Answer:
The lower limit of 95% confidence interval is 99002.
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = $100,000
Sample mean, = $111,000
Sample size, n = 36
Alpha, α = 0.05
Population standard deviation, σ = $36,730
First, we design the null and the alternate hypothesis
We have to find the lower limit of the 95% confidence interval.
95% Confidence interval:
Putting the values, we get,
The lower limit of 95% confidence interval is 99002.
Answer:
Population standard deviation, = 3683.063 .
Step-by-step explanation:
We are given that the width of the estimated confidence interval i.e. 99% is 600 and the sample size used in estimating the mean is 1000 which means ; n = 1000 and width = 600
We know that Width of confidence interval = 2 * Margin of error
<em> Margin of error</em><em> </em>= = 2.5758 *
{because at 1% significance level
z table has value of 2.5758 .}
Therefore, 600 = 2 * 2.5758 *
⇒ =
= 3683.063 .
Hence, the Population standard deviation = 3683.063 .
Answer:
Assuming that t = toppings, then the inequality is 9.25+0.75t ≤ 13.75. Solution is 6.
Step-by-step explanation:
Let's define toppings as .
If the original cost of a pizza is $9.25 and each topping costs $0.75, we can find how much the toppings would cost by multiplying 0.75 by t, the amount of toppings. Adding this to 9.25 would give us the expression needed to find the cost.
If you want to spend NO MORE than 13.75, it's ok to spend $13.75, so we write the equation with a ≤ symbol (less than or equal to).
9.25+0.75t ≤ 13.75
Now lets solve the equation.
We can subtract 9.25 from both sides, which gets us
0.75t ≤ 4.5
Now we can divide both sides by 0.75 to isolate t.
t ≤ 6
So, the maximum number of toppings you can order is 6.
Hope this helped!