Answer:
The 85% onfidence interval for the population proportion of new car buyers who prefer foreign cars over domestic cars is (0.151, 0.205).
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of 
, and a confidence level of 
, we have the following confidence interval of proportions.
In which
z is the zscore that has a pvalue of 
.
For this problem, we have that:
Sample of 421 new car buyers, 75 preferred foreign cars. So 
85% confidence level
So 
, z is the value of Z that has a pvalue of 
, so 
.
The lower limit of this interval is:
The upper limit of this interval is:
The 85% onfidence interval for the population proportion of new car buyers who prefer foreign cars over domestic cars is (0.151, 0.205).
Answer:
h = 3v/b
Step-by-step explanation:
Here, we want to select the equation that solves for h
What this simply means is that we should make h the subject of the formula
Thus, we have 3 * v = bh
3v = bh
h = 3v/b
Answer:
The probability that the product will be successfully launched given that the market test result comes back negative is 0.30.
Step-by-step explanation:
Denote the events provided as follows:
<em>S</em> = a product is successfully launched
<em>P</em> = positive test market result
The information provided is:
P (S) = 0.60
P (P | S) = 0.80
P (P | S') = 0.30
Then,
P (P' | S) = 1 - P (P | S) = 1 - 0.80 = 0.20
P (P' | S') = 1 - P (P | S') = 1 - 0.30 = 0.70
Compute the probability of positive test market result as follows:
The probability of positive test market result is 0.60.
Then the probability of negative test market result is:
P (P') = 1 - P (P)
= 1 - 0.60
= 0.40
Compute the probability that the product will be successfully launched given that the market test result comes back negative as follows:
Thus, the probability that the product will be successfully launched given that the market test result comes back negative is 0.30.
Answer:
The margin of error at 90% confidence is 0.5133 gram.
Step-by-step explanation:
Margin of error (E) = (critical value × sample standard deviation) ÷ sqrt(n)
sample standard deviation = 1.5 grams
confidence level = 90% = 0.9
significance level = 1 - C = 1 - 0.9 = 0.1 = 10%
n (sample size) = 25
degree of freedom = n - 1 = 25 - 1 = 24
critical value (t) corresponding to 24 degrees of freedom and 10% significance level is 1.711
E = (1.711×1.5) ÷ sqrt(25) = 2.5665 ÷ 5 = 0.5133 gram
Well that would equate to 3136. So I would guess grams.
