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 <em>mean</em>
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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,
is found as follows;
= (70 + 45 + 55 + 60 + 65 + 55 + 55 + 60 + 50 + 55)/10 = 57
= 57 hours
The Z-score is therefore;

Z = (57 - 63)/5 = -1.2
From the Z-table, we have;
The p-value for P(
≤ 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(
≤ 57) = 11.5%
d. Given that the percent percentage chance (11.5%) that the mean could be as low as hers (
= 57 hours) is higher than 10%, she should accept the claim.