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

The confidence interval would be given by this formula

`\hat p \pm z_(\alpha/2) \sqrt{(\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

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{(p(1-p))/(n)})`

The confidence interval would be given by this formula

`\hat p \pm z_(\alpha/2) \sqrt{(\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