1. Past experience indicates that the variance, ¦Ò2x, of the scores, X, obtained on the verbal portion of a test is 5,625. Similarly, the variance ¦Ò2y, of the scores, Y, on the mathematical portion is 2,500.

(a) A random sample of size n = 64 yields a sample mean verbal score of 600. Find a 95% confidence interval on the true average score obtained on the verbal aptitude test.
(b) A random sample of size n = 80 yields a sample mean mathematics score of 500. Find a 95% confidence interval on the true average score obtained on the mathematical aptitude test.
(c) Find a 90% confidence interval on ¦Ìy based on the data from (b). Compare the two (95% and 90%) intervals obtained. What does this result imply about the relationship between interval length and degree of confidence?
(d) Ten years prior to the year in which the above data were obtained, ¦Ìy was found to be 520. Can we safely conclude that the average score has declined over the past 10 years? Explain your answer briefly, on the basis of the confidence intervals constructed in (b) and (c)

(a) and (b) Variance = SD^2

95% = mean ± 1.96 SEm

SEm = SD/√(n-1)

(c) 90% = mean ± 1.645 SEm

(d) Z = (mean1 - mean2)/standard error (SE) of difference between means

SEdiff = √(SEmean1^2 + SEmean2^2)

SEm = SD/√(n-1)

If only one SD is provided, you can use just that to determine SEdiff.

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to the Z score.

I leave the calculations to you.

(a) To find the 95% confidence interval for the true average score obtained on the verbal aptitude test, we can use the formula:

CI = X̄ ± Z * (√(σ^2 / n))

Where:
CI = Confidence Interval
X̄ = Sample mean verbal score (600)
Z = Z-score for 95% confidence level (from Z-table, for 95% confidence level, Z = 1.96)
σ^2 = Variance of the scores (5,625)
n = Sample size (64)

Plugging in the values:

CI = 600 ± 1.96 * (√(5,625 / 64))

Calculating:

CI ≈ 600 ± 1.96 * (√(88.08))
CI ≈ 600 ± 1.96 * 9.378
CI ≈ 600 ± 18.359

The 95% confidence interval for the true average score obtained on the verbal aptitude test is approximately (581.641, 618.359).

(b) Similarly, for the mathematical aptitude test, the formula for the 95% confidence interval is:

CI = X̄ ± Z * (√(σ^2 / n))

Where:
CI = Confidence Interval
X̄ = Sample mean mathematics score (500)
Z = Z-score for 95% confidence level (from Z-table, for 95% confidence level, Z = 1.96)
σ^2 = Variance of the scores (2,500)
n = Sample size (80)

Plugging in the values:

CI = 500 ± 1.96 * (√(2,500 / 80))

Calculating:

CI ≈ 500 ± 1.96 * (√(31.25))
CI ≈ 500 ± 1.96 * 5.590
CI ≈ 500 ± 10.949

The 95% confidence interval for the true average score obtained on the mathematical aptitude test is approximately (489.051, 510.949).

(c) To find a 90% confidence interval on σy based on the data from (b), we can use the formula:

CI = X̄ ± Z * (√(σ^2 / n))

Where:
CI = Confidence Interval
X̄ = Sample mean mathematics score (500)
Z = Z-score for 90% confidence level (from Z-table, for 90% confidence level, Z = 1.645)
σ^2 = Variance of the scores (2,500)
n = Sample size (80)

Plugging in the values:

CI = 500 ± 1.645 * (√(2,500 / 80))

Calculating:

CI ≈ 500 ± 1.645 * (√(31.25))
CI ≈ 500 ± 1.645 * 5.590
CI ≈ 500 ± 9.193

The 90% confidence interval for σy is approximately (490.807, 509.193).

Comparing the two intervals (95% and 90%), we can see that the 90% confidence interval (490.807, 509.193) is narrower than the 95% confidence interval (489.051, 510.949). This implies that as the degree of confidence decreases (from 95% to 90%), the interval length gets smaller.

(d) Based on the confidence intervals constructed in (b) and (c), we cannot safely conclude that the average score has declined over the past 10 years. This is because the confidence intervals do not overlap and the true average score could fall within either interval, depending on the degree of confidence.

To find the confidence intervals in this situation, we can use the formula:

Confidence Interval = Sample Mean ± (Critical Value * Standard Deviation / sqrt(n))

where:
- Sample Mean is the mean score obtained in the sample.
- Critical Value is the value from the standard normal distribution corresponding to the desired confidence level.
- Standard Deviation is the square root of the variance.

Now, let's solve the problems step by step:

(a) To find the 95% confidence interval on the true average verbal score, we need to calculate the critical value for a 95% confidence level and use the given sample mean and the variance. The critical value for 95% confidence level is approximately 1.96 (obtained from a standard normal distribution table).

The formula for the confidence interval is:
Confidence Interval = Sample Mean ± (Critical Value * Standard Deviation / sqrt(n))

Plugging in the values, we have:
Confidence Interval = 600 ± (1.96 * sqrt(5,625) / sqrt(64))

Calculating this expression, we can find the confidence interval for the true average score on the verbal aptitude test.

(b) Similarly, for the mathematics portion, we can use the same formula but with the given sample mean, variance, and sample size (n = 80). The critical value for 95% confidence level is still 1.96.

Confidence Interval = 500 ± (1.96 * sqrt(2,500) / sqrt(80))

Calculating this expression, we can find the confidence interval for the true average score on the mathematical aptitude test.

(c) To find a 90% confidence interval on ¦Ìy based on the data from (b), we use the same formula but with a different critical value. The critical value for a 90% confidence level is approximately 1.645 (obtained from a standard normal distribution table).

Confidence Interval = 500 ± (1.645 * sqrt(2,500) / sqrt(80))

Comparing the 95% and 90% confidence intervals obtained, we can see that the 90% confidence interval is narrower than the 95% confidence interval. This implies that as the degree of confidence decreases, the interval length decreases as well.

(d) To determine if the average score has declined over the past 10 years, we need to compare the confidence intervals obtained in (b) and (c) with the value of ¦Ìy from 10 years prior (520).

If the value 520 falls within the confidence intervals obtained in (b) and (c), we can conclude that there is no significant change in the average score. However, if 520 falls outside the confidence intervals, we may conclude that the average score has declined.

By comparing the confidence intervals and determining if 520 is within those intervals, we can determine whether the average score has declined over the past 10 years.