Question

The heights and weights of 15 randomly selected adult men are shown in the table. Use α = .05 .

Ht (in.)    73  70  73  72  68  72  68  70  68  67  72  70  65  74  70

Wt (lbs.)  185 155 195 164 145 185 170 150 180 175 175 147 140 210 200

State the conclusion in words.

1. There is not sufficient sample evidence to support the claim of linear correlation.
2. The sample data support the claim of linear correlation.

The heights and weights of 15 randomly selected adult men are shown in the table. Use α = .05 .

3. Ht (in.)    73 70 73 72 68 72 68 70 68 67 72 70 65 74 70
4. Wt (lbs.) 185 155 195 164 145 185 170 150 180 175 175 147 140 210 200

What weight do you predict for a man who is 66 inches tall? (Round to the nearest hundredth.)

A video game player insists that the longer he plays a certain computer game, the higher his scores are. The table shows the total number of minutes played and the high score (in thousands of points) achieved after each 5-minute interval. Use α = .01 .

# Min   5   10    15   20    25    30    35    40    45   50   55   60

Score  48.0 53.3 101.9 72.5 121.5 146.0 196.1 118.5 150.5 80.7 36.0 64.8

Find the linear correlation coefficient, r. (Round to the nearest ten-thousandth.)

Answer 1

Answers: The null and alternative hypotheses for the given problem is H0: rho = 0 vs Ha: rho not equal to 0.

Where rho is the population correlation coefficient of the heights and weights of the adult men. The test statistic used for this problem is t = r*sqrt(n-2)/ sqrt(1-(r*r)). Where r is the sample correlation coefficient, n=total no. of observations.

We reject null hypothesis, H0, if |t(obs)| > t(alpha/2,(n-2)) where t(alpha/2,(n-2)) is the upper alpha/2 point of the Student's t distribution with (n-2) degrees of freedom.

r = n*sum(x*y) - (sum(x)*sum(y)) / sqrt[(n*sum(x*x) - sum(x)^2)*(n*sum(y*y) - sum(x)^2)]

Here t(obs) =  2.797683 and t(alpha/2,(n-2)) =  2.160369.

Thus, we reject H0 and conclude at a 5% level of significance that the sample data support the claim of linear correlation.

Option (b) is correct.

We obtain the least squares regression equation of y on x as: y_hat = a + bx where a and b are known as y-intercept and slope of the regression equation and

b = sum((xi - xbar) *(yi-ybar))/ sum((xi - xbar)^2) and a = ybar - (b * xbar) where xbar,ybar are the sample means

The least squares regression equation is obtained as y_hat = -187.241 + 5.118 x. Where y_hat is the predicted value of weight in lbs and x is the given height in inches.

We obtain the predicted weight for a man who is 66 inches tall by putting x = 66 in the above equation. It is obtained to be 150.55 lbs (Rounded to the nearest hundredth).

2) The null and alternative hypotheses for the given problem is H0: rho = 0 vs Ha: rho > 0

Where rho is the population correlation coefficient of the minutes played and scores received. The test statistic used for this problem is t = r*sqrt(n-2)/ sqrt(1-(r*r)).

We reject null hypothesis, H0, if t(obs) > t(alpha,(n-2)) where t(alpha,(n-2)) is the upper alpha point of the Student's t distribution with (n-2) degrees of freedom.

Here t(obs) = 0.2562709 and t(alpha,(n-2)) =  2.650309.

Thus we do not reject H0 and conclude on the basis of the given sample at 1% level of significance that there is no evidence that the longer he plays a certain computer game, the higher his scores are.

The linear correlation coefficient is r = 0.0808 (Rounded to the nearest ten-thousandth.)

[The answers are obtained using probability tables f Student's T distribution]