The drama club is selling tickets to their play to raise money for the show's expenses. Each student ticket sells for $7.50 and
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Answer:
(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].
(2) We can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval
(3) A survey should include at least 3002 people if we wanted the margin of error for the 90% confidence level to be about 1.5%.
Step-by-step explanation:
We are given that a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in school, 48% said they decided not to go to college because they could not afford school.
Firstly, the pivotal quantity for finding the confidence interval for the population proportion is given by;
P.Q. = ~ N(0,1)
where, = sample proportion of Americans who decide to not go to college = 48%
n = sample of American adults = 331
p = population proportion of Americans who decide to not go to
college because they cannot afford it
<em>Here for constructing a 90% confidence interval we have used a One-sample z-test for proportions.</em>
<em />
<u>So, 90% confidence interval for the population proportion, p is ;</u>
P(-1.645 < N(0,1) < 1.645) = 0.90 {As the critical value of z at 5% level
of significance are -1.645 & 1.645}
P(-1.645 < < 1.645) = 0.90
P( <
<
) = 0.90
P( < p <
) = 0.90
<u>90% confidence interval for p</u> = [ ,
]
= [ ,
]
= [0.4348, 0.5252]
(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].
(2) The interpretation of the above confidence interval is that we can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval.
3) Now, it is given that we wanted the margin of error for the 90% confidence level to be about 1.5%.
So, the margin of error =
= 54.79
n =
n = 3001.88 ≈ 3002
Hence, a survey should include at least 3002 people if we wanted the margin of error for the 90% confidence level to be about 1.5%.
Let x represent the number of quarts of Tuscan sauce and
y represent the number of quarts of marinara sauce Tiffany makes.
A quart of Tuscan sauce requires 6 tomatoes and 1 cup of oil
x quarts requires 6x tomatoes and 1x cups of oil
A quart of her marinara sauce requires 5 tomatoes and 1.25 cups of oil
y quarts requires 5y tomatoes and 1.25 y cups of oil
She has 45 tomatoes and 10 cups of oil on hand.
So the constraints are
x>=0 and y>=0