(a)
83.5

Formula: \( \bar{x} = \frac{\sum x}{n} \)

Given: \( \sum x = 2506 \), \( n = 30 \)

Calculate: \( \bar{x} = \frac{2506}{30} \approx 83.5333333333 \)

Round to one decimal place: 83.5

Result: 83.5 [1]

(b)
13.9

Formula: \( S^2_{n-1} = \frac{\sum x^2 – \frac{(\sum x)^2}{n}}{n-1} \)

Given: \( \sum x = 2506 \), \( \sum x^2 = 209,738 \), \( n = 30 \)

Calculate: \( (\sum x)^2 = 2506^2 = 6,280,036 \)
\( \frac{(\sum x)^2}{n} = \frac{6,280,036}{30} \approx 209,334.5333333333 \)
\( \sum x^2 – \frac{(\sum x)^2}{n} = 209,738 – 209,334.5333333333 \approx 403.4666666667 \)
\( S^2_{n-1} = \frac{403.4666666667}{30-1} = \frac{403.4666666667}{29} \approx 13.9126436782 \)

Round to one decimal place: 13.9

Result: 13.9 [2]

(c)
(82.1, 84.9)

Formula: \( \bar{x} \pm t \cdot \frac{s}{\sqrt{n}} \)

Given: \( \bar{x} = 83.5333333333 \), \( S^2_{n-1} \approx 13.9126436782 \), \( n = 30 \), \( t \approx 2.045 \) (for 95%, \( df = 29 \))

Calculate standard deviation: \( s = \sqrt{13.9126436782} \approx 3.7297766971 \)
Standard error: \( \frac{s}{\sqrt{n}} = \frac{3.7297766971}{\sqrt{30}} \approx 0.6808278147 \)
Margin of error: \( 2.045 \times 0.6808278147 \approx 1.392292881 \)
Interval: \( 83.5333333333 – 1.392292881 \approx 82.1410404523 \)
\( 83.5333333333 + 1.392292881 \approx 84.9256262143 \)

Round to one decimal place: (82.1, 84.9)

Result: (82.1, 84.9) [2]

(d)
The manufacturer’s claim is incorrect because 85 is outside the 95% confidence interval (82.1, 84.9)

Manufacturer’s claim: \( \mu = 85 \)

Compare: 95% confidence interval is (82.1, 84.9)
\( 85 > 84.9 \), so 85 is not within the interval

Evidence suggests the true mean differs from 85

Result: The manufacturer’s claim is likely incorrect [1]