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Answer:
See steps below
Step-by-step explanation:
a)
The <em>null hypothesis is what is already established by previous researches</em>
Business students average 6.8 hours/week studying.
<em>the alternative hypothesis tries to see how accurate is this based on new evidence.</em> Since the sample averaged 7.2 hours studying, this gives us a base to contrast the hypothesis.
Business students now average more than 6.8 hours/week studying.
As we can see, this is a right-tailed test.
b)
For a 0.05 significance level, the critical value is the value such that the area under the Normal curve N(0;1) to the right of
is less than 0.05. This is a very well-known value that can be found by looking up a table or with the computer.
=1.64
If our z-statistic is greater or equal to 1.64 we have reasons to reject the null and accept the new average.
i.
Our z-statistic is given by the formula
where
<em> is the mean of the sample
</em>
<em> is the mean of the null hypothesis
</em>
<em> is the standard deviation
</em>
<em>n is the sample size </em>
Our z-statistic is then
ii.
As we already said, our critical value is 1.64
iii.
<em>Since our z-statistic is greater than the critical value, we reject the null hypothesis and accept 7.2 hours/week as the new average. </em>
c)
The<em> p-value for our statistic is the area under the Normal N(0;1) to the right of our z-statistic</em>. (see picture attached).
This area can be calculated either with tables or computer-assisted and is 0.0345.
Comparing our p-value 0.0345 against the level of significance we see that
<em>p-value < level of significance </em>
Conclusion: when the p-value is less than the level of significance we can reject the null.
<u><em>Remark: </em></u>
Rejecting the null does not mean that the null is false or the alternative is true. It only means that the probability we had made a mistake by a random sampling error or other reason is less or equal than the p-value.
