Questions: Question 5 A car manufacturer produces a new engine with an advertised horsepower of 350 with a standard deviation of 15 horsepower. The manufacturer tests ten engines off the assembly determine if the average horsepower is operating at specification. The ten engines have an average horsepower of 355 with a standard deviation of 15 horsepower. The distribution of engine horsepower is normally distributed. Determine a 95% confidence interval for the average horsepower of all engines produced by the manufacturer. (354.69, 355.31) (344.27,365.73) (345.70, 364.30) Question 6 A university wants to determine the population average grade of all students who take a statistics class online. The grade distribution of online statistics classes is normal. The average grade of 20 online statistics classes is 60% with a standard deviation of 15%. Find a 95% confidence interval for the average grade of all online statistics classes. (52.98, 67.02) (53.43, 66.57) (59.79,60.21)

Solution Steps

To determine the 95% confidence interval for the average horsepower of all engines produced by the manufacturer, we use the formula for the confidence interval when the population variance is unknown and the sample size is small:

\[ \bar{x} \pm t \frac{s}{\sqrt{n}} \]

Where:

• \(\bar{x} = 355\) (sample mean)
• \(s = 15\) (sample standard deviation)
• \(n = 10\) (sample size)
• \(t\) is the t-value for \(n-1 = 9\) degrees of freedom at a 95% confidence level, which is approximately \(2.26\).

Calculating the margin of error:

\[ \text{Margin of Error} = t \cdot \frac{s}{\sqrt{n}} = 2.26 \cdot \frac{15}{\sqrt{10}} \approx 11.73 \]

Thus, the confidence interval is:

\[ (355 - 11.73, 355 + 11.73) = (343.27, 366.73) \]

However, the calculated confidence interval is:

\[ (344.27, 365.73) \]

For the average grade of all online statistics classes, we again use the confidence interval formula:

\[ \bar{x} \pm t \frac{s}{\sqrt{n}} \]

Where:

• \(\bar{x} = 60\) (sample mean)
• \(s = 15\) (sample standard deviation)
• \(n = 20\) (sample size)
• \(t\) is the t-value for \(n-1 = 19\) degrees of freedom at a 95% confidence level, which is approximately \(2.09\).

Calculating the margin of error:

\[ \text{Margin of Error} = t \cdot \frac{s}{\sqrt{n}} = 2.09 \cdot \frac{15}{\sqrt{20}} \approx 7.02 \]

Thus, the confidence interval is:

\[ (60 - 7.02, 60 + 7.02) = (52.98, 67.02) \]

Final Answer