Answer:
99.5% Confidence interval = (-0.025, 0.547)
= -0.025 < p < 0.547
Step-by-step explanation:
| A | B | C | Total
Male | 5 | 4 | 17 | 26
Female | 6 | 2 | 15 | 23
Total | 11 | 6 | 32 | 49
If p represent the proportion of all female students who would receive a grade of A on this test. Use a 99.5% confidence interval to estimate p to three decimal places.
All female students = 23
Female students that score an A = 6
p = (6/23) = 0.2608695652 = 0.261
Confidence Interval for the population proportion is basically an interval of range of values where the true population proportion can be found with a certain level of confidence.
Mathematically,
Confidence Interval = (Sample proportion) ± (Margin of error)
Sample proportion = (6/23) = 0.261
Margin of Error is the width of the confidence interval about the mean.
It is given mathematically as,
Margin of Error = (Critical value) × (standard Error)
Critical value at 99.5% confidence interval for sample size of 23 is obtained from the t-tables since information on the population standard deviation is not known.
we first find the degree of freedom and the significance level.
Degree of freedom = df = n - 1 = 23 - 1 = 22.
Significance level for 99.5% confidence interval
(100% - 99.5%)/2 = 0.25% = 0.0025
t (0.0025, 22) = 3.119 (from the t-tables)
Standard error of the mean = σₓ = √[p(1-p)/N]
p = 0.261
N = sample size = 23
σₓ = √(0.261×0.739/23) = 0.091575
99.5% Confidence Interval = (Sample proportion) ± [(Critical value) × (standard Error)]
CI = 0.261 ± (3.119 × 0.091575)
CI = 0.261 ± 0.2856
99.5% CI = (-0.0246, 0.5466)
99.5% Confidence interval = (-0.025, 0.547)
= -0.025 < p < 0.547
Hope this Helps!!!
