Someone please help lols

1) If the alpha level is changed from α = .05 to α = .01, what happens to boundaries for the critical region?

Alexxandr [17]

Answer:

1) a. Move farther into the tails

2) a. Decreases

Step-by-step explanation:

Hello!

1)

Let's say for example that you are making a confidence interval for the mean, using the Z-distribution:

X[bar] ± $Z_{1-\alpha /2}$ * $\frac{Sigma}{\sqrt{n} }$

Leaving all other terms constant, this are the Z-values for three different confidence levels:

90% $Z_{0.95}= 1.648$

95% $Z_{0.975}= 1.965$

99% $Z_{0.995}= 2.586$

Semiamplitude of the interval is

d= $Z_{1-\alpha /2}$ * $\frac{Sigma}{\sqrt{n} }$

Then if you increase the confidence level, the value of Z increases and so does the semiamplitude and amplitude of the interval:

↑d= ↑ $Z_{1-\alpha /2}$ * $\frac{Sigma}{\sqrt{n} }$

They have a direct relationship.

So if you change α: 0.05 to α: 0.01, then the confidence level 1-α increases from 0.95 to 0.99, and the boundaries move farther into the tails.

2)

The significance level of a hypothesis test is the probability of committing a Type I error.

If you decrease the level from 5% to 1%, then logically, the probability decreases.

I hope this helps!

5 0

3 years ago

Consider the following incomplete deposit ticket:

boyakko [2]

Answer:

B) 105.20

Step-by-step explanation:

491.60 + 157.68 - 68.44 = 581.20

6 0

3 years ago

Read 2 more answers

Plzzzzz help I don’t understand and this question is worth 23 points that’s how much I need to know how to do this and I need to

butalik [34]

You need PEMDAS
do 3·6 and then 12/6
when you subtract the product and quoetient you get 16 for the answer

6 0

2 years ago

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The camp director , Mrs Marshall, put all 1,265 backpacks on the bus. Each backpack weighed seven pounds. How many pounds did Mr

posledela

Answer: 8855 pounds

Step-by-step explanation: Simply multiply 1265 x 7

6 0

3 years ago

Read 2 more answers

The U.S. Bureau of Economic Statistics reports that the average annual salary in the metropolitan Boston area is $50,542. Suppos

xenn [34]

Answer:

(a) P(X > $57,000) = 0.0643

(b) P(X < $46,000) = 0.1423

(c) P(X > $40,000) = 0.0066

(d) P($45,000 < X < $54,000) = 0.6959

Step-by-step explanation:

We are given that U.S. Bureau of Economic Statistics reports that the average annual salary in the metropolitan Boston area is $50,542.

Suppose annual salaries in the metropolitan Boston area are normally distributed with a standard deviation of $4,246.

Let X = annual salaries in the metropolitan Boston area

SO, X ~ Normal( $\mu=$50,542,\sigma^{2} = $4,246^{2}$ )

The z-score probability distribution for normal distribution is given by;

Z = $\frac{X-\mu}{\sigma }$ ~ N(0,1)

where, $\mu$ = average annual salary in the Boston area = $50,542

$\sigma$ = standard deviation = $4,246

(a) Probability that the worker’s annual salary is more than $57,000 is given by = P(X > $57,000)

P(X > $57,000) = P( $\frac{X-\mu}{\sigma }$ > $\frac{57,000-50,542}{4,246 }$ ) = P(Z > 1.52) = 1 - P(Z $\leq$ 1.52)

= 1 - 0.93574 = 0.0643

The above probability is calculated by looking at the value of x = 1.52 in the z table which gave an area of 0.93574.

(b) Probability that the worker’s annual salary is less than $46,000 is given by = P(X < $46,000)

P(X < $46,000) = P( $\frac{X-\mu}{\sigma }$ < $\frac{46,000-50,542}{4,246 }$ ) = P(Z < -1.07) = 1 - P(Z $\leq$ 1.07)

= 1 - 0.85769 = 0.1423

The above probability is calculated by looking at the value of x = 1.07 in the z table which gave an area of 0.85769.

(c) Probability that the worker’s annual salary is more than $40,000 is given by = P(X > $40,000)

P(X > $40,000) = P( $\frac{X-\mu}{\sigma }$ > $\frac{40,000-50,542}{4,246 }$ ) = P(Z > -2.48) = P(Z < 2.48)

= 1 - 0.99343 = 0.0066

The above probability is calculated by looking at the value of x = 2.48 in the z table which gave an area of 0.99343.

(d) Probability that the worker’s annual salary is between $45,000 and $54,000 is given by = P($45,000 < X < $54,000)

P($45,000 < X < $54,000) = P(X < $54,000) - P(X $\leq$ $45,000)

P(X < $54,000) = P( $\frac{X-\mu}{\sigma }$ < $\frac{54,000-50,542}{4,246 }$ ) = P(Z < 0.81) = 0.79103

P(X $\leq$ $45,000) = P( $\frac{X-\mu}{\sigma }$ $\leq$ $\frac{45,000-50,542}{4,246 }$ ) = P(Z $\leq$ -1.31) = 1 - P(Z < 1.31)

= 1 - 0.90490 = 0.0951

The above probability is calculated by looking at the value of x = 0.81 and x = 1.31 in the z table which gave an area of 0.79103 and 0.9049 respectively.

Therefore, P($45,000 < X < $54,000) = 0.79103 - 0.0951 = 0.6959

3 0

2 years ago