520% of 100 20 1/4% of 3 0.15% of 250 200%of 79 0.3% of 80 0.28% of 50
1 answer:
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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:
The 90% confidence interval for the mean time required by all college graduates is between 5.36 years and 5.44 years.
Step-by-step explanation:
We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:
Now, we have to find z in the Ztable as such z has a pvalue of .
So it is z with a pvalue of , so
Now, find the margin of error M as such
In which is the standard deviation of the population and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 5.4 - 0.04 = 5.36 years.
The upper end of the interval is the sample mean added to M. So it is 5.4 + 0.04 = 5.44 years.
The 90% confidence interval for the mean time required by all college graduates is between 5.36 years and 5.44 years.