The length of time required for the periodic maintenance of an automobile or another machine usually has a mound-shaped probabil
1 answer:
Solution :
Mean time for an automobile to run a 5000 mile check and service = 1.4 hours
Standard deviation = 0.7 hours
Maximum average service time = 1.6 hours for one automobile
The z - score for 1.6 hours =
= 2.02
Now checking a normal curve table the percentage of z score over 2.02 is 0.0217
Therefore the overtime that will have to be worked on only 0.217 or 2.017% of all days.
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Answer:
a) Null hypothesis:
Alternative hypothesis:
b)
c) So the p value obtained was a very low value and using the significance level given we have
so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of American families owning stocks or stock funds is significantly less than 0.53 .
Step-by-step explanation:
Data given and notation
n=300 represent the random sample taken
estimated proportion of American families owning stocks or stock funds
is the value that we want to test
represent the significance level
Confidence=99% or 0.99
z would represent the statistic (variable of interest)
represent the p value (variable of interest)
Concepts and formulas to use
Part a
We need to conduct a hypothesis in order to test the claim that proportion is less than 0.53 or 53%.:
Null hypothesis:
Alternative hypothesis:
Part b
When we conduct a proportion test we need to use the z statistic, and the is given by:
(1)
The One-Sample Proportion Test is used to assess whether a population proportion is significantly different from a hypothesized value
.
Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
Statistical decision
It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.
The significance level provided . The next step would be calculate the p value for this test.
Since is a left tailed test the p value would be:
Part c
So the p value obtained was a very low value and using the significance level given we have
so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of American families owning stocks or stock funds is significantly less than 0.53 .