Maisie has three times as many CDs as Ken. Ken has twice as many CDs as Jo. Jo has more than 10 CDs. What is the least number of
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Answer:
a) One-tailed t-test about a mean.
b) The null hypothesis would be that there is no significant evidence to conclude that the adjusted grip leads to more distance on the thrown ball.
While the alternative hypothesis is that there is significant evidence to conclude that the adjusted grip leads to more distance on the thrown ball.
Mathematically,
The null hypothesis is represented as
H₀: μ ≤ 40 yards
The alternative hypothesis is given as
Hₐ: μ > 40 yards
c) p-value = 0.00621
d) The graph is presented in the attached image. The p-value obtained is less than the significance level at which the test was performed, hence, we accept the alternative hypothesis and say that there is significant evidence to conclude that the adjusted grip leads to more distance on the thrown ball.
Step-by-step explanation:
a) This test checks if the adjusted grip leads to more distance than before on the ball throws.
So, it is a one tailed test (testing only in one direction), a t-test about a mean.
b) For hypothesis testing, the first thing to define is the null and alternative hypothesis.
The null hypothesis plays the devil's advocate and usually takes the form of the opposite of the theory to be tested. It usually contains the signs =, ≤ and ≥ depending on the directions of the test.
While, the alternative hypothesis usually confirms the the theory being tested by the experimental setup. It usually contains the signs ≠, < and > depending on the directions of the test.
Since this question aims to test if adjusting the grip leads to more distance on the ball thrown.
The null hypothesis would be that there is no significant evidence to conclude that the adjusted grip leads to more distance on the thrown ball. That is, The adjusted grip leads to a distance lesser than or equal to the previous distance on the thrown ball.
While the alternative hypothesis is that there is significant evidence to conclude that the adjusted grip leads to more distance on the thrown ball.
Mathematically,
The null hypothesis is represented as
H₀: μ ≤ 40 yards
The alternative hypothesis is given as
Hₐ: μ > 40 yards
c) To do this test, we will use the z-distribution because there is information on the population standard deviation.
So, we compute the test statistic
z = (x - μ₀)/σ
x = 45 yards
μ₀ = 40 yards
σ = standard deviation = 2 yards
z = (45 - 40)/2 = 2.50
The p-value for a one-tailed test for z-test statistic of 2.50 is 0.00621
d) The sketch of the graph is presented in the attached image.
The interpretation of p-values is that
When the (p-value > significance level), we fail to reject the null hypothesis and when the (p-value < significance level), we reject the null hypothesis and accept the alternative hypothesis.
So, for this question, significance level = 0.05
p-value = 0.00621
0.00621 < 0.05
Hence,
p-value < significance level
This means that we reject the null hypothesis & accept the alternative hypothesis and say that there is enough evidence to conclude that the adjusted grip leads to more distance on the thrown ball.
Hope this Helps!!!
Answer:
a)
b)
Step-by-step explanation:
For this case we can use a linear model to solve the problem.
s) Create an equation to express the increase on the price tickets and the number of seats sold
number of seats, if w analyze the info given the number of seats after increase the price is given by
.
And let P the price for the ticket. So after the increase in ticket price the expression for the increase is P-200.
We have an additional info, for each increase of $3 the number of setas decrease 1. And the equation that gives to us the price change in terms of the increase of price is:
So then our linear equation is given by:
b) Over a certain period, the number of seats sold for this flight ranged between 90 and 115. What was the corresponding range of ticket prices?
So for this case we just need to replace the limits into the linear equation and see what we got:
So the corresponding range of ticket prices is: