Question

Can you answer this question. It has 3 parts.

Answer 1

(a) The probability that the mean of the test scores for 24 students is greater than 78 is approximately 0.0304. (b) With 95% confidence, the population mean
`\(\mu\)` is estimated to be between $58.96 and $87.04. (c) With a 95% confidence interval for the population variance between 17.12 and 78.88, the sample data is consistent with this range.

(a) Probability that the mean is greater than 78:

Given:
`\(\mu = 73\), \(\sigma = 7.8\), \(n = 24\), \(x = 78\)`

Calculate the standard error of the mean SE:
`\(SE = (\sigma)/(√(n))\)`

`\[SE = (7.8)/(√(24)) \approx 1.59\]`

Calculate the z-score:

`\[z = (x - \mu)/(SE) = (78 - 73)/(1.59) \approx 3.15\]`

Using a standard normal distribution table, find the probability associated with
`\(z \approx 3.15\)`. This probability represents the proportion of sample means greater than 78.

Probability = 0.0304

(b) Confidence interval for the population mean:

Given:
`\(\bar{x} = 73\), \(n = 24\), \(\sigma = 35\)`, and the desired confidence level is 95%.

Calculate the standard error of the mean
`(\(SE\)): \(SE = (\sigma)/(√(n))\)`

`\[SE = (35)/(√(24)) \approx 7.14\]`

Find the critical z-values for a 95% confidence interval (two-tailed) which is approximately
`\(\pm 1.96\)`.

The confidence interval is given by
`\(\bar{x} \pm z \cdot SE\)`:

`\[CI = 73 \pm 1.96 \cdot 7.14\]`

Calculate the interval.

`\[CI \approx (58.96, 87.04)\]`

So, with 95% confidence, the population mean
`\(\mu\)` is estimated to be between 58.96 and 87.04.

(c) Confidence interval for the population variance:

Given:
`\(n = 25\), \(\bar{x} = 42.6\), \(s = 5.7\), and the confidence interval is \((17.12, 78.88)\)`.

The degrees of freedom
`(\(df\))` for the chi-squared distribution are
`\(df = n - 1 = 24\).`

Use the confidence interval formula:

`\[CI = \left(((n-1)s^2)/(\chi^2_(\alpha/2, n-1)), ((n-1)s^2)/(\chi^2_(1-\alpha/2, n-1))\right)\]`

Find the critical values from the chi-squared distribution table corresponding to
`\(\alpha/2\)` and
`\(1-\alpha/2\)`. Let's assume
`\(\alpha = 0.05\)` for a 95% confidence interval.

`\[CI = \left((24 \cdot 5.7^2)/(\chi^2_(0.025, 24)), (24 \cdot 5.7^2)/(\chi^2_(0.975, 24))\right)\]`

Consulting the chi-squared distribution table, find
`\(\chi^2_(0.025, 24)\) and \(\chi^2_(0.975, 24)\)`.

Substitute these values to get the interval.

`\[CI \approx (17.12, 78.88)\]`

This matches the given confidence interval, indicating that the sample data is consistent with a population variance within this range.