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2 years ago

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A researcher wants to estimate the mean commute time of residents in Westchester County. The researcher samples 40 commuters and

find the mean commute time to be 25 minutes. Assume the population standard deviation for commute time is 12.5 minutes. Construct a 90% confidence interval for the mean commute time of residents in Westchester County. What is the Lower Bound of the confidence interval

1 answer:

Ksenya-84 [330]2 years ago

4 0

Answer:

21.759

Step-by-step explanation:

Given that :

Mean (m) = 25

Standard deviation (s) = 12.5

Sample size (n) = 40

α = 90%

The confidence interval is obtained using the relation:

Mean ± Zcritical * s/sqrt(n)

Zcritical at 90% confidence interval = 1.64

25 ± 1.64 * (12.5/sqrt(40))

Lower boundary : 25 - 1.64(1.9764235) = 21.75866546

Upper boundary : 25 + 1.64(1.9764235) = 28.24133454

(21.759, 28.241)

Hence, lower bound of confidence interval is : 21.759

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A tire manufacturer is considering a newly designed tread pattern for its all-weather tires. Tests have indicated that these tir

Nataly_w [17]

Answer:

With outlier (102)

$t=\frac{126.2-125}{\frac{9.138}{\sqrt{10}}}=0.415$

$p_v =P(t_{9}>0.415)=0.344$

If we compare the p value and a significance level assumed $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean is not significantly higher than 125 feet at 5% of significance.

Without outlier (102)

$t=\frac{128.89-125}{\frac{3.551}{\sqrt{9}}}=3.285$

$p_v =P(t_{8}>3.285)=0.0056$

And we conclude that we reject the null hypothesis since $p_v$ . So the final conclusion would be not use the method since the value of 102 observed can be a potential outlier, removing this value we see that we reject the null hypothesis and we have a significant result that the true mean is higher than 125.

Step-by-step explanation:

Previous concepts and data given

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".

Data: 129, 128, 130, 132, 135, 123, 102, 125, 128, 130

We can calculate the mean with the following formulas:

$\bar X =\frac{\sum_{i=1}^n X_i}{n}$

$s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

$\bar X=126.2$ represent the sample mean

$s=9.138$ represent the sample standard deviation

n=10 represent the sample selected

$\alpha$ significance level

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is significantly higher than 125, the system of hypothesis would be:

Null hypothesis: $\mu \leq 125$

Alternative hypothesis: $\mu > 125$

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{126.2-125}{\frac{9.138}{\sqrt{10}}}=0.415$

P-value

First we need to calculate the degrees of freedom given by:

$df=n-1=10-1= 9$

Then since is a right tailed sided test the p value would be:

$p_v =P(t_{9}>0.415)=0.344$

Conclusion

If we compare the p value and a significance level assumed $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean is not significantly higher than 125 feet at 5% of significance.

Using all the data we see that we don;t have enough info to conclude that the true mean is higher than 125, but if we see careful 9 of the 10 values are over the limit of 125 feet. And if we repeat the procedure with the outlier of 102. We got this:

$\bar X=128.89$ represent the sample mean

$s=3.551$ represent the sample standard deviation

$t=\frac{128.89-125}{\frac{3.551}{\sqrt{9}}}=3.285$

First we need to calculate the degrees of freedom given by:

$df=n-1=9-1= 8$

Then since is a right tailed sided test the p value would be:

$p_v =P(t_{8}>3.285)=0.0056$

And we conclude that we reject the null hypothesis since $p_v$ . So the final conclusion would be not use the method since the value of 102 observed can be a potential outlier removing this value we see that we reject the null hypothesis and we have a significant result that the true mean is higher than 125.

8 0

2 years ago

An object moves in simple harmonic motion with period 8 seconds and amplitude 6 cm. At time t = 0 seconds, its displacement d fr

Vladimir79 [104]

In the equation model, the simple harmonic motion is d=6sin(πt/4)-6 if the object moves in simple harmonic motion with a period of 8 seconds and amplitude of 6 cm.

<h3>What is simple harmonic motion?</h3>

Simple Harmonic Motion is described as a motion in which the restoring force is proportionate to the body's displacement from its mean position.

We have:

Amplitude A = 6 cm

Time period = 8 seconds

T = 2π/w

8 = 2π/w

w = π/4

Its displacement d from rest is -6 cm, and initially, it moves in a positive direction.

d(0) = -6 cm at t = 0

The equation becomes:

$\rm d = 6sin(\dfrac{\pi}{4}t)-6$

Thus, in the equation model, the simple harmonic motion is d=6sin(πt/4)-6 if the object moves in simple harmonic motion with a period of 8 seconds and amplitude of 6 cm.

Learn more about the simple harmonic motion here:

brainly.com/question/17315536

#SPJ1

4 0

2 years ago

A shipping container will be used to transport several 140-kilogram crates across the

velikii [3]

Answer:

91.42 crates

Step-by-step explanation:

Let

Greatest Number of 140kg crates = x

weight of crate = 140-kilogram

Weight of other shipment = 13200 kilograms

greatest weight of container = 26000 kilograms

Weight of other shipment + (weight of crate * number of crates) ≤ greatest weight of container

13,200 + 140x ≤ 26,000

140x ≤ 26,000 - 13,200

140x ≤ 12,800

x ≤ 12,800/140

x ≤ 91.42 kilograms

Greatest Number of 140kg crates = 91.42 crates

8 0

2 years ago

Help please I’m confused

mestny [16]

Answer:

x=4 and y = 3 ...answer : 1

Step-by-step explanation:

hello :

x/8 =5/10 means : x=40/10 so: x=4

y/6 =5/10 means : y=30/10 so: y=3

5 0

2 years ago

7 less than a number k is 18

Elena-2011 [213]

Answer:

$k-7=18$ ; k=25

Step-by-step explanation:

$k(-7+7)=18+7\\18+7=25\\k=25$

7 less than 25 (or K) is 18.

4 0

2 years ago

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