Answer:
The approximate 90% confidence interval is;
70,244 to 70,732
Step-by-step explanation:
Here, we want to calculate the approximate 90% C.I for the situation
Mathematically;
CI = mean ± (z * SD)/√n
From the question;
mean = 70,438
SD = 645.3
n = 30
we can get z from the CI table
90% CI is same as; 1.645 z-score
So , substituting these values;
CI = 70,438 ± ( 1.645 * 645.3)/√30
CI = 70,438 ± 194
So the CI = 70,438 -194 to 70,438 + 194
= 70,244 to 70,732
Answer: 49 cm^2
Step-by-step explanation:
The length of the small pink square box will be:
L^2 = 9
L = sqrt( 9 )
L = 3
The length of the big pink square box will be:
L^2 = 16
L = sqrt( 16 )
L = 4
The length of the yellow square box will be:
4 + 3 = 7
The area of the third square box will be
A = 7^2
A = 49 cm^2
Answer:
There is no statistical evidence at 1% level to accept that the mean net contents exceeds 12 oz.
Step-by-step explanation:
Given that a random sample of ten containers is selected, and the net contents (oz) are as follows: 12.03, 12.01, 12.04, 12.02, 12.05, 11.98, 11.96, 12.02, 12.05, and 11.99.
We find mean = 11.015
Sample std deviation = 3.157
a) 
(Right tailed test)
Mean difference /std error = test statistic
p value =0.174
Since p >0.01, our alpha, fail to reject H0
Conclusion:
There is no statistical evidence at 1% level to accept that the mean net contents exceeds 12 oz.
Answer:
4 is the median to your problem
Each person would receive 6 chocolate truffles and 4 caramel truffles because 126÷21=6 and 84÷21=4.
