Question

It is advertised that the average braking distance for a small car traveling at 65 miles per hour equals 120 feet. A transportation researcher wants to determine if the statement made in the advertisement is false. She randomly test drives 36 small cars at 65 miles per hour and records the braking distance. The sample average braking distance is computed as 114 feet. Assume that the population standard deviation is 22 feet. Use Table 1. State the null and the alternative hypotheses for the test. H_0:mu = 120: H_A: mu notequalto 120 H_0:mu 2 greaterthanorequalto 120: H_A mu < 120 H_0: mu lessthanorequalto 120: H_A: mu > 120 Calculate the value of the test statistic and the p-value. {Negative values should be indicated by a minus sign. Round "Test statistic" to 2 decimal places 3nd "p-value" to 4 decimal places.} The p-value is: p-value < 0.01 0.01 lessthanorequalto p-value < 0.025 0.025 lessthanorequalto p-value < 0.05 0.05 lessthanorequalto p-value < 0/.10 p-value greaterthanorequalto 0.10 c. Use alpha = 0.01 to determine if the average breaking distance differs from 120 feet. Repeat the test with the critical value approach. (Negative values should be indicated by a minus sign. Round your answers to 3 decimal places.)

Answer 1

Answer:

H0: mu = 120 , Ha: : mu not equal to 120

The p-value  is greater than or equal to 0.10

z= -1.636

Since the calculated value of  z does not fall in the critical region H0 is not rejected.

Step-by-step explanation:

State the null and alternate hypothesis

H0: mu = 120

average braking distance for a small car traveling at 65 miles per hour equals 120 feet.

against the claim

Ha: : mu not equal to 120

average braking distance for a small car traveling at 65 miles per hour  does not equal 120 feet.

The p-value equals 0.101764 which is greater than 0.01 and thus supports H0.

The p-value  is greater than or equal to 0.10

The test statistic z is applied

Here

Sample Mean  x~ = 114

Population mean μ =120

Sample size n= 36

Population SD =σ= 22

Z= u- μ/σ√n

Z= 114- 120/ 22/ √36

z= -1.636

This is a two tailed test and the critical region for two tailed test at alpha = 0.01 is z ∝/2= ± 2.58

Since the calculated value of  z does not fall in the critical region H0 is not rejected.