Question

You are a social worker in an OBGYN department of a hospital. The CEO of the hospital has asked you to write a report about the length of time that women spend in the hospital after giving birth. The CEO has told you that in the thinks that insured women spend longer on average in the hospital than uninsured women after childbirth. She wants you to test this claim with data from your hospital. Two samples of 16 women were taken. Test the claim that insured women spend longer in the hospital than uninsured women using an α = .01.

Insured:

Mean = 2.3 days

Standard Deviation = 0.77

Sample Size = 16

Uninsured:

Mean = 1.9 days

Standard Deviation = 0.77

Sample Size = 16

To complete this lab exercise, you should:

Identify whether you will test this claim using a 1-tailed hypothesis or a 2-tailed hypothesis.

State the Null and Research Hypotheses

Find the Critical Value using the T-Table and interpret what you will do with the null hypothesis given that Critical Value

Identify the Correct Degrees of Freedom you’ll use

HAND CALCULATE the t-value. Show your work. Start with the formula and then plug in the correct values from there.

Make a decision regarding the null. Interpret your decision with regard to this question.

Find the p-value range for the hypothesis test using your test statistic and the t-table.

BY HAND, construct a 98% Confidence Interval for the difference between insured and uninsured women. What does this confidence interval tell you?

What are your overall conclusions? Is the confidence interval interpretation consistent with your interpretation from the t-test?

Answer 1

Solution:

Insured:

Mean = days

Standard Deviation = s₁ =0.77

Sample Size = n₁ = 16

Uninsured:

Mean = days

Standard Deviation = s₂ = 0.77

Sample Size = n₂ = 16

We have to test the claim that insured women spend longer in the hospital than uninsured women using an α = .01

Part i) This is 1-tailed hypothesis, since claim is directional.

Part ii) State the Null and Research Hypotheses

Null hypothesis:

Vs

Research hypothesis ( Alternative hypothesis)

Part iii)

Find the Critical Value using the T-Table and interpret what you will do with the null hypothesis given that Critical Value

Identify the Correct Degrees of Freedom you’ll use

df = n1 + n2 - 2 = 16 + 16 - 2 = 30

Look in t table for df = 30 and one tail area = 0.01 and find t critical value.

Thus from t table , we get: t = 2.457

Reject null hypothesis H0, if t test statistic value > t critical value = 2.457.

Part iv) t test statistic value ( assuming equal variances )

where

Thus

Part v) Make a decision regarding the null. Interpret your decision with regard to this question

Since t test statistic value = t = 1.469 is not greater than 2.457 , we fail to reject H0.

Since we failed to reject H0, we conclude that there is not sufficient evidence to support the claim that insured women spend longer in the hospital than uninsured women using an α = .01.

Part vi) Find the p-value range for the hypothesis test using your test statistic and the t-table.

df = 30

look in t table for df = 30 row and find the interval in which t = 1.469 fall and then find corresponding one tail area

that would be the p-value range.

Thus from t table, t = 1.469 fall in between 1.310 and 1.697 and corresponding one tail area is in between 0.05 and 0.10

Thus p-value range = 0.05 to 0.10

Part vii) construct a 98% Confidence Interval for the difference between insured and uninsured women. What does this confidence interval tell you?

Thus

for c = 98% , two tail area = 1 - 0.98 = 0.02

df = 30

Look in t table for df = 30 and two tail area= 0.02

t_c = 2.457

Thus

Thus

We are 98% confident that the true value of difference in population means time spend of insured women and uninsured women is in between .

Part viii) What are your overall conclusions?

Since confidence interval includes 0, we conclude that there is no difference in population means time spend of insured women and uninsured women.

Is the confidence interval interpretation consistent with your interpretation from the t-test?

Yes. Since both conclusions are same.