Question

A genetic experiment with peas resulted in one sample of offspring that consisted of 435 green peas and 154 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? a. Construct a 90​% confidence interval. Express the percentages in decimal form. nothing less than p less than nothing ​(Round to three decimal places as​ needed.)

Answer 1

Solution :

Given that,

n = 435

x = 154

Point estimate = sample proportion = = x / n = 154 / 435 = 0.354

1 - = 1 - 0.354 = 0.646

At 90% confidence level

= 1 - 90%

=1 - 0.90 = 0.10

/2 = 0.05

Z/2 = Z_{0.05 = 1.645}

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

= 1.645 (((0.354 * 0.646))/ 435 )

= 0.038

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.354 - 0.038 < p < 0.354 + 0.038

0.316 < p < 0.392

31.6% < p < 39.2%

The 90% confidence interval for the population proportion p is : 31.6% < p < 39.2%

b.

n = 435

x = 154

The null and alternative hypothesis is

H₀ : p = 0.25

H_a : p 0.25

This is the two tailed test .

= x / n = 154 / 435 = 0.354

P₀ = 25% = 0.25

1 - P₀ = 1 - 0.25 = 0.75

Test statistic = z

= - P₀ / [P_{0 *} (1 - P₀ ) / n]

= 0.354 - 0.25 / [0.25 ( 1 - 0.25) / 435 ]

= 5.01

The test statistic = 5.01

P-value = 0.0000

= 0.10

0.0000 < 0.10

P-value <

Reject the null hypothesis .

There is sufficient evidence to test the claim .