I need help with this quick

Each of 16 students measured the circumference of a tennis ball by four different methods, which were:Method A: Estimate the cir

lys-0071 [83]

Answer:

a) $\bar X_A =22.744$

$\bar X_B =20.7$

$\bar X_C =21.013$

$\bar X_D =18.306$

b) $Median_A =\frac{23+24}{2}=23.5$

$Median_B =\frac{20.4+20.4}{2}=20.4$

$Median_C =\frac{21+21}{2}=21$

$Median_D =\frac{20.7+20.7}{2}=20.7$

c) $\bar X_A =23.25$

$\bar X_B =20.7$

$\bar X_C =21.04$

$\bar X_D =20.69$

Step-by-step explanation:

Data given

Assuming the following data:

Method A: 18.0, 18.0, 18.0, 20.0, 22.0, 22.0, 22.5, 23.0, 24.0,24.0,25.0,25.0, 25.0,25.0,26.0,26.4

Method B: 18.8, 18.9, 18.9, 19.6, 20.1, 20.4, 20.4, 20.4, 20.4,20.5, 21.2. 22.0,22.0, 22.0,22.0,23.6

Method C: 20.2, 20.5, 20.5, 20.7, 20.8, 20.9, 21.0, 21.0,21.0,21.0, 21.0, 21.5,21.5,21.5,21.5,21.6

Method D: 20.0, 20.0, 20.0,-20.0, 20.2, 20.5, 20.5, 20.7, 20.7,20.7, 21.0, 21.1, 21.5, 21.6, 22.1,22.3.

Part a

The sample mean is defined as:

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

If we apply this formula for the four methods we got:

$\bar X_A =22.744$

$\bar X_B =20.7$

$\bar X_C =21.013$

$\bar X_D =18.306$

Part b

Since the total number of points for each method is 16 and this is an even number we need to calculate the median as the average between the 8th and the 9th position of the data ordered from the smallest to the largest. If we do this we have that:

$Median_A =\frac{23+24}{2}=23.5$

$Median_B =\frac{20.4+20.4}{2}=20.4$

$Median_C =\frac{21+21}{2}=21$

$Median_D =\frac{20.7+20.7}{2}=20.7$

Part c

The trimmed mean by 20% means that we need to calculate removing the 20% from each of the tails, the 20% of 16 is 3.2, so we need to remove 3 observations from both ends of the data like this:

Method A: 20.0, 22.0, 22.0, 22.5, 23.0, 24.0,24.0,25.0,25.0, 25.0

Method B: 19.6, 20.1, 20.4, 20.4, 20.4, 20.4,20.5, 21.2. 22.0,22.0

Method C: 20.7, 20.8, 20.9, 21.0, 21.0,21.0,21.0, 21.0, 21.5,21.5

Method D: 20.0 20.2, 20.5, 20.5, 20.7, 20.7,20.7, 21.0, 21.1, 21.5.

If we calculate the mean for each method we got:

$\bar X_A =23.25$

$\bar X_B =20.7$

$\bar X_C =21.04$

$\bar X_D =20.69$

3 0

1 year ago

For the given functions f (x), let x0 = 0, x1 = 0.6, and x2 = 0.9. construct interpolation polynomials of degree at most one and

barxatty [35]

Check the attached file for the solution.

6 0

2 years ago

A hair salon in Cambridge, Massachusetts, reports that on seven randomly selected weekdays, the number of customers who visited

morpeh [17]

Answer:

a: 28 < µ < 34

Step-by-step explanation:

We need the mean, var, and standard deviation for the data set. See first attached photo for calculations for these...

We get a mean of 222/7 = 31.7143

and a sample standard deviation of: 4.3079

We can now construct our confidence interval. See the second attached photo for the construction steps.

They want a 90% confidence interval. Our sample size is 7, so since n < 30, we will use a t-score. Look up the value under the 10% area in 2 tails column, and degree of freedom is 6 (degree of freedom is always 1 less than sample size for confidence intervals when n < 30)

The t-value is: 1.943

We rounded down to the nearest person in the interval because we don't want to over estimate. It said 28.55, so more than 28 but not quite 29, so if we use 29 as the lower limit, we could over estimate. It's better to use 28 and underestimate a little when considering customer flow.

5 0

2 years ago

For what values of x is x2-36=5x true?

katrin2010 [14]

Answer:

x=-12

Step-by-step explanation:

x2-36=5x

add 36 to both sides

x2=5x+36

subtract 5x from both sides

x(-3)=36

divide both sides by -3

x=-12

substitute to prove right

(-12)2-36=5(-12)

-24-36=-60

-60=-60

3 0

1 year ago

Read 2 more answers

Is it possible that<br> 1952<br> is the remainder of 375139654 - 1953?

Soloha48 [4]

No I don’t think it’s possible

7 0

2 years ago