Suppose that the random variable X represents the amount of electrical energy used (in kwH) in a month for residents in Virginia. The historical amount of electrical energy used in a month for residents is 102 kwH. The following data (in kwH per month) were recorded from a random sample of 8 residents: 111 113 145 105 90 100 150 88(a) Calculate the mean and sample variance of X. (b) What is t statistic with this problem?

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dfbustos dfbustos · Answer 1 · 2020-04-04T05:03:14+02:00

Answer:

a)[tex] \bar X = \frac{\sum_{i=1}^n X_i}{n}[/tex]

[tex]\bar X = 112.75[/tex]

[tex]s^2 = \frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}[/tex]

[tex] s^2 = 540.5[/tex]

b) [tex] t = \frac{\bar X -\mu}{\frac{s}{\sqrt{n}}}[/tex]

And if we replace we got:

[tex] t = \frac{112.75-112}{\frac{23.249}{\sqrt{8}}}= 0.0912[/tex]

Step-by-step explanation:

Part a

For this case we have the following data: 111 113 145 105 90 100 150 88

We can calculate the mean with the following formula:

[tex] \bar X = \frac{\sum_{i=1}^n X_i}{n}[/tex]

And replacing we got:

[tex]\bar X = 112.75[/tex]

And the sample variance can be calculated with this formula:

[tex]s^2 = \frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}[/tex]

And replacing we got:

[tex] s^2 = 540.5[/tex]

Part b

For this case we want to check is the true mean is equal to 102 or no, the t statistic is given by:

[tex] t = \frac{\bar X -\mu}{\frac{s}{\sqrt{n}}}[/tex]

And if we replace we got:

[tex] t = \frac{112.75-112}{\frac{23.249}{\sqrt{8}}}= 0.0912[/tex]