statistics

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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

  • statistics -

    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.

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