Question

Using a sequencing method called "RNAseq," a researcher determines the level of FOXP2 expression

Answer 1

The Z-score is 2.306 to three decimal places. However, the Z-test statistic value shows that it is greater than the Z-critical value, so we reject the null hypothesis $\mathbf{H_o}$

What is the Z-score of a test statistic?

A z-score is a measurement of how often standard deviations the score obtained from a z-test is beyond or under the mean population.

From the given data information, let us find the mean and the standard deviation.

$\mathbf{Mean (\bar x) = \dfrac{1154+5998+1007+...3987+4187+887}{10}}$

$\mathbf{Mean (\bar x) = \dfrac{24256}{10}}$

Mean (x) = 2425.6

The hypothesis testing can be computed as:

$\mathbf{H_o: \mu = 2425.6}$

$\mathbf{H_1: \mu \ne 2425.6}$

The standard deviation (σ) is:

$\mathbf{=\mathbf{ \sqrt{\dfrac{(1154-2425.6)^2+....+(887-2425.6)^2}{10-1}}}}$

$\mathbf{= \sqrt{3199489}}$

= 1788.71

Now, the Z-test statistic is determined using the formula:

$\mathbf{|Z| = |\dfrac{\bar x - \mu }{\sigma}| }$

$\mathbf{|Z| = |\dfrac{2425.6 -6550 }{1788.71}| }$

$\mathbf{|Z| = |-2.3058| }$

Z ≅ 2.306  

Therefore, at a 0.05 level of significance, the Z-critical value is 2.306. However, since the Z-test statistic value shows that it is greater than the Z-critical value, so we reject the null hypothesis $\mathbf{H_o}$

The complete question is given as:

Using a sequencing method called "RNAseq," a researcher determines the level of FOXP2 expression in 10 different samples of lung tissue, and obtains these data:

• 1154 5998 1007 2501 1000 988 2547 3987 4187 887

Scientist has previously determined that FOXP2 is expressed in brain tissue at a level of 6,550. Calculate the Z-score associated with a test to see if the average expression of FOXP in her lung tissue samples is different from the levels she measured from brain tissue.

Learn more about calculating the Z-score of a test statistics here:

brainly.com/question/15569214

#SPJ1