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Slav-nsk [51]
3 years ago
15

A random sample of computer startup times has a sample mean of x¯=37.2 seconds, with a sample standard deviation of s=6.2 second

s. Since computer startup times are generally symmetric and bell-shaped, we can apply the Empirical Rule. Between what two times are approximately 95% of the data? Round your answer to the nearest tenth
Mathematics
1 answer:
juin [17]3 years ago
3 0

Answer:

95% of the data lies between 24.8 and 49.6

Step-by-step explanation:

* Lets revise the empirical rule

- The Empirical Rule states that almost all data lies within 3

  standard deviations of the mean for a normal distribution.  

- 68% of the data falls within one standard deviation.  

- 95% of the data lies within two standard deviations.  

- 99.7% of the data lies Within three standard deviations  

- The empirical rule shows that

# 68% falls within the first standard deviation (µ ± σ)

# 95% within the first two standard deviations (µ ± 2σ)

# 99.7% within the first three standard deviations (µ ± 3σ).

* Lets solve the problem

- A random sample of computer startup times has a sample mean of

   μ = 37.2 seconds

∴ μ = 37.2

- With a sample standard deviation of σ = 6.2 seconds

∴ σ = 6.2

- We need to find between what two times are approximately 95%

  of the data

∵ 95% of the data lies within two standard deviations

∵ Two standard deviations (µ ± 2σ) are:

∵ (37.2 - 2 × 6.2) = 24.8

∵ (37.2 + 2 × 6.2) = 49.6

∴ 95% of the data lies between 24.8 and 49.6

* <em>95% of the data lies between 24.8 and 49.6</em>

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Answer:

So on this case the 95% confidence interval would be given by (-1.152;5.152).

-1.152 < \mu_{safe1}-\mu_{safe2}  

Should we conclude that the average time it takes experts to crack the safes does not differ by model at the 5% significance level?

A) Yes, the 95% confidence interval for ?D contains 0.

Step-by-step explanation:

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

Let put some notation  

x=value for Safe 1 , y = value for Safe 2

x: 103, 90, 64, 120, 104, 92, 145, 106, 76

y: 101, 94, 58, 112, 103, 90, 140, 110, 74

The first step is calculate the difference d_i=x_i-y_i and we obtain this:

d: 2, -4, 6, 8, 1, 2, 5, -4, 2

The second step is calculate the mean difference  

\bar d= \frac{\sum_{i=1}^n d_i}{n}= \frac{18}{9}=2

The third step would be calculate the standard deviation for the differences, and we got:

s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =4.093

The next step is calculate the degrees of freedom given by:

df=n-1=9-1=8

Now we need to calculate the critical value on the t distribution with 8 degrees of freedom. The value of \alpha=1-0.95=0.05 and \alpha/2=0.025, so we need a quantile that accumulates on each tail of the t distribution 0.025 of the area.

We can use the following excel code to find it:"=T.INV(0.025;8)" or "=T.INV(1-0.025;8)". And we got t_{\alpha/2}=\pm 2.31

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

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

Now we have everything in order to replace into formula (1):  

2-2.31\frac{4.093}{\sqrt{9}}=-1.152  

2+2.31\frac{4.093}{\sqrt{9}}=5.152  

So on this case the 95% confidence interval would be given by (-1.152;5.152).

-1.152 < \mu_{safe1}-\mu_{safe2}  

Should we conclude that the average time it takes experts to crack the safes does not differ by model at the 5% significance level?

A) Yes, the 95% confidence interval for D contains 0.

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3 years ago
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