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1 answer:
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Answer:
Step-by-step explanation:
Given that a dairy scientist is testing a new feed additive. She chooses 13 cows at random from a large population of cows. She randomly assigns nold = 8 to get the old diet, and nnew = 5 to get the new diet including the additive.
From the data given we get the following
N Mean StDev SE Mean
Sample 1 8 43 5.1824 1.832
Sample 2 5 73 21.0832 9.429
df = 11
Std dev for difference = 13.3689
a) Yes the two are independent. The two sets of cows randomly chosen are definitely independent. Paired means equal number should be there and homogeneous conditions should be maintained.
b) Enclosed
c) Comparison of two means is the test recommended here. Because independent samples are used.\
d) Test statistic= -3.1233
(because of unequal variances we use that method)
95% confidence interval = ( -56.6676 , -3.3324 )
p value <0.05 our alpha
So reject null hypothesis.
The two means are statistically significantly different.
Answer:
81.85%
Step-by-step explanation:
Given :
The average summer temperature in Anchorage is 69°F.
The daily temperature is normally distributed with a standard deviation of 7°F .
To Find:What percentage of the time would the temperature be between 55°F and 76°F?
Solution:
Mean =
Standard deviation =
Formula :
Now At x = 55
At x = 76
Now to find P(55<z<76)
P(2<z<-1)=P(z<2)-P(z>-1)
Using z table :
P(2<z<-1)=P(z<2)-P(z>-1)=0.9772-0.1587=0.8185
Now percentage of the time would the temperature be between 55°F and 76°F =
Hence If the daily temperature is normally distributed with a standard deviation of 7°F, 81.85% of the time would the temperature be between 55°F and 76°F.