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lawyer [7]
3 years ago
7

A university claims that the mean time professors are in their offices for students is at least 6.5 hours each week. A random sa

mple of eight professors finds that the mean time in their offices is 6.2 hours each week. With a sample standard deviation of 0.49 hours from a normally distributed data set, can the university’s claim be supported at α=0.05?
Mathematics
1 answer:
kolezko [41]3 years ago
3 0
<h2>Answer with explanation:</h2>

Let \mu be the population mean.

Null hypothesis : H_0: \mu\geq 6.5

Alternative hypothesis : H_a: \mu

Since alternative hypothesis is left-tailed , so the test is a left-tailed test.

Since n= 8 <30 , so we use t-test.

Test statistic for population mean : t=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}

t=\dfrac{6.2-6.5}{\dfrac{0.49}{\sqrt{8}}}\approx-1.73

Critical value for t=t_{n-1, \alpha}=t_{7,0.05}=1.895

Since the absolute t-value (1.73) is less than the critical t-value(1.895), it means we are fail to reject the null hypothesis.

Thus , we conclude that we have enough evidence to support the university’s claim.

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Step-by-step explanation:

1. we need an equation first. the sum of all angles (108°, n°, 2n°) is equal to 180°. we can depict this with the equation: 108°+2n°+n°=180°

2. now we can solve for the missing variable, n.

108°+3n°=180° → subtract both sides by → 3n°=72° → divide both sides by 3 → n=24°

3. now that we know that n=24°, we can solve the value of ∠DOB. we can see that ∠DOB is 2n° which we just plug the number we got for n into the equation. 2*24=48° meaning ∠DOB is 48°

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