Question

PLEASE PLEASE HELP!!! THIS IS A STATISTICS AND CODING PROBLEM AND I REALLY NEED HELP WITH THE ANSWERS!!

Parental leave is often compensated to some degree, but the amount
of compensation varies greatly. You read a research article that stated that,
"across people of all incomes, 47% of leave-takers received full pay during
their leave, 16% received partial pay, and 37% received no pay."

After reading this, you wonder what the distribution of pay is for low-income households. Suppose you conduct a survey of leave-takers within households earning less than $30,000 per year. You surveyed 225 people (selected in a random sample) and found that 51 received full pay, 33 received partial pay, and 141 received no pay.

a) You would like to investigate whether the distribution of pay for households earning < $30,000 is different from that of all income levels.
Does this correspond to (mark an x in front of the correct answer) either of the choices below:
____chi-square test of independence or
____a chi-square test for goodness of fit?

b) What are the expected counts of leave-takers among households with incomes < $30,000? State the null hypothesis under which these expected counts were computed.

Expected value for full pay:_________

Expected value for partial pay:_________

Expected value for no pay:_________

$H_{0}$: <your answer here>

c) Compute the chi-square statistic and comment on which cell contributes the most to the statistic.

chi^2= <your answer here>

The cell that contributes most to the statistic is <your answer here>

<show your work here>

d) Compute the p-value for your test statistic and conclude whether you believe there is evidence against the null hypothesis in favor of the alternative
hypothesis.

p-value: <your answer here>

Is there evidence against the null hypothesis? <yes or no>

<your work here.>

I WOULD VERY MUCH APPRECIATE EXPLANATIONS!! THANK YOU SO MUCH!!

Answer 1

Solution:-

a) a chi-square test for goodness of fit.

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: The distribution of pay for households earning < $30,000 is not different from that of all income levels.

Alternative hypothesis: The distribution of pay for households earning < $30,000 is different from that of all income levels.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a chi-square goodness of fit test of the null hypothesis.

Analyze sample data. Applying the chi-square goodness of fit test to sample data, we compute the degrees of freedom, the expected frequency counts, and the chi-square test statistic. Based on the chi-square statistic and the degrees of freedom, we determine the P-value.

DF = k - 1 = 3 - 1
D.F = 2

b)

(E_i) = n * p_i

c)

Expected Number 105.75 28.34574468 51 Full pay Partial pay No pay Total 36.00 83.25 40.06081081 141 225 225 68.65655549

X² = 68.66

Er,c

where DF is the degrees of freedom, k is the number of levels of the categorical variable, n is the number of observations in the sample, E_i is the expected frequency count for level i, O_i is the observed frequency count for level i, and X² is the chi-square test statistic.

d)

The P-value is the probability that a chi-square statistic having 2 degrees of freedom is more extreme than 68.66.

We use the Chi-Square Distribution Calculator to find P(X² > 68.66) = less than 0.0001

Interpret results. Since the P-value (almost 0) is less than the significance level (0.05), we have to reject the null hypothesis.