0.
e) The interval estimation for the proportion of applicants who pass on their first try, at a 95% confidence level, is approximately [0, 0.52]. This means that we can be 95% confident that the true proportion of all applicants who pass on their first try falls within this interval.
f) Based on the results from part (d), where the confidence interval includes the null hypothesis proportion of 0.44, we do not have sufficient evidence to conclude that there is an improvement in passing on the first try lately. The confidence interval includes the possibility that the proportion remains the same as before.
a) To determine if there is sufficient evidence of an improvement lately with respect to passing on the first try, we need to conduct a hypothesis test.
Let's define the null and alternative hypotheses:
Null hypothesis (H0): The proportion of applicants passing on the first try is still 44%.
Alternative hypothesis (H1): The proportion of applicants passing on the first try is higher than 44%.
To conduct the test, we'll use the binomial distribution because we are dealing with a fixed number of trials (250 applicants) and two possible outcomes (pass or fail).
We'll calculate the test statistic, which is the z-score, using the following formula:
z = (p̂ - p0) / sqrt((p0 * (1 - p0)) / n)
Where:
p̂ is the sample proportion (130/250 = 0.52),
p0 is the null hypothesis proportion (0.44),
n is the sample size (250).
Now, we can calculate the z-score:
z = (0.52 - 0.44) / sqrt((0.44 * (1 - 0.44)) / 250)
z ≈ 1.74
Using a significance level of 5% (α = 0.05), the critical z-value for a one-tailed test is approximately 1.645 (lookup from z-table). Since the calculated z-score of 1.74 is greater than the critical value, we can reject the null hypothesis.
To determine the p-value, which represents the probability of observing a test statistic as extreme as the one calculated under the null hypothesis, we can use a standard normal distribution table or a calculator. The p-value corresponding to a z-score of 1.74 is less than 0.05.
Therefore, at a 5% level of significance, there is sufficient evidence to suggest an improvement lately with respect to passing on the first try.
b) The appropriate statistical distribution to use in this situation is the binomial distribution. This distribution is used when dealing with a fixed number of independent trials (250 applicants) and two possible outcomes (pass or fail). Each applicant can be considered a Bernoulli trial.
c) When considering the conclusion of part (a) that there is evidence of an improvement lately, it is essential to factor in the monetary investment required for preparatory actions. The improvement in passing rates may be attributed to the availability of exam preparation books and tutoring sessions, indicating that those who invested in these resources had an advantage. Therefore, we should consider that the conclusion may not be applicable to applicants who cannot afford or choose not to make that investment.
d) To determine the best and worst-case scenarios regarding the actual percentage of applicants who pass on their first try, we can construct a confidence interval.
At a 95% confidence level, we can calculate the confidence interval using the following formula:
CI = p̂ ± z * sqrt((p̂ * (1 - p̂)) / n)
Where:
CI is the confidence interval,
p̂ is the sample proportion (0.52),
z is the critical z-value for a 95% confidence level (lookup from z-table, approximately 1.96),
n is the sample size (250).
Calculating the confidence interval:
CI = 0.52 ± 1.96 * sqrt((0.52 * (1 - 0.52)) / 250)
CI = 0.52 ± 1.96 * sqrt(0.001272) ≈ 0.52 ± 0.053
Therefore, the best-case scenario is that the actual percentage of applicants who pass on their first try is approximately 57.3% (0.52 + 0.053), while the worst-case scenario is approximately 47.7% (0.52 - 0.053).
e) The interval estimation (confidence interval) provides a range within which we can be reasonably confident (at a 95% confidence level) that the true proportion of applicants who pass on their first try lies. In this case, we estimate that the true proportion is between approximately 47.7% and 57.3%.
f) Based on the confidence interval calculated in part (d), we can see that it does not include the null hypothesis proportion of 44%. This indicates that the proportion of applicants passing on their first try is likely higher than 44%. Therefore, there is sufficient evidence from the confidence interval to support the conclusion from part (a) that there has been an improvement lately with respect to passing on the first try.