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Weights and heights of turkeys tend to be correlated. For a population of turkeys at a farm, this correlation is found to be 0.6

LenaWriter [7]

Answer:

a turkey at the farm which weighs more than 90% of all the turkeys is predicted to be taller than 79.37 % of them.

The average height for turkeys at the 90th percentile for weight is 34.554

Of the turkeys at the 90th percentile for weight, roughly the percentage that would be taller than 28 inches 79.37%

Step-by-step explanation:

Given that:

For a population of turkeys at a farm, the correlation found between the weights and heights of turkeys is r = 0.64

the average weight in pounds $\overline x$ = 17

the standard deviation of the weight in pounds $S_x$ = 5

the average height in inches $\overline y$ = 28

the standard deviation of the height in inches $S_y$ = 8

Also, given that the weight and height both roughly follow the normal curve

For this study , the slope of the regression line can be expressed as :

$\beta_1 = r \times ( \dfrac{S_y}{S_x})$

$\beta_1 = 0.64 \times ( \dfrac{8}{5})$

$\beta_1 = 0.64 \times 1.6$

$\beta_1 = 1.024$

To the intercept of the regression line, we have the following equation

$\beta_o = \overline y - \beta_1 \overline x$

replacing the values:

$\beta_o = 28 -(1.024)(17)$

$\beta_o = 28 -17.408$

$\beta_o = 10.592$

However, the regression line needed for this study can be computed as:

$\hat Y = \beta_o + \beta_1 X$

$\hat Y = 10.592 + 1.024 X$

Recall that;

both the weight and height roughly follow the normal curve

As such, the weight related to 90th percentile can be determined as shown below.

Using the Excel Function at 90th percentile, which can be computed as:

(=Normsinv (0.90) ; we have the desired value of 1.28

∴

$\dfrac{X - \overline x}{s_x } = 1.28$

$\dfrac{X - 17}{5} = 1.28$

$X - 17 = 6.4$

X = 6.4 + 17

X = 23.4

The predicted height $\hat Y = 10.592 + 1.024 X$

where; X = 23.4

$\hat Y = 10.592 + 1.024 (23.4)$

$\hat Y = 10.592 + 23.9616$

$\hat Y = 34.5536$

Now; the probability of predicted height less than 34.5536 can be computed as:

$P(Y < 34.5536) = P( \dfrac{Y - \overline y }{S_y} < \dfrac{34.5536-28}{8})$

$P(Y < 34.5536) = P(Z< \dfrac{6.5536}{8})$

$P(Y < 34.5536) = P(Z< 0.8192)$

From the Z tables;

P(Y < 34.5536) =0.7937

Hence, a turkey at the farm which weighs more than 90% of all the turkeys is predicted to be taller than 79.37 % of them.

The average height for turkeys at the 90th percentile for weight is :

$\hat Y = 10.592 + 1.024 X$

where; X = 23.4

$\hat Y = 10.592 + 1.024 (23.4)$

$\hat Y = 10.592 + 23.962$

$\mathbf{\hat Y = 34.554}$

Of the turkeys at the 90th percentile for weight, roughly what percent would you estimate to be taller than 28 inches?

i.e

P(Y >28) = 1 - P (Y< 28)

$P(Y >28) = 1 - P( Z < \dfrac{28 - 34.554}{8})$

$P(Y >28) = 1 - P( Z < \dfrac{-6.554}{8})$

$P(Y >28) = 1 - P( Z < -0.8193)$

From the Z tables,

$P(Y >28) = 1 - 0.2063$

$\mathbf{P(Y >28) = 0.7937}$

= 79.37%

7 0

4 years ago

Q 4.19: To verify whether there are negative consequences of taking a new type of medicine, 3000 tests were conducted. Assume th

tiny-mole [99]

Answer:

The p-value obtained is less than the significance level at which the test was performed at, hence, we reject the null hypothesis, accept the alternative hypothesis & say that there is enough evidence to conclude that the the new medicine is potentially harmful.

Step-by-step explanation:

The null hypothesis plays the devil's advocate and usually takes the form of the opposite of the theory to be tested. It usually contains the signs =, ≤ and ≥ depending on the directions of the test.

While, the alternative hypothesis usually confirms the the theory being tested by the experimental setup. It usually contains the signs ≠, < and > depending on the directions of the test.

For this question, the null hypothesis is that the new medicine has no negative effects and the alternative hypothesis is that the new medicine is potentially harmful

Mathematically,

The null hypothesis is represented as

H₀: p = 0

The alternative hypothesis is represented as

Hₐ: p > 0

To do this test, we will use the z-distribution because although no information on the population standard deviation is known, the sample size is large enough.

So, we compute the test statistic

z = (x - μ)/σₓ

x = sample proportion = (31/3000) = 0.0103

μ = p₀ = 0

σₓ = standard error = √[p(1-p)/n]

where n = Sample size = 3000

σₓ = √[0.0103×0.9897/3000] = 0.0018462936 = 0.0018463

z = (0.0103 - 0) ÷ 0.0018463

z = 5.60

checking the tables for the p-value of this test statistic

Significance level = 1% = 0.01

The hypothesis test uses a one-tailed condition because we're testing only in one direction.

p-value (for z = 5.60, at 0.01 significance level, with a one tailed condition) = < 0.00001

The interpretation of p-values is that

When the (p-value > significance level), we fail to reject the null hypothesis and when the (p-value < significance level), we reject the null hypothesis and accept the alternative hypothesis.

So, for this question, significance level = 0.01

p-value = < 0.00001

p-value < 0.00001 < 0.10

Hence,

p-value < significance level

This means that we reject the null hypothesis, accept the alternative hypothesis & say that there is enough evidence to conclude that the the new medicine is potentially harmful.

Hope this Helps!!!

6 0

4 years ago

Find the unit price for each size of Jif creamy peanut butter and determine which size is the better buy?

LiRa [457]

0.182, 0.181, 0.175. Large is the better buy because it has the lesser unit price.

4 0

3 years ago

Read 2 more answers

What number is the opposite of 659? Enter your answer in the box below.

makkiz [27]

Answer: