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sertanlavr [38]
4 years ago
6

What is the standard deviation of the following data set rounded to the nearest tenth 3,17,18,15,12,21,9

Mathematics
2 answers:
Anuta_ua [19.1K]4 years ago
6 0
Hey,
To find the Standard deviation of the data set, there are 4 steps:
<span>1. Work out the Mean (the simple average of the numbers)
2. Then for each number: subtract the Mean and square the result.
3. Then work out the mean of those squared differences.
4. T<span>ake the square root of that and we are done!

Here is your answer:
6.1

Is there a typo? I have checked my answer multiple times. Perhaps are you missing a number in the data set?

Cheers,
Izzy</span></span>
Alenkinab [10]4 years ago
4 0
I think that it is c or d but correct me if i am wrong
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Assume the general population gets an average of 7 hours of sleep per night. You randomly select 45 college students and survey
tekilochka [14]

Answer:

a) ii. This is a left-tailed test.

b) -1.59

c) -1.301

d) i. reject null hypothesis

e) Option i) The data supports the claim that college students get less sleep than the general population.

Step-by-step explanation:

We are given the following in the question:  

Population mean, μ = 7 hours

Sample mean, \bar{x} = 6.87 hours

Sample size, n = 45

Alpha, α = 0.10

Sample standard deviation, s =  0.55 hours

First, we design the null and the alternate hypothesis

H_{0}: \mu = 7\text{ hours}\\H_A: \mu < 7\text{ hours}

a) We use one-tailed(left) t test to perform this hypothesis.

b) Formula:

t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }

Putting all the values, we have

t_{stat} = \displaystyle\frac{6.87 - 7}{\frac{0.55}{\sqrt{45}} } =-1.59

c) Now,

t_{critical} \text{ at 0.10 level of significance, 44 degree of freedom } = -1.301

Since,                    

t_{stat} < t_{critical}

d) We fail to accept the null hypothesis and reject it.

We accept the alternate hypothesis and conclude that  mean number of hours of sleep for all college students is less than 7 hours.

e) Option i) The data supports the claim that college students get less sleep than the general population.

8 0
3 years ago
​ A sample of n = 9 college students is used to evaluate the effectiveness of a new Study Skills Workshop. Each student’s grade
forsale [732]

Answer:

B) μD = 0.60 ± 0.10(1.397)

Step-by-step explanation:

The confidence interval is given by:

MD±t_{\alpha/2, n-1} \frac{s}{\sqrt{n} }

Where

MD=60

n=9

df=n-1=8

t_{\alpha/2, n-1}=1.397

s=\sqrt{0.09} =0.3

Then the confidence interval is

μD=0.60±1.397*(0.3/√9)

μD=0.60±0.10*(1.397)

6 0
3 years ago
Plz help ill give you brainlist
Sloan [31]
The constant is 3 because it has no variable. 
The coefficient is 7b because it has a variable. 
7 0
4 years ago
Suppose that a manager is interested in estimating the average amount of money customers spend in her store. After sampling 36 t
musickatia [10]

Answer:

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

The 90% confidence interval for this case would be (38.01, 44.29) and is given.

The best interpretation for this case would be: We are 90% confident that the true average is between $ 38.01 and $ 44.29 .

And the best option would be:

The store manager is 90% confident that the average amount spent by all customers is between S38.01 and $44.29

Step-by-step explanation:

Assuming this complete question: Which statement gives a valid interpretation of the interval?

The store manager is 90% confident that the average amount spent by the 36 sampled customers is between S38.01 and $44.29.

There is a 90% chance that the mean amount spent by all customers is between S38.01 and $44.29.

There is a 90% chance that a randomly selected customer will spend between S38.01 and $44.29.

The store manager is 90% confident that the average amount spent by all customers is between S38.01 and $44.29

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Solution to the problem

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

The 90% confidence interval for this case would be (38.01, 44.29) and is given.

The best interpretation for this case would be: We are 90% confident that the true average is between $ 38.01 and $ 44.29 .

And the best option would be:

The store manager is 90% confident that the average amount spent by all customers is between S38.01 and $44.29

8 0
3 years ago
Can anyone help me with 16,18,and 19 please
jenyasd209 [6]
16-A 18-B 19-D (not 100% sure on 16)
7 0
3 years ago
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