Question

The World Bank collected data on the percentage of GDP that a country spends on health expenditures ("Health expenditure," 2013) and also the percentage of woman receiving prenatal care ("Pregnant woman receiving," 2013). The data for the countries where this information is available for the year 2011 are in table #10.1.8.

a.) Test at the 5% level for a correlation between percentages spent on health expenditure and the percentage of woman receiving prenatal care.

b.) Find the standard error of the estimate.

c.) Compute a 95% prediction interval for the percentage of woman receiving prenatal care for a country that spends 5.0 % of GDP on health expenditure.

HEALTH EXPENDITURE (% of GDP)	Prenatal Care (%)
9.6	47.9
3.7	54.6
5.2	93.7
5.2	84.7
10.0	100.0
4.7	42.5
4.8	96.4
6.0	77.1
5.4	58.3
4.8	95.4
4.1	78.0
6.0	93.3
9.5	93.3
6.8	93.7
6.1	89.8

Answer 1

Result Details & Calculation

X Values
∑ = 91.9
Mean = 6.127
∑(X - M_x)² = SS_x = 56.729

Y Values
∑ = 1198.7
Mean = 79.913
∑(Y - M_y)² = SS_y = 5318.417

X and Y Combined
N = 15
∑(X - M_x)(Y - M_y) = 94.205

R Calculation
r = ∑((X - M_y)(Y - M_x)) / √((SS_x)(SS_y))

r = 94.205 / √((56.729)(5318.417)) = 0.1715

a) The hypotheses are'

H_o: Β₁ = 0

H_a: Β₁ ≠ 0

The null hypothesis states that the slope is equal to zero, and the alternative hypothesis states that the slope is not equal to zero.

Standard error is calculated as

SE = sqrt [ Σ(y_i - M_y)² / (n - 2) ] / sqrt [ Σ(x_i - M_x)² ]

SE=2.6854

Degree of fredom (DF)=n-2=15-2=13

t calculated as

t=B₁/SE

And B₁ is calculated as 1.6606

Hence t value calculated as'

t=0.62

hence p value assosiated with t value 0.62 as we use 2 tail test.

The P-Value is 0.545973 >0.05(significance level).

b). The standard error has been calculated as

SE = sqrt [ Σ(y_i - M_y)² / (n - 2) ] / sqrt [ Σ(x_i - M_x)² ]

=2.6854

c) The confidence interval calculated as'

95% CI is from -0.37 to 0.63