Answer:
21
Step-by-step explanation:
Using<em> Simple Random Sampling</em>, we can estimate the sample size by the formula
where
n = sample size
Z = the z-score corresponding to the confidence level 99.5%
S = the assumed standard deviation = 3 seconds
e = margin of error = 2 seconds
<em>It is worth noticing that the higher the confidence level, the larger the sample should be.
</em>
The z-score corresponding to a confidence level of 99.5% can be obtained either with a table or the computer and equals
Z = 3.023
Replacing the values in our formula
So the size of the sample should be at least 21.
Answer:
StartFraction 9 Over 64 EndFraction
Step-by-step explanation:
He must add the square of half the x coefficient. That coefficient is 3/4, so half of it is 3/8 and the square of that is ...
(3/8)^2 = 9/64
Brian mus add 9/64 to boths sides of the equation.
Answer:
We accept the null hypothesis that the breaking strength mean is less and equal to 1750 pounds and has not increased.
Step-by-step explanation:
The null and alternative hypotheses are stated as
H0: u ≥ 1750 i.e the mean is less and equal to 1750
against the claim
Ha: u > 1750 ( one tailed test) the mean is greater than 1750
Sample mean = x`= 1754
Population mean = u = 1750
Population deviation= σ = 65 pounds
Sample size= n = 100
Applying the Z test
z= x`- u / σ/ √n
z= 1754- 1750 / 65/ √100
z= 4/6.5
z= 0.6154
The significance level alpha = 0.1
The z - value at 0.1 for one tailed test is ± 1.28
The critical value is z > z∝.
so
0.6154 is < 1.28
We accept the null hypothesis that the breaking strength mean is less and equal to 1750 pounds and has not increased.
this makes no sense so maybe try something better
0.125, it's a terminating decimal because it has a finite number of digits.
