Complete Question
The option to the blank space are shown on the first uploaded image
Answer:
Neal's decision is to <u>fail to reject </u> the <u>null hypothesis</u> (p 0.066). There is <u>no sufficient </u> evidence to <u>prove</u> the claim that the average electricity consumption of <u>all Ledee household </u>is <u>greater than</u> , <u>854.28 kWh</u>
Step-by-step explanation:
From the question we are told that
The population mean is ![\mu = 854.11](https://tex.z-dn.net/?f=%5Cmu%20%20%3D%20%20854.11)
The sample size is ![n = 65](https://tex.z-dn.net/?f=n%20%3D%20%2065)
The sample mean is ![\= x = 879.28 \ kWh](https://tex.z-dn.net/?f=%5C%3D%20x%20%20%3D%20%20879.28%20%5C%20%20kWh)
The standard deviation is ![\sigma = 133.29 \ kWh](https://tex.z-dn.net/?f=%5Csigma%20%20%3D%20%20133.29%20%5C%20%20kWh)
The level of significance is ![\alpha = 0.05](https://tex.z-dn.net/?f=%5Calpha%20%20%3D%20%200.05)
The null hypothesis is ![H_o: \mu = 854.11](https://tex.z-dn.net/?f=H_o%3A%20%20%5Cmu%20%3D%20854.11)
The alternative hypothesis is ![H_a : \mu > 854.11](https://tex.z-dn.net/?f=H_a%20%20%3A%20%20%5Cmu%20%3E%20854.11)
The t-statistics is ![t = 1.522](https://tex.z-dn.net/?f=t%20%20%3D%20%201.522)
The p-value is ![p-value = 0.066](https://tex.z-dn.net/?f=p-value%20%20%3D%20%200.066)
Now from the given data we can see that
![p-value < \alpha](https://tex.z-dn.net/?f=p-value%20%3C%20%20%5Calpha)
Generally when this is the case , we fail to reject the null hypothesis
So
Neal's decision is to <u>fail to reject </u> the <u>null hypothesis</u> (p 0.066). There is <u>no sufficient </u> evidence to <u>prove</u> the claim that the average electricity consumption of <u>all Ledee household </u>is <u>greater than</u> , <u>854.28 kWh</u>