Proportions and Performing of Hypothesis Tests

PART

1. An opinion poll asks a random sample of 100 college juniors how they view their job prospects once they graduate. Out of the 100 students 53 said “Excellent.” Find a 95% confidence interval to estimate the proportion of college juniors who think their job prospects are excellent. Assume large samples.

The random sample is equivalent to 100 college juniors

From this sample, 53 of them considered their job prospects to be excellent, which is equivalent to 0.53.

A 95% confidence interval implies that in the event that 100 different kinds of samples are taken into consideration and a 95% confidence interval is calculated for every sample, then roughly 95 out of the 100 confidence intervals will have the true mean value, which is µ

To construct a 95% confidence interval for a population, mean µ, the correct critical value of z* (Sprinthall, 2003) is P (-1.96 < Z < 1.96) = 0.95

The test statistic is 0.7019

The confidence interval is (0.6808, 0.7019)

2. In 1999, the Bureau of Justice Statistics3 indicated that 14% of violent crimes are committed by women. Suppose that a random sample of 400 people who committed violent crimes was taken and it was determined that 65 of the 400 were women. Find a large-sample 95% confidence interval based on the sample data. Determine if the 14% is contained in the confidence interval.

65/400 = 16.25% or 0.1625

To construct a 95% confidence interval for a population, mean µ, the correct critical value of z* is P (-1.96 < Z < 1.96) = 0.95

The test statistic is 0.5636

The confidence interval is (0.5636, 0.7967)

The 14 percent which is equivalent is not contained within the confidence interval

3. The IRS is trying to determine which percentage of tax returns claim itemized deductions. A random sample of 2,000 returns was taken and the IRS found that 663 claimed itemized deductions. Find a large-sample 90% confidence interval based on the sample data.

663 claimed itemized deductions out of a random sample of 2,000 returns

This gives us

To construct a 95% confidence interval for a population, mean µ, the correct critical value of z* is P (-1.645 < Z < 1.645) = 0.90

Z from 0.3315 is 0.6293

Z from (1 -0.3315) is 0.7486

Therefore the confidence interval is: (0.6293, 0.7486)

4. What are the conditions that must be checked when performing a test of proportions?

The following are the conditions that must be checked when performing a test of proportions:

i. The samples must be independent

ii. The samples must be randomly selected

iii. The samples ought to be big enough to make use of a normal sampling distribution (Seber, 2013).

PART 2

1. Suppose that experts claim that 14% of violent crimes are committed by women. Is there enough evidence to use a benchmark of 0.05 to reject the claim, if in a sample of 400 violent crimes, 65 were committed by women? Use the four-step (state, plan, solve, conclude) process to answer this question.

State:

Null Hypothesis: 14% of violent crimes are committed by women

Alternate Hypothesis: 14% of violent crimes are not committed by women

Plan:

A sample is selected from the population, and a sample mean is measured. The random sample taken encompasses a total of 400 violent crimes. From this figure, the experts determined that 65 violent crimes were committed by women.

Solve:

The test statistic is computed as follows:

(65 / 400) * 100 = 16.25% or 0.1625

Conclude:

In this case, the benchmark used is 0.05. Taking into consideration that the test statistic is greater than the benchmark used, then we reject the null hypothesis.

2. Suppose that California Community Bank is testing the hypothesis that the proportion of deposit slips filled out incorrectly is 2%. A random sample of 1000 customers found that 28 completed the deposit slips incorrectly.

a) Identify the null and alternate hypotheses

The null hypothesis is:

The proportion of deposit slips filled out incorrectly is 2%

The alternate hypothesis is:

The proportion of deposit slips filled out incorrectly is not 2%

b) Determine if strong evidence exists to confirm that 2% of the deposit slips are filled out incorrectly?

A random sample of 1,000 customers found that 28 completed the deposit slips incorrectly.

Therefore, the proportion obtained is:

Taking this into consideration, it can be deemed that there is no strong evidence that exists to make a confirmation that 2% of the deposit slips are filled out incorrectly. This is for the reason that the proportion is greater than 2%

c) If a benchmark of 0.05 was used for the p-value, would the null hypothesis be rejected, or do we fail to reject the null?

The test statistic for this particular sample is 2.8%, which is equivalent to 0.028. Taking into consideration that the test statistic is less than 0.05, then we reject the null hypothesis.