2x+4=18 what does 'x' equal?

Members of the millennial generation are continuing to be dependent on their parents (either living with or otherwise receiving

Morgarella [4.7K]

Answer:

a)

$\bf H_0:$ The mean of adults aged 18 to 32 that continue to be dependent on their parents is 0.3

$\bf H_a:$ 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

$\bf np = 400*0.34 = 136$

and standard deviation

$\bf \sqrt{npq}=\sqrt{400*0.34*0.66}=9.4742$

Since in the current research 136 out of 400 individuals (34%) showed to be continuing dependent on their parents:

$\bf H_0:$ The mean of adults aged 18 to 32 that continue to be dependent on their parents is 0.3

$\bf H_a:$ 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 $\bf Z^*$ such that the area under the Normal curve to the right of $\bf Z^*$ is ≤ 0.05

This value can be found by using a table or the computer and is $\bf Z^*$ = 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

$\bf Z^*$ corrected is 2.145

Now we compute the z-score corresponding to the sample

$\bf z=\frac{\bar x -\mu}{s/\sqrt{n}}$