Answer: No, the data provided is not sufficient evidence to conclude that the mean height of women in the city differs from the national mean.
Step-by-step explanation:
Since we have given that
![H_0:\mu =62.7\\\\H_a:\mu \neq 62.7](https://tex.z-dn.net/?f=H_0%3A%5Cmu%20%3D62.7%5C%5C%5C%5CH_a%3A%5Cmu%20%5Cneq%2062.7)
Mean height = 61.5 inches
Standard deviation = 4.4 inches
n = 100
So, test statistic value would be
![z=\dfrac{\bar{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}\\\\z=\dfrac{61.5-62.7}{\dfrac{4.4}{\sqrt{100}}}\\\\z=\dfrac{-1.2}{0.44}\\\\z=-2.727](https://tex.z-dn.net/?f=z%3D%5Cdfrac%7B%5Cbar%7Bx%7D-%5Cmu%7D%7B%5Cdfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%7D%5C%5C%5C%5Cz%3D%5Cdfrac%7B61.5-62.7%7D%7B%5Cdfrac%7B4.4%7D%7B%5Csqrt%7B100%7D%7D%7D%5C%5C%5C%5Cz%3D%5Cdfrac%7B-1.2%7D%7B0.44%7D%5C%5C%5C%5Cz%3D-2.727)
At 10% level of significance, in two tail test ,
z = 1.28
Since 1.28 > -2.727
So, we will accept the null hypothesis.
Hence, No, the data provided is not sufficient evidence to conclude that the mean height of women in the city differs from the national mean.
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Answer:
A)
Step-by-step explanation:
Given expression:
To factor the given expression completely.
Solution:
In order to factor the expression, we will factor in pairs.
We will factor the G.C.F of the terms in the pairs.
G.C.F. of and can be given as:
Thus, G.C.F. =
G.C.F. of and can be given as:
Thus, G.C.F. =
The expression after factoring the G.C.F. pairs is given as:
Taking G.C.F. of the whole expression as is a common term.
The expression is completely factored.
Step-by-step explanation:
Answer:
P(blue then red)=5/169
Step-by-step explanation:
the probability of picking the blue is 4/26 it's replaced so the total jellybeans doesn't change therefore the sample space doesn't change. the probability of picking a red is 5/26
to find the probability of picking them in that order, you multiply the two together
(4/26)x(5/26)=20/676
reducing that we get:
P(blue then red)=5/169