Question

Health Care Knowledge Systems reported that an insured woman spends on average 2.3 days in the hospital for a routine childbirth, while an uninsured woman spends on average 1.9 days in the hospital. Stillwater Medical Center wants to prove that uninsured women spend less time on average in the hospital than insured woman. A sample of 50 insured women and 45 uninsured women was taken. The table below summarizes the sample statistics calculated. The doctor conducting the study decides based on information obtained from the article that the standard deviations of each population cannot be assumed equal. Is there significant evidence at a significance level of 0.01 to conclude the hospitals claim?

	Sample Size	Mean	Standard deviation
Insured Women	50	2.5	0.6
Uninsured Women	45	1.75	0.4

Hypotheses: (2 points)

Test Statistic:(8 points)

p-value: (3 points)

Decision/Conclusions: (2 points)

Answer 1

H₀: $\mu1 = \mu2$

H₁: $\mu1 > \mu2$

The test statistic t = ( $\small \bar x1 - \bar x2$ )/sqrt(s1^2/n1 + s2^2/n2)

                             = (2.5 - 1.75)/sqrt((0.6)^2/50 + (0.4)^2/45)

                              = 7.23

DF = (s1^2/n1 + s2^2/n2)^2/((s1^2/n1)^2/(n1 - 1) + (s2^2/n2)^2/(n2 - 1))

     = ((0.6)^2/50 + (0.4)^2/45)^2/(((0.6)^2/50)^2/49 + ((0.4)^2/45)^2/44)

     = 86

P-value = P(T > 7.23)

             = 1 - P(T < 7.23)

             = 1 - 1 = 0

Since the P-value is less than the significance level(0 < 0.01), so we should reject the null hypothesis.

So at 0.01 significance level there is sufficient evidence to conclude the hospital's claim that uninsured women spend less time on average in the hospital than the insured woman.