Question

A new golf ball design is being evaluated. A random sample of 100 pro golfers and 100 weekend golfers use the new golf ball for two months. After two months, the golfers in each sample were asked whether they preferred the ball they were testing to the balls they had played with previously. 84% of the pro sample said yes, while 65% of the weekend sample said yes. Build a 90% confidence interval estimate of the difference in the proportion of golfers in the two categories who would say that they prefer the new ball. Report the upper and lower bound for your interval.

Answer 1

The pooled sample proportion(P) = ( * n1 + * n2)/(n1 + n2)

                                                      = (0.84 * 100 + 0.65 * 100)/(100 + 100)

                                                      = 0.745

At 90% confidence interval the critical value is z_0.05 = 1.645

The 90% confidence interval is

() +/- z_0.05 * sqrt(P(1 - P)(1/n1 + 1/n2))

= (0.84 - 0.65) +/- 1.645 * sqrt(0.745 * (1 - 0.745) * (1/100 + 1/100))

= 0.19 +/- 0.1014

= 0.0886, 0.2914

Lower limit = 0.0886

Upper limit = 0.2914