Commute times in the U. S. are heavily skewed to the right. We select a random sample of 200 people from the 2000 U. S. Census who reported a non-zero commute time. In this sample the mean commute time is 27.5 minutes with a standard deviation of 18.9 minutes. Can we conclude from this data that the mean commute time in the U. S. is less than half an hour? Conduct a hypothesis test at the 5% level of significance. What is the p -value for this hypothesis test?

Step-by-step explanation:

Given that:

The population mean = 30

The sample size n = 200

The sample-mean = 27.5

The sample standard deviation s = 18.9

The level of significance ∝ = 0.05

The null hypothesis and the alternative hypothesis can be written as:

Since the alternative hypothesis indicates < sign, then this test is left-tailed

The  t-test statistics is:

t = -1.871

The p-value = P(t< -1.871)

From the standard normal t tables

The p-value = 0.0314

Thus, since the p-value is less than the level of significance ∝ = 0.05, then we reject the null hypothesis.

26 is 33.33% of 78

step-by-step explanation:

26 is what percent of 78?

26 is p% of 78

equation: y = p% * x

solving our equation for p

p% = y/x

p% = 26/78

p = 0.3333

convert decimal to percent:

p% = 0.3333 * 100 = 33.33%

constant of proportionality = 3 and the equation is c = 3p.

step-by-step explanation:

c = kp   where c = amount of caffeine in mg., p = oz of pepsi and k is the constant of proportionality.

substituting the values given:

24 = 8k

k = 3.

the same value of k will be obtained if we substitute the other given values.