Can anyone help me with this

A Type I error is committed when a. you don't reject a null hypothesis that is true. b. you don't reject a null hypothesis that

e-lub [12.9K]

Answer:

c. you reject a null hypothesis that is true

Step-by-step explanation:

We need to remember the following concepts

Error type I: Is an error associated to the probability of reject a null hypothesis when it is actually true

Error type II: Is an error associated of not rejecting a null hypothesis when the alternative hypothesis is the true

And the best answer for this case would be:

c. you reject a null hypothesis that is true

4 0

2 years ago

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.

6 0

3 years ago

Part B:

Andrew [12]

Articles of clothing that the store received last Thursday is 84

<u>Solution:</u>

Given that

A clothing donation store received numerous articles of clothing on Thursday of last week.

The number of articles of clothing that the store received last Thursday is represented by the following expression:

4 (3x − 1) − 2 (6 + 2x)

Need to calculate articles of clothing that the store received last Thursday when x = 12.5

As on Thursday x = 12.5, need to substitute x = 12.5 in above equation to get number of articles received on Thursday.

On substituting x = 12.5, we get

4 ( (3 x 12.5 ) – 1 ) – 2 ( 6 + ( 2 x 12.5 ) )

=> 4 (37.5 – 1) – 2 (6 + 25)

=> 146 – 62 = 84

On substituting x = 12.5 , we are getting value of expression as 84.

Hence articles of clothing that the store received last Thursday is 84

3 0

2 years ago

Three of four numbers have a sum of 22. If the average of the four numbers is 8, what is the fourth number?

lesantik [10]

The number is 15 wnwnjw

6 0

3 years ago

Best MEME GETS Brianlist

Sav [38]

It made me laugh i like it

8 0

2 years ago

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