# Business Statistics: Epidemiologic Statistics For Public Health Essay

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## Question:

1: The problem is about two candidates who are participating in an election. The two participants are Jones and Johns. Sara has made a prediction about the winning candidate in the election. She has asked 15 of her friends to say who will the vote. Out of 15, 10 of her friends has said that they will vote for Jones. So it is seen that more than 50% of the people have said that they will vote for Jones. This has made Sara convinced that Jones is the winner.

The problem is to test whether Jones is really the winner. The problem can be solved by performing a suitable test. The proportion of people who will vote for Jones is equal to 10/15 = 0.6666667 which is approximately equal to 0.67. It is assumed that Jones will win the election if he gets more than 50% of the votes. Therefore, the problem is to test whether the value of the proportion is equal to 0.5 or it is less than 0.5.

The hypothesis of the test is H0: p= 0.5 against H1: p < 0.5. The test is a one sided test. This is the one sample test for binomial proportion. The test statistic is given by the following:

t = (p-0.5)/sqrt((0.5(1-0.5))/n)

In the equation, p is the estimated value from the sample. The calculated value of p is 0.67. The value of the test statistic is 1.317. The test statistic is said to follow a standard normal distribution if the null hypothesis is true. The value of the standard normal variable at 5% confidence level is 1.67. Therefore, the hypothesis will be rejected if the value of the test statistic is greater than 1.67. Therefore the value of the test statistic is accepted. That is, Jones who has got more votes has a higher chance of winning.

Another method is to calculate the p-value of the test statistics. The p-value is the value which is calculated from the observation based on the alternative hypothesis (Dean, Sullivan & Soe, 2014). The number of votes obtained by Jones is supposed to be distributed as binomial with n = 15. The level of significance is ?. The p-value of the test statistic has been calculated below:

P[x<10|H0] =

The p-value of the test is 0.9407. The value is greater than alpha. Therefore, the null hypothesis of the test can be accepted. That is the winning candidate is Jones.

2. The confidence interval for the given test is:

C.I = p ± z * sqrt((0.5(1-0.5))/n)

The confidence interval for the test is (0.46725, 0.87275)

The above confidence interval is estimated by approximating the binomial distribution with normal distribution. 95% of the values of the sample lie within this range. The interval will have maximum width if the variance of the sample is 0.25n. That means the binomial distribution with parameter p=0.5 has the confidence interval of maximum width (Wallis, 2013). There is a formula for calculating the confidence interval. The formula is as follows:

N= 4/W^2 = 1/b^2.

The quantity b measures the error of the estimate. In order to have a error of 5% the sample size should be 400.

Therefore, Sara has to take a friend sample of 400 to get a 95% confidence interval.

3. The conclusion obtained from the results of the test is that the winning candidate is Jones. The data collected indicated that more than 50% of the people are voting for Jones. The proportion was 0.67 which is obviously greater than 0.5. Therefore on the basis of this data the probability has been constructed and the result of that probability said that more than 50% people are voting for Jones. The number of friends of Sara who said they are voting for Jones is ten. On the basis of this an interval has also been constructed. The values of the interval and the tests suggest that Jones have a high chance of winning the election. Therefore, Sara can conclude on the basis of her data that Jones is winning the election.

## References:

Dean, A. G., Sullivan, K. M., & Soe, M. M. (2014). OpenEpi: Open source epidemiologic statistics for public health, version.

Wallis, S. (2013). Binomial confidence intervals and contingency tests: mathematical fundamentals and the evaluation of alternative methods. Journal of Quantitative Linguistics, 20(3), 178-208.