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2 answers:
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Answer:
Step-by-step explanation:
a) H0:
Ha:
(Two tailed test at 5% significance level)
b) n=30
Mean difference =
Std error of mean =
b) Test statistic t = mean diff/std error = -1.552
df = 30-1 =29
p value=0.0657
c) Since p > 0.05 our signi. level, we accept null hypothesis.
There is no significant difference between the means.
d) Using critical value we find that test statistic is > critical value left
So accept H0
<h2>Answer with explanation:</h2>
We are asked to prove by the method of mathematical induction that:
where n is a positive integer.
- Let us take n=1
then we have:
Hence, the result is true for n=1.
- Let us assume that the result is true for n=k
i.e.
- Now, we have to prove the result for n=k+1
i.e.
<u>To prove:</u>
Let us take n=k+1
Hence, we have:
( Since, the result was true for n=k )
Hence, we have:
Also, we know that:
(
Since, for n=k+1 being a positive integer we have:
)
Hence, we have finally,
Hence, the result holds true for n=k+1
Hence, we may infer that the result is true for all n belonging to positive integer.
i.e.
where n is a positive integer.