The passing play percentages of 10 randomly selected NCAA Division 1A college football teams for home and away games in the 2020dash–2021 season are shown in the table. At alphaαequals=0.050.05, is there enough evidence to support the claim that passing play percentage is different for home and away games? Assume the samples are random and dependent, and the populations are normally distributed. Complete parts (a) through (f). College 1 2 3 4 5 6 7 8 9 10 Home passing play percentage 46.546.5 48.848.8 48.748.7 36.236.2 44.844.8 38.338.3 36.636.6 43.943.9 49.049.0 50.050.0 Away passing play percentage 42.342.3 41.841.8 50.950.9 40.240.2 46.646.6 45.145.1 37.337.3 45.245.2 48.948.9 51.851.8 Calculate d overbard. d overbardequals=enter your response here (Type an integer or a decimal. Do not round.) Part 4 Calculate s Subscript dsd. s Subscript dsdequals=enter your response here (Round to three decimal places as needed.) Part 5 (c) Find the standardized test statistic t. tequals=enter your response here (Round to two decimal places as needed.) Part 6 (d) Calculate the P-value. P-valueequals=enter your response here (Round to three decimal places as needed.) Part 7 (e) The rejection regions for this test would be tless than<minus−2.262.26 and tgreater than>2.262.26, so the null hypothesis would notwould not be rejected. Decide whether to reject or fail to reject the null hypothesis using the P-value. Compare your result with the result obtained using rejection regions. Are they the same? ▼ Fail to reject Reject the null hypothesis using the P-value. ▼ No, Yes, the results ▼ are not are the same as using the critical value approach. Part 8 (f) Interpret the decision in the context of the original claim. There ▼ is is not enough evidence to ▼ support reject the claim that the passing play percentages have ▼ decreased. not changed. changed. increased. What is the claim?What are Upper H 0 and Upper H Subscript a?d overbar(Type an integer or a decimal. Do not round.)s Subscript d(Round to three decimal places as needed.)t(Round to two decimal places as needed.)P-value(Round to three decimal places as needed.)Are they the same?the null hypothesis using the P-value.the null hypothesis using the P-value.the resultsthe resultsthe same as using the critical value approach.Thereenough evidence toenough evidence tothe claim that the passing play percentages havethe claim that the passing play percentages have

The passing play percentages of 10 randomly selected NCAA Division 1A college football teams for home and away games in the 2020dash–2021 season are shown in the table. At alphaαequals=0.050.05, is there enough evidence to support the claim that passing play percentage is different for home and away games? Assume the samples are random and dependent, and the populations are normally distributed. Complete parts (a) through (f). College 1 2 3 4 5 6 7 8 9 10 Home passing play percentage 46.546.5 48.848.8 48.748.7 36.236.2 44.844.8 38.338.3 36.636.6 43.943.9 49.049.0 50.050.0 Away passing play percentage 42.342.3 41.841.8 50.950.9 40.240.2 46.646.6 45.145.1 37.337.3 45.245.2 48.948.9 51.851.8 Calculate d overbard. d overbardequals=enter your response here (Type an integer or a decimal. Do not round.) Part 4 Calculate s Subscript dsd. s Subscript dsdequals=enter your response here (Round to three decimal places as needed.) Part 5 (c) Find the standardized test statistic t. tequals=enter your response here (Round to two decimal places as needed.) Part 6 (d) Calculate the P-value. P-valueequals=enter your response here (Round to three decimal places as needed.) Part 7 (e) The rejection regions for this test would be tless than<minus−2.262.26 and tgreater than>2.262.26, so the null hypothesis would notwould not be rejected. Decide whether to reject or fail to reject the null hypothesis using the P-value. Compare your result with the result obtained using rejection regions. Are they the same? ▼ Fail to reject Reject the null hypothesis using the P-value. ▼ No, Yes, the results ▼ are not are the same as using the critical value approach. Part 8 (f) Interpret the decision in the context of the original claim. There ▼ is is not enough evidence to ▼ support reject the claim that the passing play percentages have ▼ decreased. not changed. changed. increased. What is the claim?What are Upper H 0 and Upper H Subscript a?d overbar(Type an integer or a decimal. Do not round.)s Subscript d(Round to three decimal places as needed.)t(Round to two decimal places as needed.)P-value(Round to three decimal places as needed.)Are they the same?the null hypothesis using the P-value.the null hypothesis using the P-value.the resultsthe resultsthe same as using the critical value approach.Thereenough evidence toenough evidence tothe claim that the passing play percentages havethe claim that the passing play percentages have

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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College	1	2	3	4	5	6	7	8	9	10
Home passing play percentage	46.546.5	48.848.8	48.748.7	36.236.2	44.844.8	38.338.3	36.636.6	43.943.9	49.049.0	50.050.0
Away passing play percentage	42.342.3	41.841.8	50.950.9	40.240.2	46.646.6	45.145.1	37.337.3	45.245.2	48.948.9	51.851.8

Part 4

Calculate

s Subscript dsd.

s Subscript dsdequals=enter your response here

(Round to three decimal places as needed.)

Part 5

tequals=enter your response here

(Round to two decimal places as needed.)

Part 6

(d) Calculate the P-value.

P-valueequals=enter your response here

(Round to three decimal places as needed.)

Part 7

(e) The rejection regions for this test would be

tless than<minus−2.262.26

and

tgreater than>2.262.26,

so the null hypothesis

would notwould not

be rejected. Decide whether to reject or fail to reject the null hypothesis using the P-value. Compare your result with the result obtained using rejection regions. Are they the same?

▼

Fail to reject

Reject

the null hypothesis using the P-value.

▼

No,

Yes,

the results

▼

are not

are

the same as using the critical value approach.

Part 8

(f) Interpret the decision in the context of the original claim.

There

▼

is not

enough evidence to

▼

support

reject

the claim that the passing play percentages have

▼

decreased.

not changed.

changed.

increased.

What is the claim?What are Upper H 0 and Upper H Subscript a?d overbar(Type an integer or a decimal. Do not round.)s Subscript d(Round to three decimal places as needed.)t(Round to two decimal places as needed.)P-value(Round to three decimal places as needed.)Are they the same?the null hypothesis using the P-value.the null hypothesis using the P-value.the resultsthe resultsthe same as using the critical value approach.Thereenough evidence toenough evidence tothe claim that the passing play percentages havethe claim that the passing play percentages have

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

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