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Transcript
1. Assume that a simple random sample has been selected from a normally
distributed population. Find the test statistic, a range of numbers for the P-value,
critical value and state the final conclusion.
Claim:- The mean bottom temperature of a fish breeding pond is less than 20 degree
celcius.
Sample Data:- n=56, x-bar =19.3, s=1.6. The significance level is alpha = 0.05
Calculate the test statistic, p-value using t-table, critical value using t-table and
conclusion.
Ans.
One-Sample T
Test of H0: Mean
N
56
Mean
19.3000
= 20 vs H1 Mean
StDev
1.6000
SE Mean
0.2138
< 20
95%
Upper
Bound
T
19.6577 -3.27
P
0.001
Test Statistic = -3.27
P value = .001
Critical value = -1.673
Conclusion :reject the null hypothesis
2. A random sample of 56 measured the accuracy of their wristwatches, with positive
errors representing watches that are ahead of the correct time and negative errors
representing watches that are behin the correct time. The 56 values have a mean of 94.7 sec and a standard deviation of 152.6 sec. Use a 0.01 significance level to test
the claim that the population of all watches has a mean equal to 0.00 sec
What is the correct conclusion?
1. There is not sufficient evidence to warrant rejection of the claim that the
polpulation mean is 0.00 sec.
2. There is sufficient evidence.
Ans
One-Sample Z test
Test of Mean = 0 vs Mean > 0
The assumed standard deviation = 152.6
99%
Lower
N
Mean SE Mean
Bound
Z
P
56 94.7000 20.3920 47.2610 4.64 0.000
Conclusion :
Since p value is less than .01 reject the null hypothesis
Thus there is sufficient evidence.
3. The birth weights are recorded for a sample of male babies born to mothers taking
a special vitamin supplement. When testing the claim that the mean birth weights
for all male babies of mothers given vitamins is equal to 3.39 kg, which is the
mean for the population of all males. The significance level is alpha =0.05
N= 17.0000, sx = 0.6300, x bar = 3.3100
Ans
One-Sample T test
Test of Mean = 3.39 vs Mean < 3.39
N
17
Mean
3.31000
StDev SE Mean
0.63000 0.15280
95%
Upper
Bound
3.57677
T
-0.52
P
0.304
Since p value is grater than 0.05 , we accept the null hypothesis
4. Find the test statistic, find the critical values of chi-square and limits containing
the p-value; then determine whether there is sufficient evidence to support the
given alternative hypothesis
H1:   7,   0.02, n  16, s  12
Determine the test statistic, use chi-square table to determine the critical values
and final conclusion.
Ans
(n  1) s 2
 17.6326
Test Statistics =
2

Critical Value = 15.03
Since calculated value is greater than critical value reject the null hypothesis
5. Use the traditional method with alpha =0.05 to test the claim that a new method of
laser cutting reduces the variation in lengths of 2x4 lumber. The standard
deviation of the lengths of the sawed 2x4s is 0.33 inches. A simple random
sample of 91 lasercut 2x4s, taken from a normally distributed population, was
found to have a standard deviation of 0.3 inches. Calculate the test statistic and
the critical value to determin the correct conclusion.
Ans
Test Statistics =
(n  1) s 2
2
 74.38
Critical Value = 113.1
d.f =90
Conclusion:
Since calculated value is less than critical value accept the null hypothesis