A textbook company states that the average time a student needs to take a quiz from its book is 30 minutes with a standard devia
2 answers:
Answer:
We conclude that her students, on average, do not take 30 minutes on the quiz, contrary to what the textbook company states.
Step-by-step explanation:
We are given;
Population mean; μ = 30 minutes
Population standard deviation; σ = 3 minutes
Sample size; n = 10
Sample mean; x¯ = 25
Significance level = 5% = 0.05
Let's define the hypotheses;
Null hypothesis; H0: μ = 30
Alternative hypothesis: Ha: μ ≠ 30
Let's find the test statistic;
z = (x¯ - μ)/(σ/√n)
z = (25 - 30)/(3/√10)
z = -5.27
From online p-value from z-score calculator attached using, z = -5.27, significance level = 0.05, two tailed hypothesis, we have;
p < 0.00001
This is less than the significance level, and so we will reject the null hypothesis and conclude that Her students, on average, do not take 30 minutes on the quiz, contrary to what the textbook company stated.
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The problem can be solved step by step, if we know certain basic rules of summation. Following rules assume summation limits are identical.
Armed with the above rules, we can split up the summation into simple terms:
=> (a) f(x)=28n-n^2
=> f'(x)=28-2n
=> at f'(x)=0 => x=14
Since f''(x)=-2 <0 therefore f(14) is a maximum
(b) f(x) is a maximum when n=14
(c) the maximum value of f(x) is f(14)=196
Armed with the above rules, we can split up the summation into simple terms:
=> (a) f(x)=28n-n^2
=> f'(x)=28-2n
=> at f'(x)=0 => x=14
Since f''(x)=-2 <0 therefore f(14) is a maximum
(b) f(x) is a maximum when n=14
(c) the maximum value of f(x) is f(14)=196