Question

on occasion, medical studies need to model the proportion of the population that has a disease and compare that to observed frequencies of the disease actually occurring. suppose the end - stage renal failure in south - west wales was collected for different age groups, where individuals were randomly picked. do the data in the table show that the observed frequencies are in agreement with proportion of people in each age group? test at the 1% level.
renal failure frequencies
age group 16 - 29 30 - 44 45 - 59 60 - 74 75+ total
observed frequencies 115 120 142 131 74 582
proportion 0.23 0.25 0.22 0.21 0.09
state the hypotheses.
$h_0$: the observed frequencies select an answer in agreement with proportion of people in each age group.
$h_a$: the observed frequencies select an answer in agreement with proportion of people in each age group.
calculate the $x^2$ test statistic. round expected values to two decimal places when calculating the test statistic. round $x^2$ to three decimal places.
$x^2=$
calculate the p - value. round to four decimal places.
p - value=
state your decision.
since p - value? a, select an answer $h_0$.
interpret the results.
at the 1% significance level there select an answer enough evidence to conclude that the observed frequencies select an answer in agreement with proportion of people in each age group.

Explanation:

Step1: State the hypotheses

The null hypothesis \(H_0\) is that the observed frequencies are in agreement with the proportion of people in each age group. The alternative hypothesis \(H_a\) is that the observed frequencies are not in agreement with the proportion of people in each age group.

Step2: Calculate the expected frequencies

The formula for the expected frequency \(E_i=n\times p_i\), where \(n = 582\) (total number of observations) and \(p_i\) is the proportion for each age - group.

• For age group \(16 - 29\): \(E_1=582\times0.23 = 133.86\)
• For age group \(30 - 44\): \(E_2=582\times0.25=145.5\)
• For age group \(45 - 59\): \(E_3=582\times0.22 = 128.04\)
• For age group \(60 - 74\): \(E_4=582\times0.21=122.22\)
• For age group \(75+\): \(E_5=582\times0.09 = 52.38\)

Step3: Calculate the \(\chi^{2}\) test statistic

The formula for the \(\chi^{2}\) test statistic is \(\chi^{2}=\sum\frac{(O_i - E_i)^{2}}{E_i}\), where \(O_i\) is the observed frequency.

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Step4: Calculate the degrees of freedom and p - value

The degrees of freedom \(df=k - 1\), where \(k = 5\) (number of age groups). So \(df=5 - 1=4\)
Using a \(\chi^{2}\) - distribution table or a calculator (\(\chi^{2}\) - test function), for \(\chi^{2}=18.20\) and \(df = 4\), the p - value is \(P(\chi^{2}>18.20)\approx0.0011\)

Step5: Make a decision

Since the significance level \(\alpha = 0.01\) and the p - value \(=0.0011<0.01\)

Answer:

• \(H_0\): The observed frequencies are in agreement with proportion of people in each age group. \(H_a\): The observed frequencies are not in agreement with proportion of people in each age group.
• \(\chi^{2}=18.20\)
• \(p - value = 0.0011\)
• Since \(p - value<\alpha\), reject \(H_0\)
• At the \(1\%\) significance level there is enough evidence to conclude that the observed frequencies are not in agreement with proportion of people in each age group.