the blue mountain high school drama club is selling tickets to their spring musical. adult tickets are $4 an student tickets are
1 answer:
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h = 2
You must isolate the variable to solve using the order of operations, or PEMDAS.
Subtract 3 from both sides of the equation, and subtract 2h from both sides of the equation.
Divide both sides of the equation by 2.
Answer:
Null hypothesis:
Alternative hypothesis:
If we compare the p value and the significance level given we see that
so we can conclude that we FAIL to reject the null hypothesis, and the the actual mean is not significantly lower than 530 at 5% of significance.
Step-by-step explanation:
1) Data given and notation
represent the mean for the sample
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)
2) State the null and alternative hypotheses.
We need to conduct a hypothesis in order to determine if the claim thet the handle on average is 530 bags per hour:
Null hypothesis:
Alternative hypothesis:
We don't know the population deviation and the sample size is lees than 30, 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".
3) Calculate the statistic
We can replace in formula (1) the info given like this:
4) Calculate the critical value
We need to begin calculating the degrees of freedom
The critical value for this case would be :
The value of a that satisfy this on the normal standard distribution is a=-1.75 and would be the critical value on this case zc=-1.75
5) Calculate the P-value
Since is a one-side upper test the p value would be:
6) Conclusion
If we compare the p value and the significance level given we see that
so we can conclude that we FAIL to reject the null hypothesis, and the the actual mean is not significantly lower than 530 at 5% of significance.