A yacht is anchored 90 feet offshore
2 answers:
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Answer:
a: z = -1.936
b: 0.0265
d: z < -1.645
Reject H0 if z < -1.645
Step-by-step explanation:
We are given:
H0: µ = 20
HA: µ < 20
n = 60, sample mean: 19.6, σ = 1.6
Since the alternate hypothesis has a < sign in it, it is a left tailed test. The < or > sign in the alternate hypothesis points towards the rejection region.
For a: We need to calculate the test statistic for our situation. This is done with a z-score formula for samples.
For b: we need to use the z-score table to look up the p-value for the score we calculate in part a. The p-value is 0.0265. This means that there is only about a 2.65% chance that the sample values were a result of random chance.
For d: Since the significance level is 0.05, and this is a one tailed test, we have a critical value of z < - 1.645. This means that if the z-score we calculate in part a is less than -1.645, we will reject the null hypothesis
See attached photo for all the calculations!
Answer:
0.0159
Step-by-step explanation:
Given that a common practice of airline companies is to sell more tickets for a particular flight than there are seats on the plane, because customers who buy tickets do not always show up for the flight.
Here if X is the no of persons that do not show up, then X is binomial as each trial is independent with p = 0.04 and n =150 (no of tickets sold)
The plane is overbooked if more than 150 show up
i.e. less than 2 do not show up
Hence the probability that the airline overbooked this flight
=