Question

he mean throwing distance of a football for a Luis, a high school freshman quarterback, is 50 yards, with a standard deviation of three yards. The team coach tells Luis to adjust his grip to get more distance. The coach records the distances for 35 throws. For the 35 throws, Luis’s mean distance was 54 yards. The coach thought the different grip helped Luis throw farther than 50 yards. Conduct a hypothesis test using a preset α = 0.01. Assume the throw distances for footballs are normal. a. Determine what type of test this is: A, B or C? Blank 1 A) Single population mean with standard deviation known B) Single population mean with standard deviation not known C) Single population proportion b. Identify the null hypothesis, A, B,C or D? Blank 2 Identify the alternative hypothesis, A, B,C or D ? Blank 3 A) B) C) D) c. Find the test-statistic to 2 decimal places Blank 4 d. Find the p-value to nearest whole number Blank 5 e. State your conclusion, A, B, or C? Blank 6 A. Can't reject null. There is not sufficient evidence from this sample data to show that Luis’s mean throwing distance is greater than 50 yards. B. Reject null. There is sufficient evidence from this sample data to show that Luis’s mean throwing distance is greater than 50 yards. C. no conclusion

Answer 1

Solution:

   Given: The throw distances for footballs are normal with mean throwing distance of a football for a Luis, a high school freshman quarterback, is 50 yards, with a standard deviation of three yards.

Thus:

Sample size = n = 35

Sample mean =

Claim: The coach thought the different grip helped Luis throw farther than 50 yards.

Level of significance =

Part a) Determine what type of test this is:

A) Single population mean with standard deviation known.
Since we are gievn that population standard deviation = 3.

Part b. Identify the null hypothesis.

   

Identify the alternative hypothesis

Part c. Find the test-statistic

Part d. Find the p-value to nearest whole number

p-value= P( Z > z test statistic value)

p-value = P( Z > 7.89)

p-value = 1 - P( Z < 7.89)

z = 7.89 is very large z value and it is more than z = 3, hence area under z = 7.89 is 1.0000.

That is: P(Z < 7.89) =1.0000

Thus

p-value = 1 - 1.0000

p-value = 0.0000

Part e. State your conclusion

Since p-value = 0.0000 < 0.01 level of significance , we reject null hypothesis H0.

Thus correct option is:

B. Reject null. There is sufficient evidence from this sample data to show that Luis’s mean throwing distance is greater than 50 yards.