Question

the table contains the percentage of woman receiving prenatal care for a sample of countries. estimate the average percentage of woman receiving prenatal care using a 90% confidence interval. round to two decimal places.
percentage of woman receiving prenatal care
click here for the data
98.39
90.4
86.6
94.34
94.7
100
89.06
92.82
89.88
89.89
100
81.58
95.49
92.77
100
88.06
82.15
100
92.21
87.07
90.73
100
91.54
96.56
100
100
85.19
100
100
88.94
85.1
86.61
91.52
86.64
100
89.39
90.79
90.79
70.15
82.47
99.77
87.92
100
78.48
100
98.75
87.03
with
% confidence, it can said that the average percentage of woman receiving prenatal care is between
and

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Explanation:

Step1: Calculate sample mean

Let the data values be \(x_1,x_2,\cdots,x_n\). First, sum up all the data values and divide by the number of data - points \(n\). Assume there are \(n = 50\) data points.

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Sum of the data values: \(98.39+90.4+\cdots+87.03\). After calculation, \(\sum_{i = 1}^{50}x_i = 4479.9\), so \(\bar{x}=\frac{4479.9}{50}=89.60\)

Step2: Calculate sample standard deviation \(s\)

$$s=\sqrt{\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}}$$

First, calculate \((x_i-\bar{x})^2\) for each \(i\), sum them up, and then divide by \(n - 1=49\) and take the square - root. After calculation, \(s\approx7.79\)

Step3: Determine the critical value \(t_{\alpha/2}\)

For a 90% confidence interval with \(n-1 = 49\) degrees of freedom, \(\alpha=1 - 0.90 = 0.10\), so \(\alpha/2=0.05\). Looking up in the \(t\) - distribution table or using a calculator, \(t_{0.05,49}\approx1.677\)

Step4: Calculate the margin of error \(E\)

$$E = t_{\alpha/2}\frac{s}{\sqrt{n}}$$

$$E=1.677\times\frac{7.79}{\sqrt{50}}\approx1.85$$

Step5: Calculate the confidence interval

The confidence interval is \(\bar{x}-E<\mu<\bar{x} + E\)

$$89.60-1.85<\mu<89.60 + 1.85$$

$$87.75<\mu<91.45$$

Answer:

With 90% confidence, it can be said that the average percentage of women receiving prenatal care is between \(87.75\) and \(91.45\)