Question

A genetic experiment with peas resulted in one sample of offspring that consisted of 417 green peas and 153 yellow peas.

a. Construct a 90% confidence interval to estimate of the percentage of yellow peas.

b. It was expected that​ 25% of the offspring peas would be yellow. Given that the percentage of offspring yellow peas is not​ 25%, do the results contradict​ expectations?

Answer 1

Solution :

a ) Given that

n = 471 green peas + 153 yellow peas = 624

x = 153

   = x / n = 153 / 624 =0.245

1 - = 1 - 0.245 = 0.750

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z/2 = Z0.05 = 1.645

Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 1.645 * (((0.245 * 0.755) / 624 )

= 0.028

A 90 % confidence interval for population proportion p is ,

- E < P < + E

0.245 - 0.028 < p < 0.245 + 0.028

0.216 < p < 0.273

b ) = 0.250

1 - = 1 - 0.250 = 0.750

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z/2 = Z0.05 = 1.645

Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 1.645 * (((0.250 * 0.750) / 624)

= 0.028

A 90 % confidence interval for population proportion p is ,

- E < P < + E

0.250 - 0.028 < p < 0.250 + 0.028

0.221 < p < 0.278