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Sagot :
Answer:
a. Mean
b. μ = μ₀
c. 11.5%
d. Yes
Step-by-step explanation:
The given parameters are;
The mean work week for engineers, μ₀ = 63 hours
The standard deviation, σ = 5 hours
The number of engineers in Kara's sample, n = 10 engineers
The responses given by the 10 engineers are;
70; 45; 55; 60; 65; 55; 55; 60; 50; 55
a. The given information which the newly hired is hoping to find out that it is not is the mean
b. The null hypothesis which is that the company claim is correct, is therefore;
Null hypothesis, H₀; μ = μ₀ = 63 hours
c. Kara's mean, [tex]\overline x[/tex] is found as follows;
[tex]\overline x[/tex] = (70 + 45 + 55 + 60 + 65 + 55 + 55 + 60 + 50 + 55)/10 = 57
[tex]\overline x[/tex] = 57 hours
The Z-score is therefore;
[tex]Z=\dfrac{\overline x-\mu }{\sigma }[/tex]
Z = (57 - 63)/5 = -1.2
From the Z-table, we have;
The p-value for P([tex]\overline x[/tex] ≤ 57) = P(Z ≤ -1.2) = 0.11507
The probability as a percentage given to one decimal place, that the mean would be as low as Kara's mean, P([tex]\overline x[/tex] ≤ 57) = 11.5%
d. Given that the percent percentage chance (11.5%) that the mean could be as low as hers ([tex]\overline x[/tex] = 57 hours) is higher than 10%, she should accept the claim.
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