Answer:
[tex]\fbox{\begin{minipage}{10em}Option A is correct\end{minipage}}[/tex]
Step-by-step explanation:
Step 1: Define significance level
In this hypothesis testing problem, significance levels α is selected: [tex]0.05[/tex], the associated z-value from Laplace table:
Φ([tex]z[/tex]) = α - [tex]0.5 = 0.05 - 0.5 = -0.45[/tex]
=> [tex]z[/tex] = [tex]-1.645[/tex]
Step 2: Define null hypothesis ([tex]H_{0}[/tex]) and alternative hypothesis ([tex]H_{1}[/tex])
[tex]H_{0}[/tex] : rate of flu infection [tex]p[/tex] = 8.3% or 8.3/100 = 0.083
[tex]H_{1}[/tex] : rate of flu infection [tex]p[/tex] < 8.3% or 8.3/100 = 0.083
Step 3: Apply the formula to check test statistic:
[tex]K = \frac{f - p}{\sqrt{p(1 - p)} } * \sqrt{n}[/tex]
with [tex]f[/tex] is actual sampling percent, [tex]p[/tex] is rate of flu infection of [tex]H_{0}[/tex], [tex]n[/tex] is number of samples.
The null hypothesis will be rejected if [tex]K < z[/tex]
Step 4: Calculate the value of K and compare with [tex]z[/tex]
[tex]K = \frac{(\frac{3.5}{100}) - 0.083}{\sqrt{0.083(1 - 0.083)} } * \sqrt{200} = -2.46[/tex]
We have [tex]-2.46 < -1.645[/tex]
=>This is good evidence to reject null hypothesis.
=> The actual rate is lower. (As [tex]H_{1}[/tex] states)
Hope this helps!
:)
