Mathematics, 19.09.2019 23:50 alott1121

Consider a paint-drying situation in which drying time for a test specimen is normally distributed with σ = 7. the hypotheses h0: μ = 75 and ha: μ < 75 are to be tested using a random sample of n = 25 observations. (a) how many standard deviations (of x) below the null value is x = 72.3? (round your answer to two decimal places.) standard deviations (b) if x = 72.3, what is the conclusion using α = 0.01? (round your answers to two decimal places.) test statistic z = critical value z = what can you conclude? reject the null hypothesis. there is sufficient evidence to conclude that the mean drying time is less than 75. do not reject the null hypothesis. there is sufficient evidence to conclude that the mean drying time is less than 75. do not reject the null hypothesis. there is not sufficient evidence to conclude that the mean drying time is less than 75. reject the null hypothesis. there is not sufficient evidence to conclude that the mean drying time is less than 75. (c) what is α for the test procedure that rejects h0 when z ≤ −2.8? (round your answer to four decimal places.) α = (d) for the test procedure of part (c), what is β(70)? (round your answer to four decimal places.) β(70) = (e) if the test procedure of part (c) is used, what n is necessary to ensure that β(70) = 0.01? (round your answer up to the next whole number.) n = specimens (f) if a level 0.01 test is used with n = 100, what is the probability of a type i error when μ = 76? (round your answer to four decimal places.)

Answers: 1

Show answers

Answers

Answer from: Geo777

Part A:

The z score of the hypothesis testing of n samples of a normally distributed data set is given by:

$z= \frac{x-\mu}{\sigma/\sqrt{n}}$

Given that the population mean is 75 and the population standard deviation is 7, then the number of standard deviation of the mean for x = 72.3 is given by the z score:

$z= \frac{72.3-75}{7/\sqrt{25}} \\ \\ = \frac{-2.7}{7/5} = \frac{-2.7}{1.4} \\ \\ =-1.93$

Therefore, 72.3 is 1.93 standard deviations below the mean.

Part B:

The test statistics of the hypothesis testing of n samples of a normally distributed data set is given by:

$z= \frac{x-\mu}{\sigma/\sqrt{n}}$

Thus given that x = 72.3, μ = 75, σ = 7 and n = 25,

$z= \frac{72.3-75}{7/\sqrt{25}} \\ \\ = \frac{-2.7}{7/5} = \frac{-2.7}{1.4} \\ \\ =-1.93$

The p-value is given by:

P(-1.93)=0.03

Since α = 0.01 and p-value = 0.03, this means that the p-value is greater than the α, ant thus, we will faill to reject the null hypothesis.

Therefore, the conclussion is do not reject the null hypothesis. there is not sufficient evidence to conclude that the mean drying time is less than 75.

Part C

In order to reject the null hypothesis, the p-value must be less than or equal to α.

The p-value for z ≤ -2.8 is 0.00256.

Therefore, the α for the test procedure that rejects the null hypothesis when z ≤ −2.8 is 0.0026

Part D:

In general for the alternative hypothesis , $H_a :\mu\ \textless \ \mu_0$

$\beta(\mu') = P\left(X \ \textgreater \ \mu_0-z_{1-\alpha}\frac{\sigma}{\sqrt{n}}|\mu'\right) \\ \\ = 1-P\left(-z_{1-\alpha}+\frac{\mu_0-\mu'}{\sigma/\sqrt{n}}\right)$

So for the test procedure with α = 0.0026

$\beta(70) = 1 - P\left(-z_{0.9974}+\frac{75-70}{7/5}\right) \\ \\ =1 - P(-2.7949+3.5714)=1-P(0.7765) \\ \\ =1-0.78128\approx\bold{0.2187 }$

Part E:

For α = 0.0026, and a general sample size n we have that

$\beta(70) = 1 - P\left(-z_{0.9974}+\frac{75-70}{7/\sqrt{n}}\right) \\ \\ =1 - P\left(-2.7949+ \frac{5}{7/\sqrt{n}} \right)$

Since, we want n so that β(70) = 0.01, thus

$1 - P\left(-2.7949+ \frac{5}{7/\sqrt{n}} \right)=0.01 \\ \\ \Rightarrow P\left(-2.7949+ \frac{5}{7/\sqrt{n}} \right)=1-0.01=0.99 \\ \\ \Rightarrow P\left(-2.7949+ \frac{5}{7/\sqrt{n}} \right)=P(2.3262) \\ \\ \Rightarrow -2.7949+ \frac{5}{7/\sqrt{n}}=2.3262 \\ \\ \Rightarrow \frac{5}{7/\sqrt{n}}=5.1211 \\ \\ \Rightarrow \frac{7}{\sqrt{n}} = \frac{5}{5.1211} =0.9764 \\ \\ \Rightarrow \sqrt{n}= \frac{7}{0.9764} =7.1695 \\ \\ \Rightarrow n=(7.1695)^2=51.4$

so we need n = 52.

Part F:

p-value = $P-value=P(\bar{X}\leq\bar{x}) \\ \\ =P(\bar{X}\leq72.3)=P\left(z\leq \frac{72.3-76}{7/5} \right) \\ \\ =P\left(z\leq \frac{-3.7}{1.4} \right)=P(z\leq-2.643) \\ \\ =\bold{0.00411}$

Since p-value is larger that .01 we fail to reject µ = 76.

Answer from: cyndy50

A paint drying situation, has the standard deviation 1.93. The detailed solution is given below.

Given :

Standard Deviation, $\sigma = 7$

Mean, $\mu = 75$

Sample, n = 25

Solution :

A) We know that,

$z = \dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n} }}$ ---- (1)

Now put the values of $\rm x,\;\mu,\;\sigma \;and \;n$ in equation (1) we get,

$z = \dfrac{72.3-75}{\dfrac{7}{\sqrt{25} }}$

z = -1.93

Therefore, 72.3 is 1.93 standard deviations below the mean.

B) From the above part we know that z = - 1.93.Therefore p-value is,

P(-1.93) = 0.03

Since $\alpha$ = 0.01 and p-value = 0.03, this means that the p-value is greater than the fail to reject the null hypothesis.

C) The p-value for $z\leq -2.8$

is 0.00256

Therefore, the $\alpha$ for the test procedure that rejects the null hypothesis when $z\leq -2.8$ is 0.0026.

D) For alternate hypothesis,

$\rm \beta (\mu') = P(X\mu_0-z_1_-_\alpha \frac{\sigma}{\sqrt{n} }|\mu')$

$\rm = P(-z_1_-_\alpha +\dfrac{\mu_0-\mu'}{\dfrac{\sigma}{\sqrt{n} }})$

For $\alpha$ = 0.0026

$\rm \beta (70)=1- P(-z_0_._9_9_7_4 +\dfrac{75-70}{\dfrac{7}{\sqrt{25} }})$

$\rm \beta (70) = 1-P(-2.7949+3.5714)=1-P(0.7765)$

$\beta (70)=1-0.78128\approx 0.2187$

E) Now we want n when

$\beta (70) = 0.01$

$\rm 0.01=1- P(-z_0_._9_9_7_4 +\dfrac{75-70}{\dfrac{7}{\sqrt{n} }})$

$P(-2.7949+\dfrac{5}{\dfrac{7}{\sqrt{n} }})=0.99$

$P(-2.7949+\dfrac{5}{\dfrac{7}{\sqrt{n} }})=P(2.3262)$

$-2.7949+\dfrac{5}{\dfrac{7}{\sqrt{n} }}=2.3262$

$\rm \sqrt{n} = 7.1695$

n = 51.4

Therefore we need n = 52.

$\rm p-value = P(\bar{X}\leq \bar{x})$

$\rm p-value = P(\bar{X}\leq 72.3)=P(z\leq \dfrac{72.3-76}{\dfrac{7}{5}})$

$\rm p-value = P(z\leq -2.643)$

$\rm p-value = 0.00411$

Since p-value is larger that .01 therefore we fail to reject $\mu$ = 76.

For more information, refer the link given below

link

Answer from: Quest

3-4

step-by-step explanation:

the answer is 3-4 because both 45 and 60 has a multiple of 15.

45/3=15

60/4=15

therefore the odds against selecting a red is 4-3

Answer from: Quest

B) greg: 30 points , tremaine 38 points

Questions in other subjects:

Mathematics, 26.10.2020 17:00

Elijah drives 9 miles to his friend's house in 16 hour. How fast does Elijah drive, in miles per hour? If he drives at this rate for 2hours, how far does he drive?...

Chemistry, 26.10.2020 17:00

At what temperature would a 1.65 m NaCl solution freeze, given that the van't Hoff factor for NaCl is 1.9? Kf for water is 1.86 ∘C/m ....

Mathematics, 26.10.2020 17:00

intGiven a data tuple having the values "systems", "26. . . 30", and "46–50K" for the attributesdepartment, age, and salary, respectively, what would a na¨ıve Bayesian classificati...

Physics, 26.10.2020 17:00

Two friends, one at each end of a 4.00 m long raft over water, are playing with a 9.07-kg medicine ball. One of them throws the ball to the other. If the total mass of the people a...

Business, 26.10.2020 17:00

The demand for automobiles rises when the economy experiences an expansion. If an automobile manufacturer hires additional workers during an expansion, this would be best described...

Mathematics, 26.10.2020 17:00

An online retailer sells two packages of protein bars. 10−pack of 2.5 ounce bars 12−pack of 1.8 ounce bars $ 19.50 18.75 Which package has the better price per bar?...

Mathematics, 26.10.2020 17:00

A and B play 12 games of chess of which 6 are won by A,4 are won by B, and 2 end in a draw. They agree to play a tournament consisting of 3 games. Find the probability that (a) A w...

Mathematics, 26.10.2020 17:00

A food store makes 6- pound mixture of pecans, pistachios , and dried bananas. The cost of pecans is $1.50 per pound, pistachios cost $1.00 per pound, and dried bananas cost$1.50 p...

History, 26.10.2020 17:00

There a girl i like sofia...

English, 26.10.2020 17:00

Choose the sentence in which the word is used in the correct context. Group of answer choices The band's latest album is a worthy successor to their debut album....

More questions: Mathematics Another questions

Answers

Other questions on the subject: Mathematics

Questions in other subjects: