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A professor at a university wants to estimate the average number of hours of sleep the students get during exam week. The professor wants to find a sample mean that is within 4.266 hours of the true average for all college students at the university with 99% conconfidence. If the professor knows the standard deviation of the sleep times for all college students is 2.539, what sample size will need to be taken

Answers

Answer 1

Answer:

The sample size is $n = 2$

Step-by-step explanation:

From the question we are told that

The margin of error is E =4.266

The standard deviation is $\sigma = 2.539$

From the question we are told the confidence level is 99% , hence the level of significance is

$\alpha = (100 - 99 ) \%$

=> $\alpha = 0.01$

Generally from the normal distribution table the critical value of $(\alpha )/(2)$ is

$Z_{(\alpha )/(2) } = 2.58$

Generally the sample size is mathematically represented as

$n = [\frac{Z_{(\alpha )/(2) } * \sigma }{E} ] ^2$

=> $n = [(2.58 * 2.539 )/(4.266) ] ^2$

=> $n = 2$

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Answers

I think it is 2 hope this helped you

Answer:

1. 6 pi

2. 7 pi

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Step-by-step explanation: Just took the USATestprep test and got them all correct, sorry I dont have a proper answer on how I got them.

Please help it’s due at 5

Answers

Answer:

nei prossimi giorni, che si è svolto il convegno

A company’s total cost from manufacturing and selling x units of their product is given by: y = 2x2 – 600x + 49,000. How many units should be manufactured in order to minimize cost, and what is the minimum cost?a. (100, $9,000)b. (125, $3,250)c. (150, $4,000)d. (170, $4,800)e. (200, $9,000)