Question

Suppose a population of hearing impaired children were frtted with hearing aids. Suppose they took a multiple-choice hearing test and scored a mean of 50 and a standard deviation of 22 . Let's consider the subset of 32 children in this group who used hearing aids of brand X. Suppose this subset scored a mean of 44 and a standard deviation of 15 . Did children using brand X do significantly worse than the whole group? Test this hypothesis at the 0.05 level of significance.

a) State the hypotheses in symbols
b) Calculate the lest statistic. (Show your work)
c) Get the p-value, showing your work.
d) Interpret the p-value e) Calculate the 95% confidence interval estimale for the mean score of hearing-impaired children. (Show your work.)
f) Interpret the confidence interval
g) Calculate Cohen's d. (Show your work.)
h) Interpret Cohen's d: i) Answer the research question?

Answer 1

Children using Brand X hearing aids scored significantly lower on the hearing test than the overall population of hearing-impaired children. The P-value was much lower than the significance level of 0.05 which led us to reject the null hypothesis. Additionally, a Cohen's d calculation suggested a small to moderate effect size, reflecting a meaningful difference in test scores.

Step-by-step explanation:

Understanding Hypothesis Testing for Brand X Hearing Aids

Hypotheses Statement

• Null hypothesis (H0): μ = 50 (Brand X does not perform significantly worse than the overall mean score)
• Alternative hypothesis (Ha): μ < 50 (Brand X performs significantly worse than the overall mean score)

Test Statistic Calculation

To calculate the Z-score, use the formula: Z = (μ1 - μ0) / (σ / √n), where:

• μ1 = mean score of subset (44)
• μ0 = mean score of population (50)
• σ = standard deviation of subset (15)
• n = size of subset (32)

Z = (44 - 50) / (15 / √32) = -2.67

P-value Calculation

Use the standard normal distribution table to find the P-value corresponding to a Z-score of -2.67. The P-value is approximately 0.0038.

Interpretation of the P-value

A P-value of 0.0038 is less than the α level of 0.05, suggesting that there is significant evidence to reject the null hypothesis.

95% Confidence Interval Calculation

The 95% confidence interval = μ ± Z(α/2) * (σ / √n).

Substituting our values: 50 ± 1.96 * (22 / √32), we get the interval (45.12, 54.87).

Interpretation of the Confidence Interval

The confidence interval estimates the mean score of hearing-impaired children, indicating that we are 95% confident the true mean lies within this interval.

Cohen's d Calculation

Cohen's d = (Mean1 - Mean2) / SDpooled, where SDpooled = √[(SD12 + SD22) / 2] = √[(152 + 222) / 2] = 18.79,
Cohen's d = (44 - 50) / 18.79 = -0.32

Interpretation of Cohen's d

Cohen's d of -0.32 indicates a small to moderate effect size, suggesting that there is a small to moderate difference in test scores between Brand X users and the overall population.

Research Question Answer

Yes, children using Brand X hearing aids did perform significantly worse than the overall population on the hearing test.