Mathematics, 31.12.2019 03:31 highschoolboy

Astudy was conducted to measure the effectiveness of hypnotism in reducing pain. the measurements are centimeters on a pain scale before and after hypnosis. assume that the paired sample data are simple random samples and that the differences have a distribution that is approximately normal. construct aâ€‹ 95% confidence interval for the mean of theâ€‹"before minusâˆ’â€‹after"differences. does hypnotism appear to be effective in reducingâ€‹pain? before6.42.67.710.511.75.84.32.8aft er6.72.47.48.18.66.43.92.7

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Answer from: yen80

The 95% confidence interval for the difference would be given by (-1.776;0.376)

We are 95% confidence that the true mean difference is between $-1.776 \leq \mu_d \leq 0.376$ . Since the confidence interval contains the value 0, we don't have enough evidence to conclude that hypnotism appear to be effective in reducing pain.

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".

Let put some notation

x=test value before , y = test value after

x: 6.4 2.6 7.7 10.5 11.7 5.8 4.3 2.8

y: 6.7 2.4 7.4 8.1 8.6 6.4 3.9 2.7

The differences defined as d_i = y_i -x_i and we got:

d: 0.3, -0.2, -0.3, -2.4, -3.1, 0.6, -0.4, -0.1

We can calculate the mean and the deviation for the sample with the following formulas:

$\bar d=\frac{\sum_{i=1}^n d_i}{n}$

$s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1}$

$\bar d=-0.7$ represent the sample mean for the difference

$\mu_d$ population mean (variable of interest)

s_d =1.32 represent the sample standard deviation

n=8 represent the sample size

Confidence interval

The confidence interval for the mean is given by the following formula:

$\bar d \pm t_{\alpha/2} *\frac{s_d}{\sqrt{n}}$ (1)

In order to calculate the critical value t we need to find first the degrees of freedom, given by:

df=n-1=8-1=7

Since the Confidence is 0.95 or 95%, the value of alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,8)".And we see that $t_(\alpha/2)=2.306$

Now we have everything in order to replace into formula (1):

$-0.7-2.306\frac{1.32}{\sqrt{8}}=-1.776$

$-0.7+2.306\frac{1.32}{\sqrt{8}}=0.376$

So on this case the 95% confidence interval for the difference would be given by (-1.776;0.376)

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Answer from: Quest

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