Need help with number 2. If you don’t know it just leave it like it is (someone literally wrote “the umm” as an answer on my previous post, just don’t)

Answer:

Estimate of the population proportion [tex]\mathbf{\widehat{(P)}}[/tex] = 0.147

Step-by-step explanation:

Given -

Wiley Publications has determined that out of a sample of 7,831 of its publications for 2012, 1,157 of them had been pirated online in some form.

Let x be the no of publication had been pirated online in some form.

x = 1157

n = 7831

Population proportion [tex]\boldsymbol{\widehat{P} = \frac{x}{n}}[/tex]

                                           =  [tex]\frac{1157}{7831}[/tex]

                                         =  0.147

Answer:

25 posts

Step-by-step explanation:

So the number of fence post would be the total length of the log divided by the length of each post. As the log is 16m and is corrected to the nearest metre, it could possibly be 16.499m. As for the post that is 70 cm long and corrected to the nearest 10cm, it may as well be 65 cm (or 0.65m) each post

So the max number of fence point once can possibly cut from the log would be

16.499 / 0.65 = 25 posts

Answer:

The diameter is 3 inches

Step-by-step explanation:

If you are calculating a circle, one of the main rules you should know is that the diameter is double the radius. In this case the radius is 1.5 inches long. If you need the diameter, you need to double that figure; in this case the answer is 3 inches

Answer:

3 inches

Step-by-step explanation:

diameter is twice the length of the radius

1.5x2=3

Answer:

(a) I attached a photo with the diagram.

(b) [tex]f(x) = \big( \frac{1}{2} \big)^{x-1}[/tex]

(c) 1/4

(d) 4

(e) [tex]k^2/2[/tex]

Step-by-step explanation:

(a) I attached a photo with the diagram.

(b) The easiest way to think about this part is in terms of combinatorics. Think about it like this.

To begin with, look at the three each level of the three represents a possible outcome of throwing the coin n-times. If you throw the coin 3 times at the end in total there are 8 possible outcomes. But The favorable outcomes are just 2.

 1  - Your first outcome is HEADS and all the others are different except the last one.

 2  - Your first outcome is TAILS and all the others are different except the last one.

Therefore the probability of the event is

[tex]f(x) = P(X=x) = \frac{2}{2^x} = \frac{1}{2^{x-1}} = \big( \frac{1}{2} \big)^{x-1}[/tex]

(c)

P(X = 0) = 0 because it is not possible to have two consecutive tails or heads.

[tex]E(X > 4) = 1 - P(X \leq 3) = 1 - ( P(X = 0 ) + P(X = 1) + P(X = 2) + P(X = 3))\\= 1 - ( \frac{1}{2} + \frac{1}{4} ) = \frac{1}{4}[/tex]

(d)

Remember that this is a geometric distribution therefore

[tex]E[X] = p/(1-p)[/tex], in this case  [tex]p = 1/2[/tex] so [tex]E[X] = 1[/tex] and

[tex]E[X+1]^2 = ( E[X] +1 )^2 = (1+1)^2 = 2^2 = 4[/tex]

Also

(e)

This is a geometric distribution so its variance is

[tex]Var(X) = \frac{1-p}{p^2} = 1/2 / 1/4 = 1/2[/tex]

And using properties of variance

[tex]Var(kX - k ) = Var(kX) = k^2 var(X) = k^2 /2[/tex]

To find the required answers, we create a tree diagram to visualize the possibilities of flipping a fair coin. Then, we determine the probability mass function (pmf) of X, finding the probabilities associated with each outcome. We find that P(X=0) is not possible and P(X>4) is 0. We also calculate E((X+1)^2) and Var(kX-k) using the formulas for expected value and variance of a discrete random variable.

To answer the question, we first need to understand the possibilities when flipping a coin. Let H represent a heads and T represent a tails. The possibilities are: HT, TH, and HH. Now, let's create a tree diagram to visualize the possibilities:

(b) To find the probability mass function (pmf) of X, we need to determine the probability of each outcome. Since the coin is fair, each outcome has a probability of 1/2. Therefore, the pmf of X is: P(X=1) = 1/2, P(X=2) = 1/4, P(X=3) = 1/4.

(c) (i) P(X=0) is not possible because we need to observe tails-heads in consecutive flips. (ii) P(X>4) is the probability of observing tails-heads after at least 4 flips, which is 0 since tails-heads can occur in the 2nd or 3rd flip.

(d) To find E((X+1)^2), we need to multiply each possible outcome by its corresponding probability, square it, and then sum them up. E((X+1)^2) = (0+1)^2 * P(X=1) + (1+1)^2 * P(X=2) + (2+1)^2 * P(X=3).

(e) Var(kX-k) = k^2 * Var(X), where Var(X) is the variance of X. Since X is a discrete random variable, Var(X) = E(X^2) - (E(X))^2.

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Answer:

B is the correct answer.

Step-by-step explanation:

15 times 5 plus 450

= 525

25 times 5 plus 400

= 525

Answer:

30 hours

Step-by-step explanation:

We can use ratios to solve

5 hours        x hours

-------------- = -------------

3 walls          18 walls

Using cross products

5*18 = 3x

Divide each side by 3

5*18/3 = 3x/3

30 =x

Answer:It will take Harold 30 hours to paint all 18 of the walls

Step-by-step explanation:Your multiplying the amount of walls he is painting by 6 so you do the same to the amount of hours he is painting

Answer: inverse operation

Step-by-step explanation:

here you go

Use log definition: if logₐ(b) = c then b = a^c

log₃(x) = 6 → x = 3⁶

x = 3⁶ → x = 726

Therefore, x = 726

Best of Luck!

Answer:

Step-by-step explanation:

false

Yes, All the parabolas have the same domain.

What is Domain?

The set of all inputs of the function is called Domain of set.

Given that;

The statement is,

''All parabolas have the same domain.''

Now,

Since, The domain of the parabola will be (- ∞, ∞).

Hence,  All the parabolas have the same domain.

Thus,  All the parabolas have the same domain.

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Answer:

The probability that Susan's average score for the week is between 220 and 228 is 0.7594.

Step-by-step explanation:

Average score of Susan = u = 225

Standard deviation = [tex]\sigma[/tex] = 13

Score of Susan follow a Normal Distribution and we have the population standard deviation, as this standard deviation is of her scores of previous 14 years.

In a given week, Susan bowls 16 games. This means, our sample size is 16. So,

n = 16

We have to find the probability that her average score of the week is between 220 and 228. Since the distribution is normal and value of population standard deviation is known, we will use the concept of z-score and z-distribution to find the desired probability.

First we will convert the given numbers to their equivalent z-scores. The formula to calculate the z-scores is:

[tex]z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

x = 220 converted to z-score will be:

[tex]z=\frac{220-225}{\frac{13}{\sqrt{16}}}=-1.54[/tex]

x = 228 converted to z-score will be:

[tex]z=\frac{228-225}{\frac{13}{\sqrt{16}}}=0.92[/tex]

So, probability that Susan's score is between 220 and 228 is equivalent to probability of z score being in between - 1.54 and 0.92

i.e.

P (220 < X < 228) = P( -1.54 < z < 0.92)

From the z-table we can find the following values:

P( -1.54 < z < 0.92) = P(x < 0.92) - P(x<-1.54)

P( -1.54 < z < 0.92) = 0.8212 - 0.0618

P( -1.54 < z < 0.92) = 0.7594

Since, P (220 < X < 228) is equivalent to P( -1.54 < z < 0.92), we can conclude that the probability that Susan's average score for the week is between 220 and 228 is 0.7594.

Using the Central Limit Theorem and Z-scores, we calculate the probability that Susan's average weekly score will be between 220 and 228 from the standard normal distribution table.

This question revolves around the concept of standard deviation and probability in a normal distribution. Standard deviation measures the dispersion or variation of a set of values. In this case, her average score per week is a random variable, so Central Limit Theorem (CLT) applies because she plays a large number of games (16) every week. CLT states that the sum of a large number of independent and identically-distributed random variables has an approximately normal distribution.

We can calculate the mean (μ) and standard deviation (σ) of the weekly average:

To calculate the Z-scores for 220 and 228:

Finally, we need to find the probability P(Z₁ < Z < Z₂), which involves looking up these Z-scores in a Z-table or using statistical software or online resource that gives the values for a normal distribution.

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Answer:

The value of the test statistic is [tex]t = -2.09[/tex]

Step-by-step explanation:

The null hypothesis is:

[tex]H_{0} = 6.6[/tex]

The alternate hypotesis is:

[tex]H_{1} \neq 6.6[/tex]

Our test statistic is:

[tex]t = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis, [tex]\sigma[/tex] is the standard deviation(square roof of the variance) and n is the size of the sample.

In this problem, we have that:

[tex]X = 6.5, \mu = 6.6, \sigma = \sqrt{0.64} = 0.8, n = 280[/tex]

So

[tex]t = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

[tex]t = \frac{6.5 - 6.6}{\frac{0.8}{\sqrt{280}}}[/tex]

[tex]t = -2.09[/tex]

The value of the test statistic is [tex]t = -2.09[/tex]

Answer:

Step-by-step explanation:

CB is the radius. It is equal to BD. So, the diameter CD is ...

  6 inches + 6 inches = 12 inches

The diameter is always twice the length of the radius.

The diameter of this cone is 12 inches.

Diego’s travel distance can be determined by first subtracting the pickup fee from the total fare and then dividing this result by the cost per mile. In this case, Diego traveled a total of 11 miles.

To calculate Diego’s travel distance, begin by subtracting the pickup fee from the total fare. The pickup fee is $2.75 and the total fare is $19.25, so this calculation gives us $19.25 - $2.75 = $16.50. Next, divide this result by the cost per mile to determine the total distance traveled. As City cabs charge $1.50 per mile, this calculation is $16.50 ÷ $1.50 = 11. So, Diego traveled 11 miles in the cab.

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Diego traveled 11 miles in the city cab, as calculated by subtracting the fixed pickup fee from the total fare and then dividing by the cost per mile.

City cabs charges a $2.75 pickup fee and $1.50 per mile traveled. If Diego's fare for a cross-town cab ride is $19.25, the question asks us to calculate how far he traveled in the cab. To solve this, we need to use the equation of the total fare which is:

Total fare = Pickup fee + (Cost per mile × Number of miles)

Let 'x' be the number of miles Diego traveled. Using the above equation and substituting the given values, we have:

$19.25 = $2.75 + ($1.50 × x)

To find 'x', we will subtract the pickup fee from the total fare and then divide the result by the cost per mile:

$19.25 - $2.75 = $1.50 × x

$16.50 = $1.50 × x

x = $16.50 / $1.50

x = 11

Therefore, Diego traveled 11 miles in the cab.

Answer:

D

Step-by-step explanation:

200 men and 2,000 women

i dont know how but that was just the answer

The sampling distribution of ā-Ă will be approximately normal when both populations satisfy the conditions for np > 5 and nq > 5. The sample sizes must ensure these inequalities are met using the proportions 0.08 for men and 0.005 for women.

The sampling distribution of ā - Ă will be approximately normal when the sample sizes are large enough to satisfy the condition np > 5 and nq > 5 for both populations. Given that in the population of men, 8 percent carry the genetic trait (p1 = 0.08), and in the population of women, 0.5 percent carry the trait (p2 = 0.005), we need to find appropriate sample sizes for each population.

For the men's population:
n1p1 > 5 and n1(1 - p1) > 5.

For the women's population:
n2p2 > 5 and n2(1 - p2) > 5.

The sample sizes should be selected in a way that these inequalities are met for both populations to ensure that the sampling distribution of ā-Ă is approximately normal.

Answer:

a) The standard error would be of 8.58 pounds.

b) The margin of error is 14.11 pounds.

c) The 90% confidence interval for the population mean is between 232.89 pounds and 261.11 pounds

Step-by-step explanation:

a.What is the standard error?

The standard error is

[tex]s = \frac{\sigma}{\sqrt{n}}[/tex]

In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample. So

[tex]s = \frac{47}{\sqrt{30}} = 8.58[/tex]

The standard error would be of 8.58 pounds.

b.What is the margin of error at 90% confidence?

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

[tex]\alpha = \frac{1-0.9}{2} = 0.05[/tex]

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

So it is z with a pvalue of [tex]1-0.05 = 0.95[/tex], so [tex]z = 1.645[/tex]

Now, find the margin of error M as such

[tex]M = z*s = 1.645*8.58 = 14.11[/tex]

The margin of error is 14.11 pounds.

c. Using my sample of 30, what would be the 90% confidence interval for the population mean?

Lower bound: Sample mean subtracted by the margin of error.

247 - 14.11 = 232.89 pounds

Upper bound

247 + 14.11 = 261.11 pounds

The 90% confidence interval for the population mean is between 232.89 pounds and 261.11 pounds

Answer:

a) The standard error would be of 8.58 pounds.

b) The margin of error is 14.11 pounds.

c) The 90% confidence interval for the population mean is between 232.89 pounds and 261.11 pounds

Step-by-step explanation:

Answer:

350(1.137)^20=4563.28232

Answer:

nothing

Step-by-step explanation:

nothing

Answer: D. 7 or 12

Step-by-step explanation: on edgenuity 2020 :))

Answer:

402.12 meters

Step-by-step explanation:

to measure the circumference,

C = 2πr

2π(64) =

to run 1 full lap around the track would be 402.12 meters

i hope this helps!

:)

To find the distance of a full lap around a circular track, calculate the circumference using the formula C = 2π r. With a radius of 64 meters, the distance is approximately 402.12 meters.

To calculate the distance traveled on a full lap around a circular track, you need to find the circumference of the circle. The circumference of a circle is given by the formula C = 2π r, where π (Pi) is approximately 3.14159 and r is the radius of the circle.

Given that the radius of the track is 64 meters, we can substitute this value into the formula to find the circumference.

C = 2π(64 meters)

C = 128π meters

C = approx. 128 × 3.14159 meters

C = approx. 402.1232 meters

Therefore, if you ran one full lap around the circular track, you would travel approximately 402.12 meters.

Answer:

[tex]t=\frac{7.917-9.5}{\frac{1.501}{\sqrt{6}}}=-2.58[/tex]    

[tex]p_v =P(t_{5}<-2.58)=0.03[/tex]

If we compare the p value with a significance level for example [tex]\alpha=0.05[/tex] we see that [tex]p_v < \alpha[/tex] so we can conclude that we can reject the null hypothesis, and there is enough evidence to conclude that the true mean is significantly lower than 9.5% at 0.05 of signficance

Step-by-step explanation:

Data given and notation    

Data: 6.6% 9.1% 7.0% 6.3% 8.5% 10.0%

We can calculate the mean and the sample deviation with the following formulas:

[tex] \bar X = \frac{\sum_{i=1}^n X_i}{n}[/tex]

[tex] s = \sqrt{\frac{\sum_{i=1}^n (X-i -\bar X)^2}{n-1}}[/tex]

[tex]\bar X=7.917[/tex] represent the sample mean

[tex]s=1.501[/tex] represent the sample standard deviation

[tex]n=6[/tex] sample size    

[tex]\mu_o =9.5[/tex] represent the value that we want to test  

[tex]\alpha[/tex] represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)    

[tex]p_v[/tex] represent the p value for the test (variable of interest)

State the null and alternative hypotheses.    

We need to conduct a hypothesis in order to determine if the true mean os greater or not than 9.5%, the system of hypothesis would be:    

Null hypothesis:[tex]\mu \geq 9.5[/tex]    

Alternative hypothesis:[tex]\mu < 9.5[/tex]    

We don't know the population deviation, so for this case we can use the t test to compare the actual mean to the reference value, and the statistic is given by:    

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)    

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic    

We can replace in formula (1) the info given like this:    

[tex]t=\frac{7.917-9.5}{\frac{1.501}{\sqrt{6}}}=-2.58[/tex]    

Calculate the P-value    

The degrees of freedom are given by:

[tex] df = n-1 = 6-1=5[/tex]

Since is a lower tailed test the p value would be:    

[tex]p_v =P(t_{5}<-2.58)=0.03[/tex]

In Excel we can use the following formula to find the p value "=T.DIST(-2.58,5,TRUE)"  

Conclusion    

If we compare the p value with a significance level for example [tex]\alpha=0.05[/tex] we see that [tex]p_v < \alpha[/tex] so we can conclude that we can reject the null hypothesis, and there is enough evidence to conclude that the true mean is significantly lower than 9.5% at 0.05 of signficance

The test statistic of the sampling will be -2.58.

The following can be deduced from the information:

Mean = 7.917

Degree of freedom = 6 - 1 = 5

P value = 0.975 > 0.05

The test statistic will be:

= (7.917 - 9.5) / (1.501 / ✓56)

= -2.58

In conclusion, the test statistic is -2.58.

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Answer: the null hypothesis states that there is no difference between the number of cavities of children who had used either Toothpaste brand X or Toothpaste brand Y for a year.

Step-by-step explanation:

The null hypothesis is the hypothesis that is assumed to be true. It is an expression that is the opposite of what the researcher predicts.

The alternative hypothesis is what the researcher expects or predicts. It is the statement that is believed to be true if the null hypothesis is rejected.

Looking at the given situation,

The difference was significant at the .05 level. This means that there was enough evidence to reject null hypothesis. Since the null hypothesis contradicts the alternative hypothesis, then the null hypothesis would state that there is no difference between the number of cavities of children who had used either Toothpaste brand X or Toothpaste brand Y for a year.

The question relates to a two-sample t-test in statistics, which compares the means of two processes. It involves calculating a t-score using sample data, and comparing this score to a critical value to determine if there is a significant difference between the two processes.

This question pertains to the field of statistics within mathematics. Specifically, it involves the use of a two-sample t-test to compare the means of two different processes. The two-sample t-test is used when the population standard deviations are unknown but assumed to be equal, which is the case here.

The process involves computing a t-score using the observed data (including sample mean, standard deviation, and sample size) and then comparing this score to a critical value from the t-distribution. If the observed t-score exceeds the critical value, there is evidence to suggest that there is a significant difference between the two processes.

In this case, you would calculate the t-score for Processes A and B using their respective sample means, standard deviations, and sample sizes. If the observed t-score is greater than the critical t-value at the chosen significance level (typically 0.05 or 0.01), we reject the null hypothesis that there is no difference between the two processes.

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Answer:

[tex]28-2.064\frac{10}{\sqrt{25}}=23.872[/tex]    

[tex]28+2.064\frac{10}{\sqrt{25}}=32.128[/tex]    

So on this case the 95% confidence interval would be given by (23.9;32.1)  

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

[tex]\bar X=28[/tex] represent the sample mean

[tex]\mu[/tex] population mean (variable of interest)

s=10 represent the sample standard deviation

n=25 represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]   (1)

In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:

[tex]df=n-1=25-1=24[/tex]

Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,24)".And we see that [tex]t_{\alpha/2}=2.064[/tex]

Now we have everything in order to replace into formula (1):

[tex]28-2.064\frac{10}{\sqrt{25}}=23.872[/tex]    

[tex]28+2.064\frac{10}{\sqrt{25}}=32.128[/tex]    

So on this case the 95% confidence interval would be given by (23.9;32.1)    

The 95% confidence interval for the average age of the boutique's customers is from 24.1 to 31.9 years, calculated using the sample mean of 28 years, a standard deviation of 10 years, and a sample size of 25.

To calculate a 95% confidence interval for the average age of all customers in the boutique, we will use the sample mean, standard deviation, and the size of the sample along with the z-score corresponding to a 95% confidence level.

The formula for a confidence interval when the population standard deviation is known is:

Confidence interval = {x} (Z  {Σ}/√{n}})

In this case, we have:

Sample mean ({x}): 28 years

Standard deviation (Σ): 10 years

Sample size (n): 25

Z-score for 95% confidence: 1.96 (from z-tables)

First, we calculate the margin of error:

The margin of error = Z{Σ}/{√{n}} = 1.96  {10}/{√{25}} = 1.96 × 2 = 3.92

Then, the confidence interval is:

Lower limit = {x} - Margin of error = 28 - 3.92 = 24.08

Upper limit = {x} + Margin of error = 28 + 3.92 = 31.92

Therefore, the 95% confidence interval for the average age of the boutique's customers is from 24.1 to 31.9 years.

Answer:

Combine the like terms

Step-by-step explanation:

2 + 7 = 2x + 5 - 43

The first step is to combine the like terms

9 = 2x - 38

The sample points in the event A ∪ B are 1, 2, 3, 4, 5, 6, 8, and 10, which is determined by the union.

To identify the sample points in event A ∪ B (the union of events A and B), we need to determine the numbers that satisfy either event A or event B or both.

Event A: The number is even.

Sample points in event A are: {2, 4, 6, 8, 10}.

Event B: The number is less than 7.

Sample points in event B are: {1, 2, 3, 4, 5, 6}.

To find the sample points in the union of events A and B (A ∪ B), we combine the sample points from both events without duplication.

Thus, the sample points in A ∪ B are: {1, 2, 3, 4, 5, 6, 8, 10}.

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The sample points in the event A ∪ B, which include numbers that are either even or less than 7, are {1, 2, 3, 4, 5, 6, 8, 10}.

To find the sample points in the event A ∪ B, where event A: {The number is even} and event B: {The number is less than 7}, we first identify the numbers between 1 and 10 that satisfy each event. For event A, the even numbers between 1 and 10 are 2, 4, 6, 8, and 10. For event B, the numbers less than 7 are 1, 2, 3, 4, 5, and 6.

The union of two events, A ∪ B, includes all sample points that are in event A, event B, or in both A and B. Therefore, the union of A ∪ B includes the even numbers (2, 4, 6, 8, 10) and the numbers less than 7 (1, 2, 3, 4, 5, 6), without listing any number more than once.

Thus, the sample points in the event A ∪ B are {1, 2, 3, 4, 5, 6, 8, 10}.

Answer:

Yes, there is  enough evidence to support the claim that ionizing radiation is beneficial in terms of marketability.

Step-by-step explanation:

This is a hypothesis test for the difference between proportions.

The claim is that ionizing radiation is beneficial in terms of marketability.

Then, the null and alternative hypothesis are:

[tex]H_0: \pi_1-\pi_2=0\\\\H_a:\pi_1-\pi_2> 0[/tex]

Being π1: the true proportion of treated bulbs that can be comercialized after 240 days, and π2: the true proportion of untreated bulbs that can be comercialized after 240 days.

The significance level is 0.05.

The sample 1, of size n1=180 has a proportion of p1=0.85.

[tex]p_1=X_1/n_1=153/180=0.85[/tex]

The sample 2, of size n2=180 has a proportion of p2=0.65.

[tex]p_2=X_2/n_2=117/180=0.65[/tex]

The difference between proportions is (p1-p2)=0.2.

[tex]p_d=p_1-p_2=0.85-0.65=0.2[/tex]

The pooled proportion, needed to calculate the standard error, is:

[tex]p=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{153+117}{180+180}=\dfrac{270}{360}=0.75[/tex]

The estimated standard error of the difference between means is computed using the formula:

[tex]s_{p1-p2}=\sqrt{\dfrac{p(1-p)}{n_1}+\dfrac{p(1-p)}{n_2}}=\sqrt{\dfrac{0.75*0.25}{180}+\dfrac{0.75*0.25}{180}}\\\\\\s_{M_d}=\sqrt{0.001+0.001}=\sqrt{0.002}=0.0456[/tex]

Then, we can calculate the z-statistic as:

[tex]z=\dfrac{p_d-(\pi_1-\pi_2)}{s_{p1-p2}}=\dfrac{0.2-0}{0.0456}=\dfrac{0.2}{0.0456}=4.382[/tex]

This test is a right-tailed test, so the P-value for this test is calculated as (using a z-table):

[tex]P-value=P(t>4.382)=0.000008[/tex]

As the P-value (0.000008) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is  enough evidence to support the claim that ionizing radiation is beneficial in terms of marketability.

Final answer:

The observed marketable rate of irradiated garlic bulbs suggests that ionizing radiation is beneficial to their marketability compared to untreated bulbs, and this could be statistically tested for significance.

Explanation:

The student's data indicates that irradiated garlic bulbs have a significantly higher rate of remaining marketable (85%) after 240 days compared to unirradiated bulbs (65%). Using a significance level of α = 0.05, we can infer a positive effect of ionizing radiation on the marketability of garlic bulbs. To statistically confirm this observation, one could perform a hypothesis test (such as a chi-squared test for independence) to determine if the difference in marketability between the treated and untreated groups is significant.