Tomatoes cost $0.70 per pound. 2 6 8 Weight (lb) Price ($) 1.40 2.10
2 answers:
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Answer:
Since is a one sided test the p value would be:
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, and we can conclude that the mean for elementary school teachers is significantly higher than the mean for secondary teachers at 5% of significance
Step-by-step explanation:
Data given and notation
represent the mean of elementary teachers
represent the mean for secondary teachers
represent the sample standard deviation for elementary teacher
represent the sample standard deviation for secondary teachers
sample size selected
sample size selected
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 check if the mean of the salaries of elementary school teachers is greater than the mean of the salaries of secondary school teachers, the system of hypothesis would be:
Null hypothesis:
Alternative hypothesis:
We don't know the population deviations, so for this case is better apply a t test to compare means, 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 whether the means of two groups are equal to each other".
Calculate the statistic
We can replace in formula (1) the info given like this:
P-value
The first step is calculate the degrees of freedom, on this case:
Since is a one sided test the p value would be:
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, and we can conclude that the mean for elementary school teachers is significantly higher than the mean for secondary teachers at 5% of significance
Answer:
Those values can be helpul to find an ideal range for the number of chocolate chips per cookie.
P(5) = 19.82 chips per cookie.
P(95) = 27.38 chips per cookie.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
How might those values be helpful to the producer of the chocolate chip cookies?
Those values can be helpul to find an ideal range for the number of chocolate chips per cookie.
P(5)
5th percentile, which is the value of X when Z has a pvalue of 0.05. So it is X when Z = -1.645.
P(5) = 19.82 chips per cookie.
P(95)
95th percentile, which is the value of X when Z has a pvalue of 0.95. So it is X when Z = 1.645.
P(95) = 27.38 chips per cookie.