Could someone help me set up the confidence interval equation here?

Question

Patients recovering from an appendix operation normally spend an average of 6.3 days in the hospital. The distribution of recovery times is normal with a s.d. of 2 days. The hospital is trying a new recovery program designed to lessen the time patients spend in the hospital. The first 16 appendix patients in this new program were released from the hospital in an average of 5.3 days. On the basis of these data, can the hospital conclude that the new program has a significant reduction of recovery time? Test at the .05 level of significance.

Calculate the 95% confidence interval.

CI = x-bar ± Zcrit (standard deviation/√N)

I'm just not sure how to apply the data to the formula.

Answer 1

ANSWER: New Recovery Program reduces the average hospital stay below 6.3 days.

Why???

7 - Step Procedure for t Distributions, "one-tailed test", CONFIDENCE INTERVAL, SMALL SAMPLE SIZE

1. Parameter of interest: "μ" = population mean for days of hospital stay

2. Null hypothesis Ho: μ = 6.3

3. Alternative hypothesis Ha: μ < 6.3

4. Test statistic formula: t = (x-bar - μ)/(s/SQRT(n))

x-bar = estimate of the Population Mean (statistical mean of the sample) [5.3]

n = number of individuals in the sample [16]

s = standard deviation [2]

μ = Population Mean [6.3]

5. Computation of Test statistic formula t = -2.0

6. Determination of the P-value: The test is based on n -1 = 15 df (degrees of freedom). Table "look-up" value shows area under the 15 df curve to the left of t = -2.0 is (approximately) 0.032

7. Conclusion: with significance value α = 0.05 (1 - 0.95) confidence level, above shows P-value <= α, [0.032 <= 0.05] which means the Null hypothesis Ho: μ = 6.3 of the average days of hospital stay should be be rejected.  The Alternative Hypothesis Ha: μ < 6.3 days should be accepted stating that the true average hospital stay from the new recovery program is less than 6.3 days.

Answer 2

The hypotheses I would choose are

H0: μ1 = μ0

Ha: μ1 < μ0 (not Ha: μ1 ≠ μ0)

where μ0 = 6.3 and μ1 is not known

The critical region is xbar < μ0 - (σ/√n)z_(1-α)

= 6.3 - (2/√16)(1.645)

= 5.4775

Since the observed value of xbar (5.3) lies in the critical region, we reject the null hypothesis -- that is, the data support the hypothesis that the new program reduces recovery time.

This assumes that the variance using the new program is the same as the variance using the old program.