Nationwide, it is estimated that 40% of service stations have gas tanks that leak to some extent. A new program in California is designed to lessen the prevalence of these leaks. We want to assess the effectiveness of the program by seeing if the percentage of service stations whose tanks leak has decreased. To do this, we randomly sample 27 service stations in California and determine whether there is any evidence of leakage. In our sample, only 7 of the stations exhibit any leakage. Is there evidence that the new program is effective

Since the calculated value of z= -1.496 does not fall in the critical region z < -1.645 we conclude that the new program is effective. We fail to reject the null hypothesis .

Step-by-step explanation:

The sample proportion is p2= 7/27= 0.259

and q2= 0.74

The sample size = n= 27

The population proportion = p1= 0.4

q1= 0.6

We formulate the null and alternate hypotheses that the new program is effective

H0: p2> p1   vs  Ha: p2 ≤ p1

The test statistic is

z= p2- p1/√ p1q1/n

z= 0.259-0.4/ √0.4*0.6/27

z=  -0.141/0.09428

z= -1.496

The significance level ∝ is 0.05

The critical region for one tailed test is z ≤ ± 1.645

Since the calculated value of z= -1.496 does not fall in the critical region z < -1.645 we conclude that the new program is effective. We fail to reject the null hypothesis .

an advantage of the standard deviation is that it increases as the dispersion of the data increases. e. the interquartile range is preferred when the data are not skewed or no have outliers. an advantage of the standard deviation is that it uses all the observations in its computation.