Your local newspaper contains a large number of advertisements for unfurnished one-bedroom apartments in a particular neighborhood. you choose 36 at random and calculate that their mean monthly rent is $530 with a standard deviation of $78. you want to find out if the sample data gives a good reason to believe that the average rent for all advertised one-bedroom apartments is less than $550 per month. conduct the following hypothesis test at 0.01 level of significance.
μ < 550
Let us take,
Null Hypothesis(H₀) : μ = 550
Alternative Hypothesis(H₁) : μ < 550
Now, The Z-statistic for mean is given by,
Here, n = 36, μ = 550, = 530, s = 78
Z = (530 - 550) ÷ √(78 / 36)
Z = -20 ÷ 1.47196
Z = -13.58732
Thus, Z-value = -13.587 with 35 degree of freedom.
and α = 0.01
The value of p is < .00001.
Since, the value of p is less than α.
The result is significant at p < .01.
Thus, we reject the null-hypothesis.
Hence, μ < 550
the graph has a slope of 1/3 because each time you move up 1, you go to the right 3. so, slope = rise/run = 1/3. we can cross choice a off the list since we want a slope of 3, rather than 1/3.
the first table shown has a slope of -3. this is because the rise is -6 and the run is 2, so slope = rise/run = -6/2 = -3. note how x increases by 2 each time (run) and y drops by 6 each time (rise). choice b can be eliminated.
the second table has a slope of 3. each time x goes up by 1, the y values increase by 3. slope = rise/run = 3/1 = 3. alternatively you can use the slope formula to prove that the slope is 3 here. use any two rows of values to generate the two points. choice c is one of the answers
the third table also has a slope of 3. let's use the slope formula to prove as such. i'm going to use the first two rows of the table. so i'll use the two points (2,1) and (5,10)
m = slope
m = (y2-y1)/(x2-x1)
m = (10-1)/(5-2)
m = 9/3
m = 3 so we get a slope of 3
choice d is the other answer
where's the diagrams?