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 [tex]\mathbf{H_o}[/tex]
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.
[tex]\mathbf{Mean (\bar x) = \dfrac{1154+5998+1007+...3987+4187+887}{10}}[/tex]
[tex]\mathbf{Mean (\bar x) = \dfrac{24256}{10}}[/tex]
Mean (x) = 2425.6
The hypothesis testing can be computed as:
[tex]\mathbf{H_o: \mu = 2425.6}[/tex]
[tex]\mathbf{H_1: \mu \ne 2425.6}[/tex]
The standard deviation (σ) is:
[tex]\mathbf{=\mathbf{ \sqrt{\dfrac{(1154-2425.6)^2+....+(887-2425.6)^2}{10-1}}}}[/tex]
[tex]\mathbf{= \sqrt{3199489}}[/tex]
= 1788.71
Now, the Z-test statistic is determined using the formula:
[tex]\mathbf{|Z| = |\dfrac{\bar x - \mu }{\sigma}| }[/tex]
[tex]\mathbf{|Z| = |\dfrac{2425.6 -6550 }{1788.71}| }[/tex]
[tex]\mathbf{|Z| = |-2.3058| }[/tex]
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 [tex]\mathbf{H_o}[/tex]
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:
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:
https://brainly.com/question/15569214
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