Carmen used a random number generator to simulate a survey of how many children live in the households in her town. There
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Answer:
If we compare the p value and the significance level given we see that
so we can conclude that we have enough evidence to reject the null hypothesis.
We can say that at 5% of significance the mean average waiting time is significantly less than 200 seconds per call.
Step-by-step explanation:
Data given and notation
represent the sample mean
represent the population standard deviation
sample size
represent the value that we want to test
represent the significance level for the hypothesis test.
z would represent the statistic (variable of interest)
represent the p value for the test (variable of interest)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the population mean is less than 200, the system of hypothesis are :
Null hypothesis:
Alternative hypothesis:
Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:
(1)
z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
Calculate the statistic
We can replace in formula (1) the info given like this:
P-value
Since is a one-side left tailed test the p value would given by:
Conclusion
If we compare the p value and the significance level given we see that
so we can conclude that we have enough evidence to reject the null hypothesis.
We can say that at 5% of significance the mean average waiting time is significantly less than 200 seconds per call.
After 6 days equation would be and answer
to represent before 6 days just put 4^5=1024
to represent before 6 days just put 4^5=1024