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3 years ago

14

I need help with the hypotheses testing.

1 answer:

pentagon [3]3 years ago

6 0

Answer:

Hi hows life going?

Step-by-step explanation:

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A sample size must be determined for estimating a population mean given that the confidence level is % and the desired margin o

Sauron [17]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

$n =290$

b

$n =129$

c

The correct option is D

Step-by-step explanation:

From the question we are told that

The margin of error is $E = 0.23$

The largest value is k = 15

The smallest value is u = 7

From the question we are told the confidence level is 95% , hence the level of significance is

$\alpha = (100 - 95 ) \%$

=> $\alpha = 0.05$

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.96$

Generally population standard deviation estimate is mathematically represented as

$\sigma = \frac{Range }{4}$

Here the Range is mathematically represented as

$Range = k - u$

=> $Range = 15 - 7$

=> $Range = 8$

=> $\sigma = \frac{8 }{4}$

=> $\sigma = 2$

Generally the sample size is mathematically represented as

$n = [\frac{Z_{\frac{\alpha }{2} } * \sigma }{E} ] ^2$

=> $n = [1.96 * 2 }{0.23} ] ^2$

=> $n =290$

Generally to obtain a conservatively small sample size the population standard deviation estimate becomes

=> $\sigma = \frac{Range }{6}$

=> $\sigma = \frac{8 }{6}$

=> $\sigma =1.333$

Generally the conservatively small sample size is mathematically evaluated as

$n = [\frac{Z_{\frac{\alpha }{2} } * \sigma }{E} ] ^2$

=> $n = [1.96 * 1.333 }{0.23} ] ^2$

=> $n =129$

6 0

3 years ago

A fossil was analyzed and determined to have a carbon-14 level that is 80 % that of living organisms. the half-life of c-14 is 5

liq [111]

Answer:

The fossil is 1860 years old.

Step-by-step explanation:

The equation for the amount of fossil has the following format:

$Q(t) = Q(0)e^{rt}$

In which Q(t) is the amount after t years, Q(0) is the initial amount and r is the rate of change.

Half-life of c-14 is 5730 years.

This means that $Q(5730) = 0.5Q(0)$

So

$Q(t) = Q(0)e^{rt}$

$0.5Q(0) = Q(0)e^{5730r}$

$e^{5730r} = 0.5$

$\ln{e^{5730r}} = \ln{0.5}$

$5730r = \ln{0.5}$

$r = \frac{\ln{0.5}}{5730}$

$r = -0.00012$

So

$Q(t) = Q(0)e^{-0.00012t}$

How old is the fossil?

This is t for which

$Q(t) = 0.8Q(0)$

So

$Q(t) = Q(0)e^{-0.00012t}$

$0.8Q(0) = Q(0)e^{-0.00012t}$

$e^{-0.00012t} = 0.8$

$\ln{e^{-0.00012t}} = \ln{0.8}$

$-0.00012t = \ln{0.8}$

$t = -\frac{\ln{0.8}}{-0.00012}$

$t = 1860$

The fossil is 1860 years old.

4 0

3 years ago

A gardener is planting two types of trees: Type A is 4 feet tall and grows at a rate of 24 inches per year. Type B is 5 feet tal

goldenfox [79]

It would take 5 yrs. for both trees to be the same height.

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3 years ago

At one store, 20 percent of the pairs of shoes sold are of men's athletic shoes. Lionel rings up 8 orders which contain one or m

PIT_PIT [208]

B Explanation
Explain

7 0

3 years ago

Factor completely x2 + 8x - 20

kvv77 [185]

X-2
Hope this helps!
Vote me brainliest!

3 0

3 years ago

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