Recently, TWA reported an on-time arrival rate of 78.4%. Assume that a later random sample of 750 flights results in 630 that are on time. If TWA were to claim that its on-time arrival rate is now higher than 78.4%, would the claim be supported at the 0.02 level of significance? Please show all 4 steps of the classical approach.

Step-by-step explanation:

1. Null hypothesis: u <= 0.784

Alternative hypothesis: u > 0.784

2. Find the test statistics: z using the one sample proportion test. First we have to find the standard deviation

Using the formula

sd = √[{P (1-P)}/n]

Where P = 0.84 and n = 750

sd =√[{0.84( 1- 0.84)/750]}

sd=√(0.84 (0.16) /750)

SD =√(0.1344/750)

sd = √0.0001792

sd = 0.013

Then using this we can find z

z = (p - P) / sd

z = (0.84-0.784) / 0.013

z =(0.056/0.013)

z = 4.3077

3. Find the p value and use it to make conclusions...

The p value at 0.02 level of significance for a one tailed test with 4.3077 as z score and using a p value calculator is 0.000008254.

4. Conclusions: the results is significant at 0.02 level of significance suck that we can conclude that its on-time arrival rate is now higher than 78.4%.

c

step-by-step explanation:

it would be see because your formula is f(x)=6x, so 6 times 1 is 6 and you go on and the best answer is c.