PLEASE HELP ME !! I HAVE TO PASS THIS TEST !!

gayaneshka [121]

See attachment file below.
Answer: 11/3

Hope it helped!

6 0

4 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

4 years ago

A teacher buys packs of notebooks to give to his students. The ratio of gold notebooks to red notebooks is represented in the ta

Naddika [18.5K]

Answer:

there would be 40 gold notebooks

Step-by-step explanation:

if the ratio is 5g to 3r i took the 24 red and divided it by 3 which gave me 8. I then multiplied 8 by 5 and got 40. so 40:24 is equal to 5:3

4 0

3 years ago

5. 18 m 11 m Not drawn to scale 29 m b. 445 m 7 m d. 203 m

Luden [163]

Answer: D

Explanation: Apply the pythagorean theorem (a^2 + b^2 = c^2) to get 11^2 + b^2 = 18^2. From that you can solve for b and you get the square root of 203.

Hope this helped! :)

8 0

3 years ago

Read 2 more answers

A circle has its center at (6,3). One point on the circle is (3, 1). Which is

Ilia_Sergeevich [38]

Answer: i believe it is D

Step-by-step explanation:

:)

8 0

3 years ago

Read 2 more answers