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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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oksano4ka [1.4K]

Answer: There is a probability of 0.05 that there is neither truck is available.

Step-by-step explanation:

Since we have given that

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So, probability that either first truck or second truck is available is given by

P(A\cup B)=P(A)+P(B)-P(A\cap B)\\\\P(A\cup B)=0.75+0.50-0.30\\\\P(A\cup B)=0.95

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Rewrite in standard form
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