A major airline company is concerned that its proportion of late arrivals has substantially increased in the past month. Historical data shows that 18% of the company airplanes have arrived late. In a random sample of 1,240 airplanes, 310 airplanes have arrived late.
a. Formulate the null and the alternative hypotheses.
b. Compute the p-value.
c. Using ¦Á =.05, what is your conclusion?
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a. The null hypothesis would be that the proportion of late arrivals has not changed and remains at 18%. The alternative hypothesis would be that the proportion of late arrivals has increased from 18%.
b. To compute the p-value, we need to use a hypothesis test for a proportion. The formula for calculating the test statistic is:
z = (p - p0) / sqrt((p0*(1-p0))/n)
where:
p = proportion of late arrivals in the sample (310/1240)
p0 = proportion of late arrivals under the null hypothesis (18% or 0.18)
n = sample size (1240)
Using these values, we can calculate the test statistic:
z = (0.25 - 0.18) / sqrt((0.18*(1-0.18))/1240)
z = 0.07 / sqrt((0.18*0.82)/1240)
z = 0.07 / sqrt(0.1476/1240)
z = 0.07 / sqrt(0.000119032)
Now, we need to find the p-value associated with this test statistic. We can do this by looking up the corresponding z-value in a standard normal distribution table. In this case, the z-value is 0.07, which corresponds to a p-value of approximately 0.5240.
c. The p-value of 0.5240 is greater than the significance level of 0.05. Therefore, we do not have enough evidence to reject the null hypothesis. In other words, we cannot conclude that the proportion of late arrivals has increased from 18% based on the given sample data.