an outdoor swimming pool costs $9 per day to visit during the summer. there is also a $25 yearly registration fee. is the total
2 answers:
Answer:
no
Step-by-step explanation:
Since there is a $25 yearly registration fee, the total price will never be directly proportional to the number of days visited because if the number of days visited is multiplied by 9, it will have a leftover yearly registration fee, which cannot be accounted for unless written with a function. See the attached image for a table.
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Answer:
a) If we compare the p value and a significance level assumed we see that
so we can reject the null hypothesis, and the the actual true mean is significantly lower than 0.18 ounces.
b) The rejection region zone should be: t <-2.62
c) The rejection region in terms of the mean is: <0.168
Step-by-step explanation:
Data given and notation
represent sample mean
represent the standard deviation for the sample
sample size
represent the value that we want to test
represent the significance level for the hypothesis test.
t 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 determine if the mean is less than 0.18 ounces, the system of hypothesis would be:
Null hypothesis:
Alternative hypothesis:
We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:
(1)
t-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:
Calculate the P-value
Since is a one-side left tailed test the p value would be:
Conclusion
If we compare the p value and a significance level assumed we see that
so we can reject the null hypothesis, and the the actual true mean is significantly lower than 0.18 ounces.
Should the industrial engineer accept or reject the null hypothesis that μ>= 0.18 ounce?
We reject the null hypothesis.
Rejection region in terms of t: t <
On this case we need to find first the degrees of freedom given by:
Now since the test it's one left tailed test we need a value on the t distribution with 14 degrees of freedom such that we have 0.01 of the area on the left and 0.99 of the area on the right. For this case we can use the following excel code:
"=T.INV(0.01,14)" and we got that the rejection region zone should be: t <-2.62
Rejection region in terms of ẋ: ẋ < .
We can use th critical value founded on the last part and we can use this formula similar to the z score.
And we can solve for like this:
And the rejection region in terms of the mean is: <0.168