Answer:
a)
The mean of adults aged 18 to 32 that continue to be dependent on their parents is 0.3
The mean of adults aged 18 to 32 that continue to be dependent on their parents is greater than 0.3
b) 34%
c) practically 0
d) Reject the null hypothesis.
Step-by-step explanation:
a)
Since an individual aged 18 to 32 either continues to be dependent on their parents or not, this situation follows a Binomial Distribution and, according to the previous research, the probability p of “success” (depend on their parents) is 0.3 (30%) and the probability of failure q = 0.7
According to the sample, p seems to be 0.34 and q=0.66
To see if we can approximate this distribution with a Normal one, we must check that is not too skewed; this can be done by checking that np ≥ 5 and nq ≥ 5, where n is the sample size (400), which is evident.
<em>We can then, approximate our Binomial with a Normal </em>with mean
and standard deviation
Since in the current research 136 out of 400 individuals (34%) showed to be continuing dependent on their parents:
The mean of adults aged 18 to 32 that continue to be dependent on their parents is 0.3
The mean of adults aged 18 to 32 that continue to be dependent on their parents is greater than 0.3
So, this is a r<em>ight-tailed hypothesis testing. </em>
b)
According to the sample the proportion of "millennials" that are continuing to be dependent on their parents is 0.34 or 34%
c)
Our level of significance is 0.05, so we are looking for a value such that the area under the Normal curve to the right of
is ≤ 0.05
This value can be found by using a table or the computer and is = 1.645
<em>Applying the continuity correction factor (this should be done because we are approximating a discrete distribution (Binomial) with a continuous one (Normal)), we simply add 0.5 to this value and </em>
corrected is 2.145
Now we compute the z-score corresponding to the sample
where
= mean of the sample
= mean of the null hypothesis
s = standard deviation of the sample
n = size of the sample
The sample z-score is then
The p-value provided by the sample data would be the area under the Normal curve to the left of 33.7759 which can be considered zero.
d)
Since the z-score provided by the sample falls far to the left of we should reject the null hypothesis and propose a new mean of 34%.