The first thing we need is the sample mean xbar. Normally we add up all the values and divide by the sample size n; however, we don't have the actual data values. Instead we have class ranges with frequencies attached. There is a way to figure out the sample mean though.
Compute the midpoints of each class interval. Add up the endpoints and divide by 2, so (8.35+8.43)/2 = 8.39
Therefore the midpoint of 8.35 and 8.43 is 8.39
The midpoint of 8.44 and 8.52 is 8.48 (similar reasons as computing the previous midpoint)
The midpoint of 8.53 and 8.61 is 8.57
The midpoint of 8.62 and 8.70 is 8.66
The midpoint of 8.71 and 8.79 is 8.75
The midpoint of 8.80 and 8.80 is 8.84
The midpoints are:
8.39, 8.48, 8.57, 8.66, 8.75, 8.84
Multiply those midpoints with the corresponding frequencies:
8.39*2 = 16.78
8.48*6 = 50.88
8.57*12 = 102.84
8.66*18 = 155.88
8.75*10 = 87.5
8.84*2 = 17.68
Those products are then added up:
16.78+50.88+102.84+155.88+87.5+17.68 = 431.56
Divide this sum by the total frequency 50 (sum of the frequency values or we can use the given sample size)
431.56/50 = 8.6312
All that work led to the value of xbar
xbar = 8.6312
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We know xbar now, so we can use it to find the z score
z = (xbar - mu)/(sigma/sqrt(n))
z = (8.6312 - 8.65)/(0.105/sqrt(50))
z = -1.2660579
which is approximate
If you have a texas instruments (TI) calculator, then you'll follow these steps
Step 1) Hit the button labeled "2ND" up at the top left corner. Then hit the key labeled "VARS". This brings up the statistical distribution menu.
Step 2) Scroll down to the second item labeled "normalcdf" and hit enter
Step 3) Type in some large negative value, say -99, and then the z score we got which was approximately -1.2660579. Then close things off with a closing parenthesis. See the attached image to see what I mean.
Following these steps will compute the area under the standard normal z curve from z = -99 to z = -1.2660579. The -99 is used since its so far out to the left that its practically negative infinity.
The calculator should produce an approximate result of 0.1027 (as the attached image shows)
This represents the area to the left of the z score (the area under the curve of course). Since this is a two tail test, we double this area to get 2*0.1027 = 0.2054
The p value is approximately 0.2054 which is much larger than the alpha value 0.05
Since the p value is larger than the alpha value, we fail to reject the null hypothesis (mu = 8.65). We only reject the null if the p value is smaller than alpha.
We do not have enough statistically significant evidence to reject the null. We have no choice but to accept the null.
So to answer the question: yes the president's statement is believable. The hourly average wage at this company is $8.65
