Answer:
Step-by-step explanation:
all you need to do is use the distributive property and then you"ll get the answer
SOLUTION:
Case: Hypothesis testing
Step 1: Null and Alternative hypotheses
![\begin{gathered} H_0:\mu=P127.50 \\ H_1:\mu\leq P127.50 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20H_0%3A%5Cmu%3DP127.50%20%5C%5C%20H_1%3A%5Cmu%5Cleq%20P127.50%20%5Cend%7Bgathered%7D)
Step 2: T-test analysis
![\begin{gathered} t=\frac{\hat{x}-\mu}{\frac{s}{\sqrt{n}}} \\ t=\frac{135-127.5}{\frac{75.25}{\sqrt{450}}} \\ t=2.144 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20t%3D%5Cfrac%7B%5Chat%7Bx%7D-%5Cmu%7D%7B%5Cfrac%7Bs%7D%7B%5Csqrt%7Bn%7D%7D%7D%20%5C%5C%20t%3D%5Cfrac%7B135-127.5%7D%7B%5Cfrac%7B75.25%7D%7B%5Csqrt%7B450%7D%7D%7D%20%5C%5C%20t%3D2.144%20%5Cend%7Bgathered%7D)
Step 3: t-test with the significance level
![\begin{gathered} t_{\alpha}=? \\ \alpha=0.05 \\ From\text{ }tables \\ t_{0.05}=1.654 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20t_%7B%5Calpha%7D%3D%3F%20%5C%5C%20%5Calpha%3D0.05%20%5C%5C%20From%5Ctext%7B%20%7Dtables%20%5C%5C%20t_%7B0.05%7D%3D1.654%20%5Cend%7Bgathered%7D)
Step 4: Comparing
![t>t_{\alpha}](https://tex.z-dn.net/?f=t%3Et_%7B%5Calpha%7D)
So tail to reject the null hypothesis. There is enough evidence at a 0.05 level of significance to claim that the mean spent is greater than P127.50.
Final answer:
Yes, there is evidence sufficient to conclude that the mean amount spent is greater than P127.50 per month at a 0.05 level of significance.
The mean is the average of the numbers. It is easy to calculate: add up all the numbers, then divide by how many numbers there are. In other words it is the sum divided by the count.