Question

Heart rates for newborns born in one of two hospitals were measured to determine whether birth outcomes are comparable at the two locations. Assume the hospitals are independent.

Relationship between heart rate and hospital
Hospital	Mean heart rate (bpm)	SD	n
Riverside Methodist	127	11	216
NRV General	133	12	153

a) Test for a significant difference between mean heart rate in the two hospitals. Remember, as always, to give hypotheses, a p-value, and any other necessary information. Choose an appropriate p-value.

We compare the length of stay in days from several randomly selected patients:

Length of stay comparison
Hospital	Length of stays (days)
Riverside Methodist	21, 10, 32, 60, 7, 44, 27, 3, 16, 26, 33
NRV General	86, 27, 9, 70, 88, 75, 120, 60, 35, 73, 96, 44, 240

a) Why is a t-test inappropriate in this case?

b) Conduct an appropriate nonparametric test to determine whether the two hospitals are com- parable in length of stay. Report a conclusion and necessary work.

Answer 1

In first case we use two sample t test for summary data, using MINITAB
Using minitab commands:

The command is Stat>>>Basic Statistics >>2 sample t...

Summarized data: Choose if you have summary values for the sample size, mean, and standard deviation for each sample.

First

Sample size: Enter the value for the sample size.

Mean: Enter the value for the mean.

Standard deviation: Enter the value for the standard deviation.

Second

Sample size: Enter the value for the sample size.

Mean: Enter the value for the mean.

Standard deviation: Enter the value for the standard deviation.

then click on Option select level of confidence = 1 - alpha = (1 - 0.05)*100 = 95.0

Alternative " "

Click on Ok

Again "click on OK"

We get the following output

MTB > TwoT 216 127 11 153 133 12;
SUBC>   Pooled.

Two-Sample T-Test and CI

Sample    N Mean   StDev SE Mean
1      216    127.0   11.0     0.75
2         153     133.0   12.0     0.97

Difference = mu (1) - mu (2)
Estimate for difference: -6.00
95% CI for difference: (-8.37, -3.63)
T-Test of difference = 0 (vs not =): T-Value = -4.97 P-Value = 0.000 DF = 367
Both use Pooled StDev = 11.4248

Decision rule: 1) If p-value < level of significance (alpha) then we reject null hypothesis

2) If p-value > level of significance (alpha) then we fail to reject null hypothesis.

Here p value = 0.00 < 0.05 so we used first rule.

That is we reject null hypothesis .

Conclusion: At 5% level of significance there are sufficient evidence to say that the significant difference between mean heart rate in two hospitals.