Answer :
To find the 90% confidence interval for Steve's average golf score, we will use the formula for a confidence interval for a population mean when the population standard deviation is known:
[tex]CI = \bar{x} \pm z_{\frac{\alpha}{2}} \cdot \frac{\sigma}{\sqrt{n}}[/tex]
Where:
- [tex]\bar{x}[/tex] is the sample mean,
- [tex]z_{\frac{\alpha}{2}}[/tex] is the z-score that corresponds to the desired confidence level,
- [tex]\sigma[/tex] is the population standard deviation, and
- [tex]n[/tex] is the sample size.
Given:
- Sample mean ([tex]\bar{x}[/tex]) = 93.6
- Population standard deviation ([tex]\sigma[/tex]) = 4.2
- Sample size ([tex]n[/tex]) = 34
For a 90% confidence interval, we look up the z-score for [tex]\alpha/2 = 0.05[/tex], which is approximately 1.645 (from the z-table).
Now, we can plug these values into the formula:
[tex]CI = 93.6 \pm 1.645 \cdot \frac{4.2}{\sqrt{34}}[/tex]
First, calculate [tex]\frac{4.2}{\sqrt{34}}[/tex]:
[tex]\frac{4.2}{\sqrt{34}} \approx \frac{4.2}{5.831} \approx 0.7203[/tex]
Next, calculate the margin of error:
[tex]1.645 \times 0.7203 \approx 1.184[/tex]
Now, compute the confidence interval:
[tex]CI = 93.6 \pm 1.184[/tex]
This gives us:
- Lower limit: [tex]93.6 - 1.184 = 92.416[/tex]
- Upper limit: [tex]93.6 + 1.184 = 94.784[/tex]
Rounding the numbers to one decimal place, the 90% confidence interval is approximately [tex]92.4[/tex] to [tex]94.8[/tex].
So, the correct answer is C) 92.4 to 94.8.
