Question

1.A genetic experiment with peas resulted in one sample of offspring that consisted of 422 green peas and 165 yellow peas.

a. Construct a 95​% 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?

2.Salaries of 38 college graduates who took a statistics course in college have a​ mean, x of $62,900.Assuming a standard​ deviation,sigmaσ​,of ​$19,662​,construct a 90​% confidence interval for estimating the population mean muμ.

Answer 1

Q1:

n = 165+422 = 587

x = 165

p̂ = x/n = 0.2811

a) 95% Confidence interval :

At α = 0.05, two tailed critical value, z_c = NORM.S.INV(0.05/2) = 1.960

Lower Bound = p̂ - z_c*√(p̂ *(1- p̂)/n) = 0.2811 - 1.96 *√(0.2811*0.7189/587) = 0.245

Upper Bound = p̂ + z_c*√(p̂ *(1- p̂)/n) = 0.2811 + 1.96 *√(0.2811*0.7189/587) = 0.317

0.245 < p < 0.317

b) As the confidence interval contain 0.25, we fail to reject the null hypothesis.

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Q2:

x̅ = 62900, σ = 19662, n = 38

90% Confidence interval :

At α = 0.1 two tailed critical value, z_c = ABS(NORM.S.INV(0.1/2)) = 1.645

Lower Bound = x̅ - z_c*σ/√n = 62900 - 1.645 * 19662/√38 = 57653.58

Upper Bound = x̅ + z_c*σ/√n = 62900 + 1.645 * 19662/√38 = 68146.42

57653.58 < µ < 68146.42