Nitric oxide is sometimes given to newborns who experience respiratory failure.

Question

Nitric oxide is sometimes given to newborns who experience respiratory failure. In one experiment, nitric oxide was given to 114 infants. This group was compared to a control group of 121 infants. The length of hospitalization (in days) was recorded for each of the 235 infants. The mean in the nitric oxide sample was y_1 = 36.4; the mean in the control sample was y_2 = 29.5. A 95% confidence interval for mu_1 - mu_2 is (-2.3, 16.1), where mu_1 is the population mean length of hospitalization for infants who get nitric oxide and mu_2 is the mean length of hospitalization for infants in the control population. Assume equal variances. Which of the following conclusion is correct? (1) We are 95% confident that mu_1 is greater than mu_2 since most of the confidence interval is greater than zero. (2) We are 95% confident that the difference between mu_1 and mu_2 is between -2.3 days and 16.1 days. (3) We are 95% confident that the difference between y_1 and y_2 is between -2.3 days and 16.1 days. (4) 95% of the nitric oxide infants were hospitalized longer than the average control infant. Based on the given two-sided confidence interval, should we reject H_0 if we tested H_0: mu_1 - mu = 0 against H_1: mu_1 - mu_2 notequalto 0, using alpha = 0.05? Explain.

Explanation / Answer

Part-a

Option—2 is the answer

Part-b

We would not reject the null hypothesis as 95% confidence interval includes zero.

Part-c

95% confidence interval of diference in mean =(-2.3, 16.1)

So, Margin of error = (16.1-(-2.2))/2=9.15

Degree of freedom=n1+n2-2=114+121-2=233

At 5% level critical t=1.97

So standard error of difference of means SE= Margin of error/1.97=9.15/1.97=4.64

So test statistic t=Mean difference/SE

=(36.4-29.5)/4.64

=1.49

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