Respuesta :
Answer:
a)
[tex]\bf H_0:[/tex] The mean of adults aged 18 to 32 that continue to be dependent on their parents is 0.3
[tex]\bf H_a:[/tex] 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.
We can then, approximate our Binomial with a Normal with mean
[tex]\bf np = 400*0.34 = 136[/tex]
and standard deviation
[tex]\bf \sqrt{npq}=\sqrt{400*0.34*0.66}=9.4742[/tex]
Since in the current research 136 out of 400 individuals (34%) showed to be continuing dependent on their parents:
[tex]\bf H_0:[/tex] The mean of adults aged 18 to 32 that continue to be dependent on their parents is 0.3
[tex]\bf H_a:[/tex] 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 right-tailed hypothesis testing.
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 [tex]\bf Z^*[/tex] such that the area under the Normal curve to the right of [tex]\bf Z^*[/tex] is ≤ 0.05
This value can be found by using a table or the computer and is [tex]\bf Z^*[/tex]= 1.645
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
[tex]\bf Z^*[/tex] corrected is 2.145
Now we compute the z-score corresponding to the sample
[tex]\bf z=\frac{\bar x -\mu}{s/\sqrt{n}}[/tex]
where
[tex]\bf \bar x[/tex]= mean of the sample
[tex]\bf \mu[/tex]= mean of the null hypothesis
s = standard deviation of the sample
n = size of the sample
The sample z-score is then
[tex]\bf z=\frac{136 - 120}{9.4742/20}=16/0.47341=33.7759[/tex]
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 [tex]\bf Z^*[/tex] we should reject the null hypothesis and propose a new mean of 34%.