. A random sample of 100 children with working mothers showed that they were absent from school an average of 6 days per term wi
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Answer:
a) 0.057
b) 0.5234
c) 0.4766
Step-by-step explanation:
a)
To find the p-value if the sample average is 185, we first compute the z-score associated to this value, we use the formula
where
N = size of the sample.
So,
As the sample suggests that the real mean could be greater than the established in the null hypothesis, then we are interested in the area under the normal curve to the right of 1.5811 and this would be your p-value.
We compute the area of the normal curve for values to the right of 1.5811 either with a table or with a computer and find that this area is equal to 0.0569 = 0.057 rounded to 3 decimals.
So the p-value is
b)
Since the z-score associated to an α value of 0.05 is 1.64 and the z-score of the alternative hypothesis is 1.5811 which is less than 1.64 (z critical), we cannot reject the null, so we are making a Type II error since 175 is not the true mean.
We can compute the probability of such an error following the next steps:
<u>Step 1 </u>
Compute
So <em>we would make a Type II error if our sample mean is less than 185.3721</em>.
<u>Step 2</u>
Compute the probability that your sample mean is less than 185.3711
So, <em>the probability of making a Type II error is 0.5234 = 52.34% </em>
c)
<em>The power of a hypothesis test is 1 minus the probability of a Type II error</em>. So, the power of the test is
1 - 0.5234 = 0.4766
It is related to econometrics, however using simple algebraic logic we can solve this problem. General formula with residual is 
, where r is a residual. Plugging the pair into the formula, we obtain that 
and r=0.5