Step-by-step explanation:
Answer:
Yes, it would be statistically significant
Step-by-step explanation:
The information given are;
The percentage of jawbreakers it produces that weigh more than 0.4 ounces = 60%
Number of jawbreakers in the sample, n = 800
The mean proportion of jawbreakers that weigh more than 0.4 = 60% = 0.6 = =p
The formula for the standard deviation of a proportion is
Solving for the standard deviation gives;
Given that the mean proportion is 0.6, the expected value of jawbreakers that weigh more than 0.4 in the sample of 800 = 800*0.6 = 480
For statistical significance the difference from the mean = 2× = 2*0.0173 = 0.0346 the equivalent number of Jaw breakers = 800*0.0346 = 27.7
The z-score of 494 jawbreakers is given as follows;
Therefore, the z-score more than 2 × which is significant.
Answer:
−11
Step-by-step explanation:
−3−8
=−3−8
=−3+−8
=−11
Answer:
y² + y - 6
Step-by-step explanation:
(y - 2)(+ 1)(y + 3) simplifies to
(y - 2)(y + 3)
Each term in the second factor is multiplied by each term in the first factor, that is
y(y + 3) - 2(y + 3) ← distribute parenthesis
= y² + 3y - 2y - 6 ← collect like terms
= y² + y - 6
Answer:
0.5867 is the lower bound for the proportion of voters who favor the proposal.
Step-by-step explanation:
We are given the following in the question:
Sample size, n = 40
Number of voters who support the proposal, x = 29
95% Confidence interval:
Putting the values, we get:
0.5867 is the lower bound for the proportion of voters who favor the proposal.