Here is the information that we have:
[tex]\begin{gathered} Claim: \\ \mu>1290 \\ \alpha=0.06 \\ \sigma=197.95 \\ Sample\text{ Statistics:} \\ X=1319.39 \\ n=275 \end{gathered}[/tex]A)
According to the claim and the sample statistics, the null hypothesis and the alterative hypothesis are given by:
[tex]\begin{gathered} H_0:\mu>1290 \\ H_a:\mu\leq1290 \end{gathered}[/tex]B)
The standardized test statistic is given by:
[tex]|z|=|\frac{\mu-X}{\frac{\sigma}{\sqrt{n}}}|=|\frac{1290-1319.39}{\frac{197.95}{\sqrt{275}}}|=2.46[/tex]C)
The p-value represent the probability of obtaining a result that confronts the claim. According to a normal table, for z = -2.46, the p-value is given by p = 1 - 0.5 - 0.4931 = 0.0069
D)
Since p = 0.0069 < a = 0.06, the null hypothesis is not rejected.
