What is the measure of angle Y?

Because of safety considerations, in May 2003 the Federal Aviation Administration (FAA) changed its guidelines for how small com

Rasek [7]

Answer:

a) The 95% confidence interval for the mean summer weight (including carry-on luggage) of Frontier Airlines passengers is between 179 and 187 pounds. This means that we are 95% sure that the mean summer weight of all Frontier Airlines passengers is between these two values.

b)

The 95% confidence interval for the mean winter weight (including carry-on luggage) of Frontier Airlines passengers is between 185.4 pounds and 194.6 pounds. This means that we are 95% sure that the mean winter weight of all Frontier Airlines passengers is between these two values.

c) They are respected, as the upper bound of both intervals is below the new FAA recommendations.

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve these questions.

Question a:

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 100 - 1 = 99

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 99 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.95}{2} = 0.975$ . So we have T = 1.9842

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 1.9842\frac{20}{\sqrt{100}} = 4$

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 183 - 4 = 179 pounds.

The upper end of the interval is the sample mean added to M. So it is 183 + 4 = 187 pounds.

The 95% confidence interval for the mean summer weight (including carry-on luggage) of Frontier Airlines passengers is between 179 and 187 pounds. This means that we are 95% sure that the mean summer weight of all Frontier Airlines passengers is between these two values.

Question b:

Critical value is the same(same sample size and confidence level).

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 1.9842\frac{23}{\sqrt{100}} = 4.6$

The lower end of the interval is the sample mean subtracted by M. So it is 190 - 4.6 = 185.4 pounds.

The upper end of the interval is the sample mean added to M. So it is 190 + 4.6 = 194.6 pounds.

The 95% confidence interval for the mean winter weight (including carry-on luggage) of Frontier Airlines passengers is between 185.4 pounds and 194.6 pounds. This means that we are 95% sure that the mean winter weight of all Frontier Airlines passengers is between these two values.

c. The new FAA recommendations are 190 pounds for summer and 195 pounds for winter. Comment on these recommendations in light of the confidence interval estimates from Parts (a) and (b).

They are respected, as the upper bound of both intervals is below the new FAA recommendations.

7 0

3 years ago

In the figure, BP is an angle bisector of ∠CBD.

telo118 [61]

Possible answer is A)5

8 0

3 years ago

I need heeelllppp please

barxatty [35]

$\frac{ { - 5}^{2} + {10}^{2} }{(2 \times ( - 5) \times 3) + {3}^{3} } = \frac{ - 25 + 100}{ - 30 + 27} = \\ = \frac{75}{ - 3} = - 25$

5 0

3 years ago

Read 2 more answers

What is science??ㅤㅤㅤㅤㅤ

Darina [25.2K]

Answer:

the study in which observations are made, experiments are done and logical conclusions are drawn in order to understand the principles of nature.

8 0

3 years ago

The principal at Crest Middle School, which enrolls only sixth-grade students and seventh-grade students, is interested in deter

AlekseyPX

Answer:

a) [ -27.208 , -12.192 ]

b) New procedure is not recommended

Step-by-step explanation:

Solution:-

- It is much more common for a statistical analyst to be interested in the difference between means than in the specific values of the means themselves.

- The principal at Crest Middle School collects data on how much time students at that school spend on homework each night.

- He/She takes a " random " sample of n = 20 from a sixth and seventh grades students from the school population to conduct a statistical analysis.

- The summary of sample mean ( x1 & x2 ) and sample standard deviation ( s1 & s2 ) of the amount of time spent on homework each night (in minutes) for each grade of students is given below:

<u>Mean ( xi )</u> <u> Standard deviation ( si )</u>

Sixth grade students 27.3 10.8

Seventh grade students 47.0 12.4

- We will first check the normality of sample distributions.

We see that sample are "randomly" selected.
The mean times are independent for each group
The groups are selected independent " sixth " and " seventh" grades.

The means of both groups are conforms to 10% condition of normality.

Hence, we will assume that the samples are normally distributed.

- We are to construct a 95% confidence interval for the difference in means ( u1 and u2 ).

- Under the assumption of normality we have the following assumptions for difference in mean of independent populations:

Population mean of 6th grade ( u1 ) ≈ sample mean of 6th grade ( x1 )

Population mean of 7th grade ( u2 ) ≈ sample mean of 6th grade ( x2 )

Therefore, the difference in population mean has the following mean ( u ^ ):

u^ = u1 - u2 = x1 - x2

u^ = 27.3 - 47.0

u^ = -19.7

- Similarly, we will estimate the standard deviation (Standard Error) for a population ( σ^ ) represented by difference in mean. The appropriate relation for point estimation of standard deviation of difference in means is given below:

σ^ = √ [ ( σ1 ^2 / n1 ) + ( σ2 ^2 / n2 ) ]

Where,

σ1 ^2 : The population variance for sixth grade student.

σ2 ^2 : The population variance for sixth grade student.

n1 = n2 = n : The sample size taken from both populations.

Therefore,

σ^ = √ [ ( 2*σ1 ^2 / n )].

- Here we will assume equal population variances : σ1 ≈ σ2 ≈ σ is "unknown". We can reasonably assume the variation in students in general for the different grade remains somewhat constant owing to other reasons and the same pattern is observed across.

- The estimated standard deviation ( σ^ ) of difference in means is given by:

σ^ =

$s_p*\sqrt{\frac{1}{n_1} + \frac{1}{n_2} } = s_p*\sqrt{\frac{1}{n} + \frac{1}{n} } = s_p*\sqrt{\frac{2}{n}}\\\\\\s_p = \sqrt{\frac{(n_1 - 1 )*s_1^2 + (n_2 - 1 )*s_2^2}{n_1+n_2-2} } = \sqrt{\frac{(n - 1 )*s_1^2 + (n - 1 )*s_2^2}{n+n-2} } = \sqrt{\frac{(n - 1 )*s_1^2 + (n - 1 )*s_2^2}{2n-2} } \\\\s_p = \sqrt{\frac{(20 - 1 )*s_1^2 + (20 - 1 )*s_2^2}{2(20)-2} } \\\\s_p = \sqrt{\frac{19*10.8^2 + 19*12.4^2}{38} } = \sqrt{135.2} \\\\s_p = 11.62755$

σ^ = 11.62755*√2/20

σ^ = 3.67695

- Now we will determine the critical value associated with Confidence interval ( CI ) which is defined by the standard probability of significance level ( α ). Such that:

Significance Level ( α ) = 1 - CI = 1 - 0.95 = 0.05

- The reasonable distribution ( T or Z ) would be determined on the basis of following conditions:

The population variances ( σ1 ≈ σ2 ≈ σ ) are unknown.
The sample sizes ( n1 & n2 ) are < 30.

Hence, the above two conditions specify the use of T distribution critical value. The degree of freedom ( v ) for the given statistics is given by:

v = n1 + n2 - 2 = 2n - 2 = 2*20 - 2

v = 38 degrees of freedom

- The t-critical value is defined by the half of significance level ( α / 2 ) and degree of freedom ( v ) as follows:

t-critical = t_α / 2, v = t_0.025,38 = 2.024

- Then construct the interval for 95% confidence as follows:

[ u^ - t-critical*σ^ , u^ + t-critical*σ^ ]

[ -19.7 - 2.042*3.67695 , -19.7 + 2.042*3.67695 ]

[ -19.7 - 7.5083319 , -19.7 + 7.5083319 ]

[ -27.208 , -12.192 ]

- The principal should be 95% confident that the difference in mean times spent of homework for ALL 6th and 7th grade students in this school (population) lies between: [ -27.208 , -12.192 ]

- The procedure that the matched-pairs confidence interval for the mean difference in time spent on homework prescribes the integration of time across different sample groups.

- If we integrate the times of students of different grades we would have to make further assumptions like:

The intelligence levels of different grade students are same
The aptitude of students from different grades are the same
The efficiency of different grades are the same.

- We have to see that both samples are inherently different and must be treated as separate independent groups. Therefore, the above added assumptions are not justified to be used for the given statistics. The procedure would be more bias; hence, not recommended.

8 0

3 years ago