To determine whether we reject the null hypothesis [tex]\( H_0: \mu \leq 12 \)[/tex] in favor of the alternative hypothesis [tex]\( H_a: \mu > 12 \)[/tex], we will follow the outlined steps.
### Given Values:
- Sample size ([tex]\( n \)[/tex]): 25
- Sample mean ([tex]\( \bar{x} \)[/tex]): 14
- Sample standard deviation ([tex]\( s \)[/tex]): 4.32
- Population mean ([tex]\( \mu \)[/tex]): 12
### a. Compute the value of the test statistic
To compute the test statistic for a one-sample t-test, use the formula:
[tex]\[
t = \frac{\bar{x} - \mu}{s / \sqrt{n}}
\][/tex]
Substituting the values:
[tex]\[
t = \frac{14 - 12}{4.32 / \sqrt{25}}
\][/tex]
[tex]\[
t = \frac{2}{4.32 / 5}
\][/tex]
[tex]\[
t = \frac{2}{0.864} \approx 2.31
\][/tex]
The value of the test statistic is [tex]\( t \approx 2.31 \)[/tex].
### b. Use the [tex]\( t \)[/tex] distribution table to compute a range for the [tex]\( p \)[/tex]-value
To find the range for the [tex]\( p \)[/tex]-value, we need the degrees of freedom (df), calculated as:
[tex]\[
df = n - 1 = 25 - 1 = 24
\][/tex]
Looking at the [tex]\( t \)[/tex]-distribution table for 24 degrees of freedom, we find the range that includes the test statistic [tex]\( t = 2.31 \)[/tex]:
- For a one-tailed test with [tex]\( t \)[/tex] around 2.31, the corresponding [tex]\( p \)[/tex]-value falls between 0.01 and 0.025.
Thus, the [tex]\( p \)[/tex]-value is between 0.01 and 0.025.
### c. Using [tex]\( \alpha = 0.05 \)[/tex], what is the critical value for the test statistic?
For [tex]\( \alpha = 0.05 \)[/tex] in a one-tailed test with 24 degrees of freedom, we find the critical value from the [tex]\( t \)[/tex]-distribution table:
- The critical value for [tex]\( \alpha = 0.05 \)[/tex] and [tex]\( df = 24 \)[/tex] is approximately 1.711.
### Summary:
a. The value of the test statistic is [tex]\( 2.31 \)[/tex].
b. The [tex]\( p \)[/tex]-value is between 0.01 and 0.025.
c. The critical value for the test statistic at [tex]\( \alpha = 0.05 \)[/tex] is [tex]\( 1.711 \)[/tex].
