Information on a packet of seeds claims that 93% of them will germinate. Of the 220...

Information on a packet of seeds claims that 93% of them will germinate. Of the 220 seeds that I planted, only 187 germinated.

Compute a 95% two-sided Agresti-Coull CI on the proportion of seeds that germinate. Use as a point estimator p^ the proportion of seeds that germinated during the experiment.
Round your answers to two decimal places

Solutions

Expert Solution

Level of Significance,   α =    0.05
Number of Items of Interest,   x =   187
Sample Size,   n =    220

Sample Proportion ,    p̂ = x/n =    0.8500
z -value =   Zα/2 =    1.960   [excel formula =NORMSINV(α/2)]

Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0241
margin of error , E = Z*SE =    1.960   *   0.0241   =   0.0472

95%   Confidence Interval is
Interval Lower Limit = p̂ - E =    0.850   -   0.0472   =   0.803
Interval Upper Limit = p̂ + E =   0.850   +   0.0472   =   0.897

95%   confidence interval is (   80.28% < p < 89.72% )

orchestra answered 1 year ago

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