Answer:
Step-by-step explanation:
to use the elimination you have to make one of the numbers same
so
and you would have to change the sign so that both of the 3x could cancel out.
New system of equation
so we know the y value so now you could find the x value take one equation
you could try it with both of the equations
Answer:
1. ( 19.1416, 23.5384,)
2. (0.276348, 0.46651)
3. the sample size = 170.73 approximately 171
4. sample size = 334.07 approximately 334
5. sample of 334 should be taken by manager
Step-by-step explanation:
mean = bar x = 21.34 dollars
size of sample n = 70
standard deviation of sample = 9.22
we use t distribution as the population standard deviation is not known.
95% Confidence interval
1-α = 0.95
α = 0.05
degree of freedom = 70-1 = 69
α/2 = 0.025
using the t distribution tsble,
= 1.9949
confidence interval =
= 21.34 +- (1.9949*1.10200)
= 21.34 + 2.1984, 21.34 - 2.1984
= (23.5384, 19.1416)
the confidence interval of the mean amount spent at the supply store can be written as 19.1416<u<23.5384
2. sixe of those who only have a cat
p = 26/70 = 0.371429
at 90 % confidence interval,
1-α = 0.90
α = 0.10
we use the z table here
z(0.10/2) = Z(0.05)
= 1.645
= 0.371429 +-( 1.645 x 0.0578)
= 0.371429 + 0.095081, 0.371429 - 0.095081
= (0.276348, 0.46651)
3. sd = 10$
margin of erro,r e = 1.50$
α = 0.05
using z table
α/2 = Z0.025
= 1.96
sample size = 1.96² * 10² / 1.50²
= 3.8416 * 100/ 2.25
= 170.73
the sample size is approximately 171
d. we have 0.5 as sample proportion now
margin of error = 0.045
α = 0.10
Zα/2 = 0.05
= 1.645
sample size = 1.645²x0.5(1-0.5) / 0.045²
= 0.676506/0.002025
= 334. 07
sample size = 334
5. sample of 334 should be taken by manager
Answer:
We conclude that the mean weight of coffee in its containers is at least 32 ounces which means that the data support the claim.
Step-by-step explanation:
We are given that a coffee packaging plant claims that the mean weight of coffee in its containers is at least 32 ounces.
A random sample of 15 containers were weighed and the mean weight was 31.8 ounces with a sample standard deviation of 0.48 ounces.
Let = <u><em>mean weight of coffee in its containers.</em></u>
SO, Null Hypothesis, :
32 ounces {means that the mean weight of coffee in its containers is at least 32 ounces}
Alternate Hypothesis, :
< 32 ounces {means that the mean weight of coffee in its containers is less than 32 ounces}
The test statistics that would be used here <u>One-sample t-test statistics</u> as we don't know about population standard deviation;
T.S. = ~
where, = sample mean weight = 31.8 ounces
s = sample standard deviation = 0.48 ounces
n = sample of containers = 15
So, <u><em>the test statistics</em></u> = ~
= -1.614
The value of t test statistics is -1.614.
<u>Now, at 0.01 significance level the t table gives critical value of -2.624 at 14 degree of freedom for left-tailed test.</u>
Since our test statistic is more than the critical value of t as -1.614 > -2.624, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <u><em>we fail to reject our null hypothesis</em></u>.
Therefore, we conclude that the mean weight of coffee in its containers is at least 32 ounces which means that the data support the claim.