Sam makes $15 for every lawn that he mows. He

babunello [35]

44x10^-2
or in other words
44 times 10 to the -2nd power

3 0

3 years ago

A study was conducted to measure the effectiveness of hypnotism in reducing pain. The measurements are centimeters on a pain sca

saul85 [17]

Answer:

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)

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.

6 0

3 years ago

Use the listing method to represent the following set.

Dafna11 [192]

The answer is c bc x is 3 and more

7 0

3 years ago

Read 2 more answers

A randomized trial tested the effectiveness of diets on adults. Among 36 subjects using Diet 1, the mean weight loss after a yea

seropon [69]

Answer:

The 95% confidence interval is

$0.45 < \mu_1 - \mu_2 < 5.35$

Step-by-step explanation:

From the question we are told that

The first sample size is $n_1 = 36$

The first sample mean is $\= x_1 = 3.5$

The first standard deviation is $\sigma_1 = 5.9 \ pounds$

The second sample size is $n_2 = 36$

The second sample mean is $\= x_2 = 0.6$

The second standard deviation is $\sigma = 4.4$

Generally the degree of freedom is mathematically represented as

$df = \frac{ [ \frac{s_1^2 }{n_1 } + \frac{s_2^2 }{n_2} ]^2 }{ \frac{1}{(n_1 - 1 )} [ \frac{s_1^2}{n_1} ]^2 + \frac{1}{(n_2 - 1 )} [ \frac{s_2^2}{n_2} ]^2 }$

=> $df = \frac{ [ \frac{5.9^2 }{34 } + \frac{4.4^2 }{34} ]^2 }{ \frac{1}{(34 - 1 )} [ \frac{5.9^2}{34} ]^2 + \frac{1}{(34- 1 )} [ \frac{4.4^2}{ 34} ]^2 }$

=> $df =63$

Generally the standard error is mathematically represented as

$SE = \sqrt{ \frac{s_1 ^2 }{n_1} + \frac{s_2^2 }{ n_2 } }$

=> $SE = \sqrt{ \frac{ 5.9 ^2 }{ 36 } + \frac{ 4.4^2 }{36} }$

=> $SE = 1.227$

From the question we are told the confidence level is 95% , hence the level of significance is

$\alpha = (100 - 95 ) \%$

=> $\alpha = 0.05$

Generally from the t distribution table the critical value of at a degree of freedom of is

$t_{\frac{\alpha }{2}, 63 } = 1.998$

Generally the margin of error is mathematically represented as

$E = t_{\frac{\alpha }{2}, 63 } * SE$

=> $E = 1.998 * 1.227$

=> $E = 2.45$

Generally 95% confidence interval is mathematically represented as

$(\= x_1 - \x_2) -E < \mu$

=> $(3.5 - 0.6) - 2.45 < \mu_1 - \mu_2 < ( 3.5 - 0.6) + 2.45$

=> $0.45 < \mu_1 - \mu_2 < 5.35$

8 0

2 years ago

What is the median ?

mel-nik [20]

The median value of a range of values.

denoting or relating to a value or quantity lying at the midpoint of a frequency distribution of observed values or quantities, such that there is an equal probability of falling above or below it.

a straight line drawn from any vertex of a triangle to the middle of the opposite side.

7 0

3 years ago