Answer to Question #348567 in Statistics and Probability for Kheti

Question

Your given ∑ 𝑥 = 44 , ∑ 𝑥2 = 174, ∑ 𝑥𝑦 = 1324, in addition you also given the values of y

as:

Y 26 28 24 18 35 24 36 25 31 37 30 32

3a. calculate the Pearson correlation coefficient [7]

3b. estimate the y value associated with x=4 [8].

3c. You are given the mean of 20.3 for a random sample of 90 observations from a normal distribution population with a standard deviation of 3.9. Construct a 95% confidence level and interpret your answer. [3]

Answer 1

$$\bar{Y}=\dfrac{1}{n}\sum _{i}Y_i=\dfrac{1}{12}(26 +28+ 24+ 18 +35 +24$$ $$+ 36 +25 +31+ 37+ 30+ 32=\dfrac{173}{6}$$

$$SS_{XX}=\sum_iX_i^2-\dfrac{1}{n}(\sum _{i}X_i)^2$$ $$=174-\dfrac{44^2}{12}=\dfrac{38}{3}$$

$$\sum_iY_i^2=26^2 +28^2+ 24^2+ 18^2 +35^2 +24^2$$

$$+ 36^2 +25^2 +31^2+ 37^2+ 30^2+ 32^2=10336$$

$$SS_{YY}=\sum_iY_i^2-\dfrac{1}{n}(\sum _{i}Y_i)^2$$ $$=10336-\dfrac{346^2}{12}=\dfrac{1079}{3}$$

$$SS_{XY}=\sum_iX_iY_i-\dfrac{1}{n}(\sum _{i}X_i)(\sum _{i}Y_i)$$ $$=1324-\dfrac{44(346)}{12}=\dfrac{166}{3}$$

3a.

$$r=\dfrac{SS_{XY}}{\sqrt{SS_{XX}SS_{YY}}}=\dfrac{166}{\sqrt{38(1079)}}=0.8198$$

Strong positive correlation.

3b.

$$B=\dfrac{SS_{XY}}{SS_{XX}}=\dfrac{166}{38}=4.368421$$

$$A=\bar{Y}-B\bar{X}=\dfrac{173}{6}-4.368421(\dfrac{44}{12})=12.815790$$

$$=12.815790$$

$$y=12.815790+4.368421x$$

$$y(4)=12.815790+4.368421(4)=30.29$$

3c. The critical value for $\alpha = 0.05$ is $z_c = z_{1-\alpha/2} = 1.96.$

The corresponding confidence interval is computed as shown below:

$$CI=(\bar{x}-z_c\times\dfrac{\sigma}{\sqrt{n}},\bar{x}+z_c\times\dfrac{\sigma}{\sqrt{n}})$$

$$=(20.3-1.96\times\dfrac{3.9}{\sqrt{90}},20.3+1.96\times\dfrac{3.9}{\sqrt{90}})$$

$$=(19.494, 21.106)$$

Therefore, based on the data provided, the 95% confidence interval for the population mean is $19.494 < \mu < 21.106,$ which indicates that we are 95% confident that the true population mean $\mu$ is contained by the interval $(19.494, 21.106).$