The weight of Kyle's dog increased from 29.7 pounds to 48.9 pounds in an 8 month period. What was the average weight increase pe
2 answers:
You might be interested in
Answer:
We can therefore conclude that the geographical distribution of hotline callers could be the same as the U.S population distribution.
Step-by-step explanation:
The null Hypothesis: Geographical distribution of hotline callers could be the same as the U.S. population distribution
Alternative hypothesis: Geographical distribution of hotline callers could not be the same as the U.S. population distribution
The populations considered are the Midwest, South, Northeast, and west.
The number of categories, k = 4
Number of recent calls = 200
Let the number of estimated parameters that must be estimated, m = 0
The degree of freedom is given by the formula:
df = k - 1-m
df = 4 -1 - 0 = 3
Let the significance level be, α = 5% = 0.05
For α = 0.05, and df = 3,
from the chi square distribution table, the critical value = 7.815
<u>Observed and expected frequencies of calls for each of the region:</u>
<u>Northeast</u>
Observed frequency = 39
It contains 18.1% of the US Population
The probability = 0.181
Expected frequency of call = 0.181 * 200 = 36.2
<u>Midwest</u>
Observed frequency = 55
It contains 21.9% of the US Population
The probability = 0.219
Expected frequency of call = 0.219 * 200 =43.8
<u>South</u>
Observed frequency = 60
It contains 36.7% of the US Population
The probability = 0.367
Expected frequency of call = 0.367 * 200 = 73.4
<u>West</u>
Observed frequency = 46
It contains 23.3% of the US Population
The probability = 0.233
Expected frequency of call = 0.233 * 200 = 46
Where observed frequency
Expected frequency
Calculate the test statistic value, x²
Since the test statistic value, x²= 5.535 is less than the critical value = 7.815, the null hypothesis will not be rejected, i.e. it will be accepted. We can therefore conclude that the geographical distribution of hotline callers could be the same as the U.S population distribution.
Answer:
a) 94.06% probability that a water specimen collected from the river has an alkalinity level exceeding 45 milligrams per liter.
b) 94.06% probability that a water specimen collected from the river has an alkalinity level below 55 milligrams per liter.
c) 50.98% probability that a water specimen collected from the river has an alkalinity level between 48 and 52 milligrams per liter.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
a. exceeding 45 milligrams per liter.
This probability is 1 subtracted by the pvalue of Z when X = 45. So
has a pvalue of 0.0594.
1 - 0.0594 = 0.9406
94.06% probability that a water specimen collected from the river has an alkalinity level exceeding 45 milligrams per liter.
b. below 55 milligrams per liter.
This probability is the pvalue of Z when X = 55.
has a pvalue of 0.9604.
94.06% probability that a water specimen collected from the river has an alkalinity level below 55 milligrams per liter.
c. between 48 and 52 milligrams per liter.
This is the pvalue of Z when X = 52 subtracted by the pvalue of Z when X = 48. So
X = 52
has a pvalue of 0.7549
X = 48
has a pvalue of 0.2451
0.7549 - 0.2451 = 0.5098
50.98% probability that a water specimen collected from the river has an alkalinity level between 48 and 52 milligrams per liter.