Question

(a) A random sample of 10 houses in a particular area, each of which is heated with natural gas, is selected and the amount of gas (therms) used during the month of January is determined for each house. The resulting observations are 153, 103, 125, 149, 118, 109, 86, 122, 138, 99. Let μ denote the average gas usage during January by all houses in this area. Compute a point estimate of μ.therms(b) Suppose there are 25,000 houses in this area that use natural gas for heating. Let τ denote the total amount of gas used by all of these houses during January. Estimate τ using the data of part (a).therms(c) Use the data in part (a) to estimate p, the proportion of all houses that used at least 100 therms.(d) Give a point estimate of the population median usage (the middle value in the population of all houses) based on the sample of part (a).therms

Answer 1

Answer:

a) $\hat \mu = \bar x =\sum_{i=1}^{10} \frac{x_i}{10}=120.2$

b) $\hat \tau =n\hat \mu =25000x120.2=3005000$

c) $\hat p= \hat \theta =\frac{8}{10}=0.8$

d) Median =120

Step-by-step explanation:

1) Some important concepts

The mean refers to the "average that is used to derive the central tendency of data analyzed. It is determined by adding all the data points in a population and then dividing the total by the number of points".

Method of moments "involves equating sample moments with theoretical moments". For example the first sample moment about the origin is defined as $M_1=\frac{1}{n} \sum_{i=1}^{n}x_i =\bar X$

The median is "the middle number in a sorted, ascending or descending, list of numbers and can be more descriptive of that data set than the average".

When we are trying to estimate the population proportion, p.

All estimation is based on the fact that the normal can be used to approximate the binomial distribution when np and nq are both at least 5. Where p is the probability of success and q the probability of failure.

2) Part a

Using the method of the moments a point of estimate for the $\mu$ is:

$\hat \mu = \bar x =\sum_{i=1}^{10} \frac{x_i}{10}=120.2$

3) Part b

If $\hat \mu$ is an individual estimate for the average gas usage during January and $\tau$ represent "the total amount of gas used by all of these houses during January" then the estimation for the total would be given by:

$\hat \tau =n\hat \mu =25000x120.2=3005000$

3) Part c

For this part we want to estimate p ="the proportion of all houses that used at least 100 therms". If X is the random variable who represent the number of houses that exceed the usage of 100, we see that 8 out of 10 values are above 100, so the random variable X would be distributed binomial

$X \sim Bin(10,0.8)$ where n=10 and

$\hat p= \hat \theta =\frac{8}{10}=0.8$

4) Part d

In order to find the median we need to put the data in order first, like this:

86,99,103,109,118,122,125,138,149,153

Since we have 10 observations and this number is even the procedure that we need to use in order to find the median is:

a) Find the value at position[n/2]=[10/2]=[5] on the data set ordered. For this case the value at position [5] is 118

b) Find the value at position[n/2 +1]=[10/2 +1]=[6] on the data set ordered. For this case the value at position [6] is 122

c) Find the average from the values obtained on steps a) and b). for this case (118+122)/2=120

So the Median = 120

Answer 2

The mean of the data set is 120.2

The total gas used =  3005000

The proportion of at least therms = 0.8

The median of the set = 120

a. To get the average gas usage, we are asked to calculate the mean of the observation.

Average

$\frac{ 153+103+125+149+118+109+86+122+138+99}{10}$

= 120.2

b. The question says that 25000 houses make use of natural gas, then

total gases used by these houses =

120.2 * 25000

= 3005000

c. The proportion of the houses that used above 100 therms,

The house above 100 here are, 125, 149, 118, 109, 122, 138

They are 8 in number.

8/10 = 0.8

0.8 is the proportion that used at least 100 therms.

d. We have to find the median for the set here. We arrange the details in ascending order.

86,99,103,109,118,122,125,138,149,153

Median = 118+122/2

= 240/2

= 120