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4 years ago

The table below summarizes baseline characteristics on patients participating in a clinical trial

1 answer:

8 0

The best and most correct answer among the choices provided by the question is the fifth choice which is both A and B.

Age and total cholesterol are baseline characteristics are significantly different (at the 0.05 level of significance) between treatment groups.
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Paired t-test Application (Part 1)An environmental engineer is tasked with determining whether a power plant cooling system is h

qaws [65]

Complete Question

The complete question is shown on the first and second uploaded image

Answer:

Part 1

The correct option is B

Part 11

The correct option is H

Step-by-step explanation:

From the question we are told that

The sample size is n = 15

Generally the sample mean for the input temperature is mathematically represented as

$\= x _1 = \frac{\sum x_i}{n}$

$\= x _1 = \frac{57.6 + 68.9 \cdots +60.4 }{15}$

$\= x _1 = 62.57$

Generally the sample mean for the output temperature is mathematically represented as

$\= x _2 = \frac{\sum x_i}{n}$

$\= x _2 = \frac{65.1 + 74.4 \cdots +67.3 }{15}$

$\= x _2 = 55.97$

Generally the difference between the mean of the input temperature and that of the output temperature is

$d = \= x_1 - \= x_2$

=> $d = 62.57 - 55.97$

=> $d = 6.6$

Generally the standard deviation of the difference between the input temperature and the output temperature is mathematically represented as

$s_d = \sqrt {\frac{1}{n-1 } \sum [d_i - d]^2}$

=> $s_d = \sqrt{\frac{[(57.6 - 65.1) - 6.6]^2+[(68.9 - 74.4) - 6.6]^2+ \cdots +[(68.1 - 74.7) - 6.6]^2 }{15-1} }$

=> $s_d = 1.732$

The null hypothesis is $H_o : \mu_1 -\mu_2 = 6$

The alternative hypothesis is $H_a : \mu _1 - \mu_2\ne 6$

Generally the test statistics is mathematically represented as

$t = \frac{d - 6}{ \frac{s_d}{ \sqrt{n} } }$

$t = \frac{6.6- 6}{ \frac{1.732}{ \sqrt{15} } }$

$t = 1.342$

Generally the p-value is mathematically represented as

$p-value = 2 * P(t > 1.342)$

From the student t distribution table( reference - danielsoper(dot)com(slash)statcalc(slash)calculator) at a degree of freedom of df = n-1 = 15-1 = 14

$P(t > 1.342) = t_{1.342 , 14} = 0.100478$

$p-value = 2 * 0.100478)$

$p-value = 0.201$

From the values obtained we see that the $p-value > \alpha$ hence the decision rule is fail to reject the null hypothesis

The conclusion is

The cooling system changes the temperature of the water by 6 degrees.

5 0

4 years ago

Birth weights at a local hospital have a Normal distribution with a mean of 110 oz and a standard deviation of 15 oz. The propor

monitta

Answer:

c. 0.136.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 110, \sigma = 15$