Mathematics, 31.03.2020 03:24 dirtridersteve65

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.

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Answer from: jwbri

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".

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

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

Answer from: Quest

Google it that might work

Answer from: Quest

well

44-3-3-3=35

35+4+4+4+4=51

so the ansewr would be 51 degreesat 7: 00

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