A 10-year-old survey found that the average net income for self-employed accountants was $74,914. A researcher wants to determine whether this figure has changed( can be changed in both direction ) and takes a random sample of 112 self-employed accountants, producing a sample mean of $78,695. Assume the population standard deviation of net incomes for self-employed accountants is $14,530. Is there any evidence to suggest that the average net income has changed at a 5% level of significance?

Determine the statistical test

Question

A 10-year-old survey found that the average net income for self-employed accountants was $74,914. A researcher wants to determine whether this figure has changed( can be changed in both direction )  and takes a random sample of 112 self-employed accountants, producing a sample mean of $78,695. Assume the population standard deviation of net incomes for self-employed accountants is $14,530. Is there any evidence to suggest that the average net income has changed at a 5% level of significance?

Determine the statistical test

BlackTom AI · Accepted Answer

We start by restating the key elements of the problem to ensure understanding:
- Population mean under H0: μ0 = 74,914
- Sample size: n = 112
- Sample mean: x̄ = 78,695
- Population standard deviation: σ = 14,530
- Significance level: α = 0.05, two-sided test (because the question asks whether the average has changed in either direction)

Next, we determine the appropriate statistical test given the information provided. The hallmark for choosing a Z test is knowing the population standard deviation σ and having a reasonably large sample size, which justifies using the standard normal distribution for the test statistic. Here, σ is provided as 14,530 and n = 112, so the sampling distribution of x̄ under H0 is normal with standard error SE = σ/√n.

Compute the standard error:
- √n = √112 ≈ 10.583
- SE = 14,530 / 10.583 ≈ 1,373 (approximately)

Compute the z statistic:
- Difference between observed and hypothesized means: x̄ − μ0 = 78,695 − 74,914 = 3,781
- z = (x̄ − μ0) / SE ≈ 3,781 / 1,373 ≈ 2.75

Now, assess the p-value for a two-sided test with z ≈ 2.75. A z-score of 2.75 corresponds to a two-tailed p-value around 0.006 (well below 0.05), indicating strong evidence against H0 and in favor of a change in the average net income.

Option-by-option analysis:
- Option: Z test
  This option aligns with the scenario: known population standard deviation and a sufficiently large sample, which supports using a Z test for the mean.
- Option: T Test
  A T test would be more appropriate if σ were unknown and we were estimating the standard deviation from the sample. Since σ is given as known, employing a T test is unnecessary here.
- Option: F test
  The F test is used for comparing variances or in the context of ANOVA, not for testing a single mean against a specified value when σ is known. It is not appropriate for this one-mean mean-shift problem.
- Option: Alpha Test
  There is no standard statistical procedure called an Alpha Test in this context; this option conflates the significance level with a test name, which is incorrect.

From these considerations, the test statistic is based on the standard normal distribution due to the known σ and the modest-to-large sample size, yielding a z statistic of about 2.75 and a p-value well below α = 0.05, indicating evidence that the average net income has changed (in either direction).

BSTA011 Week 8 post lecture practice quiz

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