The mean throwing distance of a football for a Marco, a high school freshman quarterback, is 40 yards, with a standard deviation of two yards. The team coach tells Marco to adjust his grip to get more distance. The coach records the distances for 20 throws. For the 20 throws, Marco’s mean distance was 45 yards. The coach thought the different grip helped Marco throw farther than 40 yards. Test the coach’s claim at α = 0.05. Assume the throw distances for footballs are normal.

Required:
Determine:

a. What type of test this is?
b. Set up the hypothesis test.
c. Find the p-value.
d. Sketch the graph and state your conclusion.

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!!!

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step-by-step explanation:

d 25 pi/6

step-by-step explanation:

the diameter is 10 inches

the radius is 1/2 of the diameter or 5 inches

the area of the circle is

a = pi r^2

  = pi 5^2

  = 25 pi

the slice is 60 degrees of the entire circle.

60 degrees out of 360   60/360 = 1/6

the area of the slice is 1/6 of the circle

a slice = 1/6 a circle

a slice = 1/6 * 25 pi