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Zinaida [17]
2 years ago
9

Regress smoker on cubic polynomials of age, using a linear probability model. What is the p-value for testing the hypothesis tha

t the probability model is linear in age? (two decimal places)
Mathematics
1 answer:
Vlada [557]2 years ago
8 0

The statement " only the estimate intercept is statistically significant at the 5% level is wrong" is wrong about the estimate.

<h3>What is probit regression?</h3>

A probit model is a type of regression in statistics where the dependant variable can only take two values.

We have:

Regress smoker on cubic polynomials of age

If we use linear probability model.

Here the data are missing, but we can say about the estimate that:

Only the estimate intercept is statistically significant at the 5% level is wrong,

Thus, the statement " only the estimate intercept is statistically significant at the 5% level is wrong" is wrong about the estimate.

Learn more about the probit regression here:

brainly.com/question/23389011

#SPJ4

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Monica paid sales tax of $1.80 when she bought a new bike helmet. If the sales tax rate was 9%, how much did the store charge fo
Anna11 [10]

Answer:

$20 dollars

Step-by-step explanation:

1.80 / .09 (9% in decimal form)

= 20

8 0
3 years ago
A random sample of 500 registered voters in Phoenix is asked if they favor the use of oxygenated fuels year-round to reduce air
Stells [14]

Answer:

a) 0.0853

b) 0.0000

Step-by-step explanation:

Parameters given stated that;

H₀ : <em>p = </em>0.6

H₁ : <em>p  = </em>0.6, this explains the acceptance region as;

p° ≤ \frac{315}{500}=0.63 and the region region as p°>0.63 (where p° is known as the sample proportion)

a).

the probability of type I error if exactly 60% is calculated as :

∝ = P (Reject H₀ | H₀ is true)

   = P (p°>0.63 | p=0.6)

where p° is represented as <em>pI</em><em> </em>in the subsequent calculated steps below

   

    = P  [\frac{p°-p}{\sqrt{\frac{p(1-p)}{n}}} >\frac{0.63-p}{\sqrt{\frac{p(1-p)}{n}}} |p=0.6]

    = P  [\frac{p°-0.6}{\sqrt{\frac{0.6(1-0.6)}{500}}} >\frac{0.63-0.6}{\sqrt{\frac{0.6(1-0.6)}{500}}} ]

    = P   [Z>\frac{0.63-0.6}{\sqrt{\frac{0.6(1-0.6)}{500} } } ]

    = P   [Z > 1.37]

    = 1 - P   [Z ≤ 1.37]

    = 1 - Ф (1.37)

    = 1 - 0.914657 ( from Cumulative Standard Normal Distribution Table)

    ≅ 0.0853

b)

The probability of Type II error β is stated as:

β = P (Accept H₀ | H₁ is true)

  = P [p° ≤ 0.63 | p = 0.75]

where p° is represented as <em>pI</em><em> </em>in the subsequent calculated steps below

  = P [\frac{p°-p} \sqrt{\frac{p(1-p)}{n} } }\leq \frac{0.63-p}{\sqrt{\frac{p(1-p)}{n} } } | p=0.75]

  = P [\frac{p°-0.6} \sqrt{\frac{0.75(1-0.75)}{500} } }\leq \frac{0.63-0.75}{\sqrt{\frac{0.75(1-0.75)}{500} } } ]

  = P[Z\leq\frac{0.63-0.75}{\sqrt{\frac{0.75(1-0.75)}{500} } } ]

  = P [Z ≤ -6.20]

  = Ф (-6.20)

  ≅ 0.0000 (from Cumulative Standard Normal Distribution Table).

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Find the missing length indicated.​
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Answer:12

Step-by-step explanation:

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