Question

Sleep. In $1908$W. S. Gosset published the article "The Probable Error of a Mean" (Biometrika, Vol, 6, pp. $1-25$). In this pioneering paper, written under the pseudonym " "Student," Gosset introduced what later became known as Student's $t-$distribution. Gosset used the following data set, which gives the additional sleep in hours obtained by a sample of $10$patients using laevohysocyamine hydrobromide.

a. Obtain and interpret a $95\%$confidence interval for the additional sleep that would be obtained on average for all people using laevohysocyamine hydrobromide. (Note: $\tilde{\mathrm{x}}=2.33\mathrm{hr};s=2.002\mathrm{hr}$)

b. Was the drug effective in increasing sleep? Explain your answer.

Short answer

Part (a) We are $95\%$certain that the mean additional hours of sleep for all patients using laevohysocymine hydrobromide is between $0.90$and $3.76$hours.

Part (b) we are at least $95\%$confident that an increase in mean hours of sleep is beneficial. As a result, we can conclude that the medicine was successful in improving sleep.

Step-by-step solution

Part (a) Step 1: Given information

1.9	0.8	1.1	0.1	-0.1
4.4	5.5	1.6	4.6	3.4

Part (a) Step 2: Concept

The formula used: The confidence interval$\left(\overline{x}-t_{\frac{a}{2}}\times \frac{s}{\sqrt{n}},\overline{x}+t_{\frac{a}{2}}\times \frac{s}{\sqrt{n}}\right)$

Part (a) Step 3: Calculation

Let $\mu$population s.d. be the population mean additional sleep hours, and $\sigma$be unknown.

Here the sample size$n=10$

We will utilize the $t-$interval approach to find $95\%\mathrm{Cl}$of $\mu$as population s.d., $\sigma$is unknown.

According to one mean t-interval technique, the percent $\mathrm{Cl}$of $\mu$is $100(1-\alpha )\%$

$\left(\overline{x}-t_{\frac{a}{2}}\times \frac{s}{\sqrt{n}},\overline{x}+t_{\frac{a}{2}}\times \frac{s}{\sqrt{n}}\right)$

Where $t_{\frac{\alpha }{2}}$is the $t-$value with area $\frac{\alpha }{2}$to its right, $\frac{\alpha }{2}$is the sample size, and$n$ is the sample size.

Here confidence level $=95\%$

$$\begin{gathered} =100\times 0.95\% \\ \therefore 1-\alpha =0.95 \\ \Rightarrow \alpha =1-0.95 \\ \Rightarrow \alpha =0.05 \\ \Rightarrow \frac{\alpha }{2}=0.025 \\ df=n-1 \\ =10-1=9 \\ \therefore \text{For}df=9,t_{\alpha }=t_{0025}=2.262 \end{gathered}$$

Part (a) Step 4: Calculation

Given that sample mean $\overline{x}=2.33$

Sample S.D $s=2.002$

$$\begin{gathered} \therefore 95\%\mathrm{Cl}\text{of}\mu \\ =\left(\overline{x}-t_{0.025}\frac{s}{\sqrt{n}},\overline{x}+t_{0.025}\frac{s}{\sqrt{n}}\right) \\ =\left(2.33-2.262\times \frac{2.002}{\sqrt{10}},2.33+2.262\times \frac{2.002}{\sqrt{10}}\right) \\ =(2.33-1.43,2.33+1.43) \\ =(0.8980,3.7620) \end{gathered}$$

i.e., we are $95\%$certain that the mean additional hours of sleep for all patients using laevohysocymine hydrobromide is between $0.90$ and $3.76$ hours.

Part (b) Step 1: Explanation

We discovered that the mean hour of increased sleep for all patients using laevohysocymine bromide lies between two positive numbers $0.90$ hrs and $3.76$ hrs based on the $95\%$ confidence interval. As a result, we are at least $95\%$ confident that an increase in mean hours of sleep is beneficial. As a result, we can conclude that the medicine was successful in improving sleep.