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fenix001 [56]
3 years ago
9

Supposethat z is a normally distributed variable with variance 4. You collect a sample of size n for which you get a sample aver

age value of 3. You are asked to test the null hypothesis that the mean is smaller than 2. Find the sample size n above which you can reject the null hypothesiswith a 95% of confidence.
Mathematics
1 answer:
Nikitich [7]3 years ago
7 0

Answer:

The sample size is 4

Step-by-step explanation:

Null hypothesis: The mean is 2

Alternate hypothesis: The mean is less than 2

Mean = 3

sd = sqrt(variance) = sqrt(4) = 2

At 95% confidence level, t-value is 1.960

Assuming the lower bound of the mean is 1.04

Lower bound = mean - (t×sd/√n)

1.04 = 3 - (1.96×2/√n)

3.92/√n = 3 - 1.04

3.92/√n = 1.96

√n = 3.92/1.96

√n = 2

n = 2^2

n = 4

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Someone helppp!!!!!!
Yakvenalex [24]
Mother correct answer is -3,5
8 0
3 years ago
In a random sample of 400 residents of Boston, 320 residents indicated that they voted for Obama in the last presidential electi
coldgirl [10]

Answer:

C.I =  0.7608   ≤ p ≤   0.8392

Step-by-step explanation:

Given that:

Let consider a  random sample n = 400 candidates where  320 residents indicated that they voted for Obama

probability \hat p = \dfrac{320}{400}

= 0.8

Level of significance ∝ = 100 -95%

= 5%

= 0.05

The objective is to  develop a 95% confidence interval estimate for the proportion of all Boston residents who voted for Obama.

The confidence internal can be computed as:

=\hat p  \pm Z_{\alpha/2} \sqrt{\dfrac{ p(1-p)}{n } }

where;

Z_{0.05/2} = Z_{0.025} = 1.960

SO;

=0.8  \pm 1.960 \sqrt{\dfrac{ 0.8(1-0.8)}{400 } }

=0.8  \pm 1.960 \sqrt{\dfrac{ 0.8(0.2)}{400 } }

=0.8  \pm 1.960 \sqrt{\dfrac{ 0.16}{400 } }

=0.8  \pm 1.960 \sqrt{4 \times 10^{-4}}

=0.8  \pm 1.960 \times 0.02}

=0.8  \pm 0.0392

= 0.8 - 0.0392     OR   0.8 + 0.0392  

= 0.7608    OR    0.8392

Thus; C.I =  0.7608   ≤ p ≤   0.8392

3 0
3 years ago
The diameters of ball bearings are distributed normally. The mean diameter is 107 millimeters and the population standard deviat
tensa zangetsu [6.8K]

Answer:

0.6710

Step-by-step explanation:

The diameters of ball bearings are distributed normally. The mean diameter is 107 millimeters and the population standard deviation is 5 millimeters.

Find the probability that the diameter of a selected bearing is between 104 and 115 millimeters. Round your answer to four decimal places.

We solve using z score formula

z = (x-μ)/σ, where

x is the raw score

μ is the population mean = 107 mm

σ is the population standard deviation = 5 mm

For x = 104 mm

z = 104 - 107/5

z = -0.6

Probability value from Z-Table:

P(x = 104) = 0.27425

For x = 115 mm

z = 115 - 107/5

z = 1.6

Probability value from Z-Table:

P(x = 115) = 0.9452

The probability that the diameter of a selected bearing is between 104 and 115 millimeters is calculated as:

P(x = 115) - P(x = 104)

0.9452 - 0.27425

= 0.67095

Approximately = 0.6710

8 0
3 years ago
Samantha drew Triangle JKL with the coordinates (2,3), (4, 3) and (5,2). She reflected this image over the x-axis to create an i
zavuch27 [327]

Answer:

J'(2,-3), K'(4, -3) and L'(5,-2)

Step-by-step explanation:

The given triangle has vertices at  J(2,3), K(4, 3) and L(5,2).

The transformation rule for a reflection in the x-axis is:

(x,y)\to (x,-y)

We substitute the points to obtain the coordinates  of the image triangle  J'K'L'

J(2,3)\to J'(2,-3)

K(4,3)\to K'(4,-3)

L(5,2)\to L'(5,-2)

In other words, we negate the y-coordinates of triangle JKL to obtain the coordinates of J'K'L'

See attachment for graph.  

3 0
3 years ago
Nvm bro i got the answer
KIM [24]
Ok and have a nice day
6 0
2 years ago
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