Elective Class Percentage of Students Wanting to Take the Class

Quantitative noninvasive techniques are needed for routinely assessing symptoms of peripheral neuropathies, such as carpal tunne

Pavlova-9 [17]

Answer:

$t=\frac{2.35-1.83}{\sqrt{\frac{0.88^2}{10}+\frac{0.54^2}{7}}}}=1.507$

$p_v =P(t_{(15)}>1.507)=0.076$

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and the group CTS, NOT have a significant higher mean compared to the Normal group at 1% of significance.

Step-by-step explanation:

1) Data given and notation

$\bar X_{CTS}=2.35$ represent the mean for the sample CTS

$\bar X_{N}=1.83$ represent the mean for the sample Normal

$s_{CTS}=0.88$ represent the sample standard deviation for the sample of CTS

$s_{N}=0.54$ represent the sample standard deviation for the sample of Normal

$n_{CTS}=10$ sample size selected for the CTS

$n_{N}=7$ sample size selected for the Normal

$\alpha=0.01$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean for the group CTS is higher than the mean for the Normal, the system of hypothesis would be:

Null hypothesis: $\mu_{CTS} \leq \mu_{N}$

Alternative hypothesis: $\mu_{CTS} > \mu_{N}$

If we analyze the size for the samples both are less than 30 so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{\bar X_{CTS}-\bar X_{N}}{\sqrt{\frac{s^2_{CTS}}{n_{CTS}}+\frac{s^2_{N}}{n_{N}}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{2.35-1.83}{\sqrt{\frac{0.88^2}{10}+\frac{0.54^2}{7}}}}=1.507$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n_{CTS}+n_{N}-2=10+7-2=15$

Since is a one side right tailed test the p value would be:

$p_v =P(t_{(15)}>1.507)=0.076$

We can use the following excel code to calculate the p value in Excel:"=1-T.DIST(1.507,15,TRUE)"

Conclusion

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and the group CTS, NOT have a significant higher mean compared to the Normal group at 1% of significance.

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4 years ago

Does Anybody Know The Answer?

Taya2010 [7]

Yes. When a graph has more than one line it is a piece wise graph.

5 0

3 years ago

Mr. Taylor raised all his students’ scores on a recent test by five

Darina [25.2K]

2. The mean is the average so it phase shifted up five points where as the range is the difference between the highest and lowest # which wouldn't change if all the numbers were adjusted.

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-4(-3x - 7) whats the answer

kiruha [24]

12x+28 is the simplified expression

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Read 2 more answers

What is the probability that a randomly selected member of a normally distributed population will lie more than 1.8 standard dev

Helen [10]

Answer:

$P(X> \mu +1.8\sigma)=P(\frac{X-\mu}{\sigma}>\frac{\mu +1.8\sigma-\mu}{\sigma})=P(Z>1.8)$

And we can find this probability using the complement rule:

$P(z>1.8)=1-P(z$

And using the normal standard distirbution table or excel we got:

$P(z>1.8)=1-P(z$

Step-by-step explanation:

Previous concepts

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".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the variable if interest of a population, and for this case we know the distribution for X is given by:

$X \sim N(\mu,\sigma)$

We are interested on this probability

$P(X>\mu +1.8 \sigma)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X> \mu +1.8\sigma)=P(\frac{X-\mu}{\sigma}>\frac{\mu +1.8\sigma-\mu}{\sigma})=P(Z>1.8)$

And we can find this probability using the complement rule:

$P(z>1.8)=1-P(z$

And using the normal standard distirbution table or excel we got:

$P(z>1.8)=1-P(z$

5 0

4 years ago