Answer:
<em>Test statistic </em>
<em> </em>
t = <em>1.076</em>
Step-by-step explanation:
<u><em>Step(i):-</em></u>
Given Mean of the Population (μ) = 8.0
<em>Mean of the sample (x⁻) = 8.25</em>
Given data
8,9,9,8,8,9,8,7
Given sample size n= 8
Given sample standard deviation(S) = 0.661
<u><em>Step(ii):-</em></u>
<em>Null hypothesis : H: (μ) = 8.0</em>
<em>Alternative Hypothesis :H:(μ) > 8.0</em>
<em>Degrees of freedom = n-1 = 8-1=7</em>
<em>Test statistic </em>
<em> </em>
<em></em>
<em> </em>
<em></em>
<em> t = 1.076</em>
<em>Critical value </em>
<em> t₍₇,₀.₀₅₎ = 2.3646</em>
<em>The calculated value t = 1.076 < 2.3646 at 0.05 level of significance</em>
<em>Null hypothesis is accepted</em>
<em>Test the hypothesis that the true mean quiz score is 8.0 against the alternative that it is not greater than 8.0</em>
<em></em>
Answer:
the answer is c
Step-by-step explanation:
Ignore the scribbles
Search up Elimination and substitution
I believe the answer is A, the standard deviation is preferable to the range as a measure of variation because the standard deviation takes into account all of the observations, whereas the range considers only the largest and the smallest. Range gives an overall spread of data from the lowest to the largest and thus can be influenced by anomalies, standard deviation on the other hand, takes into account the variable data/spread about the mean and allows for statistical use so inferences can be made.<span />