Question

A medical research group is recruiting people to complete short surveys about their medical history. For example, one survey asks for information on a person’s family history in regards to cancer. Another survey asks about what topics were discussed during the person’s last visit to a hospital. So far, as people sign up, they complete an average of just 4 surveys, and the standard deviation of the number of surveys is about 2.2. The research group wants to try a new interface that they think will encourage new enrollees to complete more surveys, where they will randomize each enrollees to either get the new interface or the current interface. How many new enrollees do they need for each interface to detect an effect size of 0.5 surveys per enrollee, if the desired power level is 80%

Answer 1

Answer: They need 32 new enrollees .

Step-by-step explanation:

The formula to find the sample size :

$n=(\dfrac{z_{\alpha/2}\ \sigma}{E})^2$

, where $\sigma$ = Population Standard deviation

E = Margin of error

$z_{\alpha/2}$ is the two-tailed z-value as per confidence level or significance level $(\alpha)$ .

Given : Standard deviation: $\sigma=2.2$

To find : Minimum sample size of enrollee with margin of error 0.5 , if the desired power level is 80%.

Margin of error : E= 0.5

Two-tailed z value for 80% confidence level i.e. significance level of $\alpha=0.20$

$z_c=z_{\alpha/2}=z_{0.10}=1.2816$

Then , the formula to find the sample size :

$n=(\dfrac{(1.2816) \times2.2}{0.5})^2=31.7987721216\approx32$

Hence, required sample size : 32