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:
D
Step-by-step explanation:
If there are 20 students, and each team has 5 students, then there will be 4 teams of third graders that will play kickball.
20/5 = 4
Answer:
140.37 in²
Step-by-step explanation:
C = 2πr
42 = 2πr
21 = πr
r = 21/π
A = πr²
A = π × (21/π)²
≈ 140.37 in²
Answer:
9, -3
Step-by-step explanation:
Answer:
1/2 or 0.5
Step-by-step explanation:
Within 1 to 6,
Even numbers are 2, 4, 6
While odd numbers are 1, 3, 5
Assuming cards are mixed up properly and the probabilities of drawing a card with any number from 1 to 6 is constant,
Probability of drawing even number cards = number of units of even number cards ÷ total number of units of cards from 1 to 6 aka total number of cards
= 3 / 6
= 1 / 2