Question

The U.S infant mortality rates (IMR) for both sexes and all races for the years 1981-1990 ( coded as years 1-10) are in the table.

Year	1	2	3	4	5	6	7	8	9	10
IMR	11.9	11.5	11.2	10.8	10.6	10.4	10.1	10.0	9.8	9.2

a) Calculate a 95% confidence interval for the mean IMR for 1985

b) Suppose that for Australia in 1985 , the IMR was 9.3 deaths per 1000 live births. Was the U.S IMR significantly higher than that of Australia in 1985? us α = 0.01, use statistic t = (a + bx* - hypothesized value)/(S (a+bx*))

Answer 1

a) 95 % CI

This simple confidence interval calculator uses a t statistic and sample mean (M) to generate an interval estimate of a population mean (μ).

The formula for estimation is:

μ = M ± t(s_M)

where:

M = sample mean
t = t statistic determined by confidence level
s_M = standard error = √(s²/n)

Calculation

M = 10.55
t = 2.26
s_M = √(0.82²/10) = 0.26

μ = M ± t(s_M)
μ = 10.55 ± 2.26*0.26
μ = 10.55 ± 0.5866

Result

M = 10.55, 95% CI [9.9634, 11.1366].

You can be 95% confident that the population mean (μ) falls between 9.9634 and 11.1366.

b) Suppose that for Australia in 1985 , the IMR was 9.3 deaths per 1000 live births. Was the U.S IMR significantly higher than that of Australia in 1985? us α = 0.01, use statistic t = (a + bx* - hypothesized value)/(S (a+bx*))

To calculate a+bx, fit the regression model on the data

Sum of X = 55
Sum of Y = 105.5
Mean X = 5.5
Mean Y = 10.55
Sum of squares (SS_X) = 82.5
Sum of products (SP) = -22.25

Regression Equation = ŷ = bX + a

b = SP/SS_X = -22.25/82.5 = -0.2697

a = M_Y - bM_X = 10.55 - (-0.27*5.5) = 12.03333

ŷ = -0.2697X + 12.03333

a+bx= -0.2697X + 12.03333

Test statistics as,

t = (a + bx* - hypothesized value)/(S (a+bx*))= (9.52-9.3)/(0.82(9.52)= 0.02

P value for the statistics is as 0.97

that means there no significant value and he current value and 9.3

thanks