A certain flight arrives on time 80 percent of the time. Suppose 132 flights are randomly selected. Use the normal approximation
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Answer:
So we reject the null hypothesis and accept the alternate hypothesis that rats learn slower with sound.
Step-by-step explanation:
In this data we have
Mean= u = 18
X= 38
Standard deviation = s= 6
1) We formulate the null and alternate hypothesis as
H0: u = 18 against Ha : u > 18 One tailed test .
2) The significance level alpha = ∝= 0.05 and Z alpha has a value ± 1.645 for one tailed test.
3)The test statistics used is
Z= X- u / s
z= 38-18/6= 3.333
4) The calculated value of z = 3.33 is greater than the z∝ = 1.645
5) So we reject the null hypothesis and accept the alternate hypothesis that rats learn slower with sound.
First we set the criteria for determining the true of value of the variable. That whether the rats learn in less or more than 18 trials.
Then we find the value of z for the given significance value given and the test about to be checked.
Then the test statistic is determined and calculated.
Then both value of z and z alpha re compared. If the test statistics falls in the rejection region reject the null hypothesis and conclude alternate hypothesis is true.
The figure shows that the calulated z value lies outside the given z values
