Answer:
0.04
Step-by-step explanation:
In this case we must use the following formula to be able to find the standard error, which would be the square root, of the probability for its complement divided by the population:
SE=
Now, knowing that p = 0.552 and n = 160 we replace and we are left with:
SE =
SE =
SE = 0.039
Which means that the standard error of the proportion is 0.04
Answer:
Following are the responses to the given question:
Step-by-step explanation:
Please find the table in the attached file.
mean and standard deviation difference:
For point a:
hypotheses are:
degree of freedom:
From t table, at, reject null hypothesis if
.
test statistic:
because the , removing the null assumption. Data promotes a food product manufacturer's assertion with a likelihood of Type 1 error of 0.05.
For point b:
From t table, at , removing the null hypothesis if
.
because , fail to removing the null hypothesis.
The data do not help the foodstuff producer's point with the likelihood of a .01-type mistake.
For point c:
Hypotheses are:
Degree of freedom:
From t table, at , removing the null hypothesis if
.
test statistic:
Since , The null hypothesis fails to reject. The results do not support the packaged food producer's claim with a Type 1 error probability of 0,05.
From t table, at, reject null hypothesis if
.
Since , fail to reject null hypothesis.
Data do not support the claim of the producer of the dietary product with the probability of Type 1 error of .01.
Answer:
There will be total 2048 parts of the given paper if James if able to fold the paper eleven times.
The needed function is
Step-by-step explanation:
Let us assume the piece of paper James decides to fold is a SQUARE.
Now, let us assume:
n : the number of times the paper is folded.
y : The number of parts obtained after folds.
Now, if the paper if folded ONCE ⇒ n = 1
Also, when the pap er is folded once, the parts obtained are TWO equal parts.
⇒ for n = 1 , y = 2 ..... (1)
Similarly, if the paper if folded TWICE ⇒ n = 2
Also, when the paper is folded twice, the parts obtained are FOUR equal parts.
⇒ for n = 2 , y = 4 ..... (2)
⇒
Continuing the same way, if the paper is folded SEVEN times ⇒ n = 7
So,
⇒ There are total 128 equal parts.
Lastly, if the paper is folded ELEVEN times ⇒ n = 11
So,
⇒ There are total 2048 equal parts.
Hence, there will be total 2048 parts of the given paper if James if able to fold the paper eleven times.
And the needed function is