Question

Melatonin pills are a popular remedy for providing insomnia relief. A company developed a new formulation of melatonin pill and conducted a trial on a random sample of 2,000 American adults who were already experiencing insomnia and using melatonin pills. A 95% confidence interval for mean increase in sleep hours per typical night was calculated based on data from the trial: (0.10, 0.15) hours. Assess whether the evidence suggests the new formulation is a substantial improvement over melatonin pills currently available on the market.

Answer 1

Answer:

The confidence interval would be given by this formula  

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$  

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$, with that value we can find the quantile required for the interval in the normal standard distribution.  

$z_{\alpha/2}=1.96$  

And for this case the confidence interval obtained is (0.10, 0.15)

So then we can conclude that this result shows a significant increase in the sleeping hours since the lower limit for the confidence interval is higher than 0 and at 5% of significance we can conclude that the new treatment increase the sleeping hours

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

$p$ represent the real population proportion of interest   "increase in sleep hours"

$\hat p$ represent the estimated proportion for the sample  

n=2000 is the sample size required  

$z$ represent the critical value for the margin of error  

Solution to the problem

The population proportion have the following distribution  

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$  

The confidence interval would be given by this formula  

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$  

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$, with that value we can find the quantile required for the interval in the normal standard distribution.  

$z_{\alpha/2}=1.96$  

And for this case the confidence interval obtained is (0.10, 0.15)

So then we can conclide that this result shows a significant increase in the sleeping hours since the lower limit for the confidence interval is higher than 0 and at 5% of significance we can conclude that the new treatment increase the sleeping hours