Question

the table contains the percentage of women receiving prenatal care for a sample of countries. estimate the average percentage of women receiving prenatal care using a 99% confidence interval. round to two decimal places.
percentage of women receiving prenatal care
click here for the data
100
80.84
89.81
91.12
83.35
87.08
87.34
84.14
88.02
81.21
88.18
93.41
87.48
99.24
92.93
91.25
80.99
95.18
84.57
87.84
97.84
100
93.17
75.63
81.29
96.37
100
99.17
87.83
89.44
80.86
88.86
91.4
87.73
100
89.66
89.21
93.97
97.03
93.39
99.57
82.2
82.22
87.61
94.54
90.11
86.03
with % confidence, it can said that the average percentage of women receiving prenatal care is between and

Explanation:

Step1: Calculate sample mean

Let the data - points be \(x_1,x_2,\cdots,x_n\). Here \(n = 45\).
The sample mean \(\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}\).
\(\sum_{i=1}^{45}x_i=100 + 80.84+89.81+\cdots+86.03\)
\(\sum_{i = 1}^{45}x_i = 3977.34\)
\(\bar{x}=\frac{3977.34}{45}\approx88.39\)

Step2: Calculate sample standard - deviation

The sample standard - deviation \(s=\sqrt{\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}}\)
First, calculate \((x_i-\bar{x})^2\) for each \(i\).
\(\sum_{i = 1}^{45}(x_i - 88.39)^2=3399.97\)
\(s=\sqrt{\frac{3399.97}{44}}\approx8.79\)

Step3: Determine the critical value

For a 99% confidence interval with \(n-1 = 44\) degrees of freedom, using a t - distribution table (or a calculator with t - distribution functions), the critical value \(t_{\alpha/2}\) where \(\alpha=1 - 0.99 = 0.01\) and \(\alpha/2=0.005\).
Since \(n = 45\) is relatively large, we can also approximate using the standard normal distribution. For a 99% confidence interval, \(z_{\alpha/2}=2.576\) (using the standard normal distribution table).
The margin of error \(E = z_{\alpha/2}\frac{s}{\sqrt{n}}\)
\(E = 2.576\times\frac{8.79}{\sqrt{45}}\approx2.576\times\frac{8.79}{6.7082}\approx2.576\times1.31\approx3.38\)

Step4: Calculate the confidence interval

The 99% confidence interval is \(\bar{x}-E<\mu<\bar{x} + E\)
\(88.39-3.38<\mu<88.39 + 3.38\)
\(85.01<\mu<91.77\)

Answer:

With 99% confidence, it can be said that the average percentage of women receiving prenatal care is between \(85.01\) and \(91.77\)