Sure, let's solve this step-by-step.
Step 1: State the given information.
- Sample size ([tex]\( n \)[/tex]): 14
- Sample mean ([tex]\( \bar{x} \)[/tex]): 1194
- Sample standard deviation ([tex]\( s \)[/tex]): 166.6
- Population mean ([tex]\( \mu \)[/tex]): 1023
- Confidence level: 0.95
Step 2: Calculate the t-value for the sample mean.
The formula for the t-value is:
[tex]\[
t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}}
\][/tex]
Plugging in the values:
[tex]\[
t = \frac{1194 - 1023}{\frac{166.6}{\sqrt{14}}}
\][/tex]
To find the denominator:
[tex]\[
\frac{166.6}{\sqrt{14}} \approx 44.5
\][/tex]
Now, solving for [tex]\( t \)[/tex]:
[tex]\[
t = \frac{171}{44.5} \approx 3.84
\][/tex]
Step 3: Determine the critical value [tex]\( t_{0.95} \)[/tex] for a 95% confidence level.
For a 95% confidence level with [tex]\( df = n - 1 = 14 - 1 = 13 \)[/tex], the critical t-value can be obtained from t-distribution tables or statistical software. For a two-tailed test, we find [tex]\( t_{0.975} \)[/tex] because [tex]\(0.95\)[/tex] confidence level corresponds to [tex]\(0.025\)[/tex] in each tail.
The critical t-value [tex]\( t_{0.975} \approx 2.16 \)[/tex].
Step 4: Compare the calculated t-value with the critical t-value.
The calculated t-value is:
[tex]\[
t = 3.84
\][/tex]
The critical t-value [tex]\( t_{0.95} \)[/tex] for 13 degrees of freedom is approximately 2.16.
Since [tex]\( 3.84 \)[/tex] is greater than [tex]\( 2.16 \)[/tex], the t-value does not fall between [tex]\( -2.16 \)[/tex] and [tex]\( 2.16 \)[/tex].
Therefore, we can state:
The [tex]\( t \)[/tex]-value of [tex]\( t = 3.84 \)[/tex] does not fall between [tex]\( -t_{0.95} \)[/tex] and [tex]\( t_{0.95} \)[/tex] because [tex]\( t_{0.95} = 2.16 \)[/tex].
