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3 years ago

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HELP PLSSSSSSSSSSSSSSSSSSSSSSS

1 answer:

Allushta [10]3 years ago

7 0

Answer:

3) not equivalent

4) equivalent multiply by 2

5) equivalent multiply by 2

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A phone manufacturer wants to compete in the touch screen phone market. He understands that the lead product has a battery life

alisha [4.7K]

Answer:

a)

The null hypothesis is $H_0: \mu \leq 10$

The alternative hypothesis is $H_1: \mu > 10$

b-1) The value of the test statistic is t = 1.86.

b-2) The p-value is of 0.0348.

Step-by-step explanation:

Question a:

Test if the battery life is more than twice of 5 hours:

Twice of 5 hours = 5*2 = 10 hours.

At the null hypothesis, we test if the battery life is of 10 hours or less, than is:

$H_0: \mu \leq 10$

At the alternative hypothesis, we test if the battery life is of more than 10 hours, that is:

$H_1: \mu > 10$

b-1. Calculate the value of the test statistic.

The test statistic is:

We have the standard deviation for the sample, so the t-distribution is used to solve this question

$t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, s is the standard deviation and n is the size of the sample.

10 is tested at the null hypothesis:

This means that $\mu = 10$

In order to test the claim, a researcher samples 45 units of the new phone and finds that the sample battery life averages 10.5 hours with a sample standard deviation of 1.8 hours.

This means that $n = 45, X = 10.5, s = 1.8$

Then

$t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}$

$t = \frac{10.5 - 10}{\frac{1.8}{\sqrt{45}}}$

$t = 1.86$

The value of the test statistic is t = 1.86.

b-2. Find the p-value.

Testing if the mean is more than a value, so a right-tailed test.

Sample of 45, so 45 - 1 = 44 degrees of freedom.

Test statistic t = 1.86.

Using a t-distribution calculator, the p-value is of 0.0348.

5 0

2 years ago

Given the Arithmetic series A 1 + A 2 + A 3 + A 4 A1+A2+A3+A4 14 + 18 + 22 + 26 + . . . + 86 What is the value of sum?

postnew [5]

Answer:

950

Step-by-step explanation:

The common difference is 4, so the general term can be written:

... an = 14 + 4(n -1)

The value of n for the last term is ...

... 86 = 14 + 4(n -1) . . . . . the computation for the last term, 86

... 72 = 4(n -1) . . . . . . . . . subtract 14

... 18 = n -1 . . . . . . . . . . . divide by 4

... 19 = n . . . . . . . . . . . . . add 1

Your series has 19 terms. The first term is 14 and the last is 86, so the average term is (14+86)/2 = 50. Since there are 19 terms, the sum of them is ...

... 19×50 = 950

7 0

2 years ago

Can someone help me really please

JulsSmile [24]

Answer:

26

Step-by-step explanation:

Let x represent the missing number

$8\frac{11}{15} = 7\frac{x}{15}$

Convert both fractions to improper fractions

$\frac{131}{15} = \frac{105 + x}{15}$

Cross multiply

$15(105 + x) = 15(131)$

$1,575 + 15x = 1,965$

Subtract both sides by 1,575

$15x = 1,965 - 1,575$

$15x = 390$

Divide both sides by 15

$x = \frac{390}{15}$

x = 26

4 0

2 years ago

A student S is suspected of cheating on exam, due to evidence E of cheating being present. Suppose that in the case of a cheatin

Kay [80]

Answer:

Step-by-step explanation:

Suppose we think of an alphabet X to be the Event of the evidence.

Also, if Y be the Event of cheating; &

Y' be the Event of not involved in cheating

From the given information:

$P(\dfrac{X}{Y}) = 60\% = 0.6$

$P(\dfrac{X}{Y'}) = 0.01\% = 0.0001$

$P(Y) = 0.01$

Thus, $P(Y') \ will\ be = 1 - P(Y)$

P(Y') = 1 - 0.01

P(Y') = 0.99

The probability of cheating & the evidence is present is = P(YX)

$P(YX) = P(\dfrac{X}{Y}) \ P(Y)$

$P(YX) =0.6 \times 0.01$

$P(YX) =0.006$

The probabilities of not involved in cheating & the evidence are present is:

$P(Y'X) = P(Y') \times P(\dfrac{X}{Y'})$

$P(Y'X) = 0.99 \times 0.0001 \\ \\ P(Y'X) = 0.000099$

(b)

The required probability that the evidence is present is:

P(YX or Y'X) = 0.006 + 0.000099

P(YX or Y'X) = 0.006099

(c)

The required probability that (S) cheat provided the evidence being present is:

Using Bayes Theorem

$P(\dfrac{Y}{X}) = \dfrac{P(YX)}{P(Y)}$

$P(\dfrac{Y}{X}) = \dfrac{P(0.006)}{P(0.006099)}$

$P(\dfrac{Y}{X}) = 0.9838$

5 0

2 years ago

A sample size of n = 64 is drawn from a population whose standard deviation is o = 5.6.

AlekseyPX

Answer:

The margin of error for a 99% confidence interval for the population mean is 1.8025.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.99}{2} = 0.005$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.005 = 0.995$ , so $z = 2.575$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

In this problem:

$\sigma = 5.6$

So

$M = 2.575*\frac{5.6}{\sqrt{64}} = 1.8025$

The margin of error for a 99% confidence interval for the population mean is 1.8025.

5 0

2 years ago