Yes, since p of 0.037 is less than 0.05, reject the null. Claim is alternative, so is supported
Step-by-step explanation:
[tex]\µ = 6[/tex]
Sample size = 16
S = 7.3
[tex]\bar{x}=9.5[/tex]
Significance level ( ∝) =0.05
[tex]H_{0}[/tex] : µ = 6
[tex]H_{a}[/tex] : µ > 6
The sample size n < 30 and the standard population deviation is uncertain, we will be using one t-test
Test Statistic :
[tex]t=\bar{X}-\mu / s[/tex]
[tex]t=\frac{9.5-6}{7.3 / \sqrt{16}}=1.9178[/tex]
Finding the p value,
P value = 0.0371