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Vinil7 [7]
3 years ago
9

A student group claims that first-year students at a university must study 2.5 hours (150 minutes) per night during the school w

eek. A skeptic suspects that they study less than that on the average. A class survey finds that the average study time claimed by 272 students is xbar= 141 minutes, with a sample standard deviation of 66.
Regard these students as a random sample of all first-year students and suppose we know that the study times follow a Normal distribution.

What is the standard error of the sample mean?

Consider the hypotheses H0: µ = 150 against Ha: µ < 150.

What is the value of the test statistic? Give your answer to two decimal places.

The p-value for the correct test statistic is 0.0126. What do you conclude at the a = 0.05 level?

We do not have enough evidence to say that students are not studying 2.5 hours per night during the school week.
We have strong evidence that, on average, students study less than 2.5 hours per night during the school week.
Mathematics
1 answer:
AlekseyPX3 years ago
7 0

Answer:

We have strong evidence that on average, students study less than 150 minutes per night during the school week

Step-by-step explanation:

Normal distribution:

mean     μ₀ = 150

Sample:

Sample size    n = 272

Sample mean   x = 141

Sample standard deviation  s  = 66

The standard error of the sample mean  SE = σ /√n

SE = 66/√272

SE = 66 / 16,49

SE = 4

Test Hypothesis:

Null hypothesis                            H₀             x  =   μ₀

Alternative hypothesis                Hₐ            x <    μ₀

z(s)  test statistics is:

z(s)  =  ( x  -  μ₀ ) / s/√n

z(s) = - 9 /4

z(s) =  -  2,25

p-value  for that z(s)      p-value  = 0,0122

Then for α =  0,05 p-value < 0,05

We are in the rejection region we need to reject H₀

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Rockwell hardness of pins of a certain type is known to have a mean value of 50 and a standard deviation of 1.3. (Round your ans
Maurinko [17]

Answer:

0.0039 is the probability that the sample mean hardness for a random sample of 12 pins is at least 51.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 50

Standard Deviation, σ = 1.3

Sample size, n = 12

We are given that the distribution of hardness of pins is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

Standard error due to sampling =

=\dfrac{\sigma}{\sqrt{n}} = \dfrac{1.3}{\sqrt{12}} = 0.3753

P(sample mean hardness for a random sample of 12 pins is at least 51)

P( x \geq 51) = P( z \geq \displaystyle\frac{51 - 50}{0.3753}) = P(z \geq 2.6645)

= 1 - P(z < 2.6645)

Calculation the value from standard normal z table, we have,  

P(x \geq 51) = 1 - 0.9961= 0.0039

0.0039 is the probability that the sample mean hardness for a random sample of 12 pins is at least 51.

3 0
3 years ago
Gus is making a Chilli recipe that calls 3 parts beans to five parts ground beef .if he is using 8 cups of ground beef for a big
masya89 [10]
First you need to find the percent equivalent to that of the ratio. The ratio is 3:5. From there you would take the 3, and divide it by 5. 3/5=0.6. Apply that to the information that you are given: 8(0.6)=4.8.
4 0
3 years ago
Four tickets to the baseball game and 4 meal tickets cost $144.
Doss [256]

Answer: $108

Step-by-step explanation:

what is 144 / 4 ? 36 = one meal ticket

what is 36 x 3 ? 108

each game ticket costs 108 dollars

6 0
2 years ago
By what number we multiply -7/16 so that the product may be 46/28
Aliun [14]

Answer:

The eq will be=-7/16 × x=46/28

x=46/28 ×-16/7

x=-192/49

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3 years ago
A bike and skate shop rents bike for $21 per day and pairs of skates for $20 per day. To remain viable, the shop needs to make a
Leya [2.2K]
Let the number of bike be x and the number of skates be y, then
21x + 20y ≥ 362 . . . (1)
2y = x . . . (2)

Putting (2) into (1), then
21(2y) + 20y ≥ 362
42y + 20y ≥ 362
62y ≥ 362
y ≥ 5.84

The least number of pairs of skates they need to rent each day to make their minimum is 6.
7 0
3 years ago
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