The U.S. Department of Transportation, National Highway Traffic Safety Administration, reported that 77% of all fatally injured automobile drivers were intoxicated. A random sample of 53 records of automobile driver fatalities in a certain county showed that 35 involved an intoxicated driver. Do these data indicate that the population proportion of driver fatalities related to alcohol is less than 77% in Kit Carson County? Use α = 0.05.

Answer: The population proportion is between 0.34 and 0.46 .

Step-by-step explanation:

Interpretation of 95% confidence interval : A person can be about 95% confident that the true population parameter lies in the interval.

A confidence interval is constructed for an unknown population proportion, p. A sample is collected, and the 95% confidence interval is calculated to be 0.40 ± 0.06.

i.e. Lower limit = [tex]0.40-0.06=0.34[/tex]

Upper limit =  [tex]0.40+0.06=0.46[/tex]

i.e. 95% confidence interval = (0.34, 0.46)

i.e. A person can be about 95% confident that the true population proportion (p) lies in the interval  (0.34, 0.46).

⇒ It is most accurate to say that there is approximately 95% confidence in the assertion that:

The population proportion is between 0.34 and 0.46 .

Answer:

Null hypothesis:[tex]p=0.1[/tex]  

Alternative hypothesis:[tex]p \neq 0.1[/tex]  

[tex]z=\frac{0.087 -0.1}{\sqrt{\frac{0.1(1-0.1)}{367}}}=-0.83[/tex]  

[tex]p_v =2*P(z<-0.83)=0.41[/tex]  

The p value obtained was a very high value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of orders that were not accurate is not significant different from 0.1 or 10% .  

and the accuracy of the test yes is acceptable since the p value obtained is large enough to fail to reject the null hypothesis.

Step-by-step explanation:

Data given and notation n  

n=367 represent the random sample taken

X=32 represent the orders that were not accurate

[tex]\hat p=\frac{32}{367}=0.087[/tex] estimated proportion of orders that were not accurate

[tex]p_o=0.1[/tex] is the value that we want to test

[tex]\alpha=0.05[/tex] represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

[tex]p_v{/tex} represent the p value (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the rate of inaccurate orders is equal to 10%:  

Null hypothesis:[tex]p=0.1[/tex]  

Alternative hypothesis:[tex]p \neq 0.1[/tex]  

When we conduct a proportion test we need to use the z statisitc, and the is given by:  

[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)  

The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

[tex]z=\frac{0.087 -0.1}{\sqrt{\frac{0.1(1-0.1)}{367}}}=-0.83[/tex]  

Statistical decision  

It's important to refresh the p value method or p value approach . "This methos is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.  

Since is a bilateral test the p value would be:  

[tex]p_v =2*P(z<-0.83)=0.41[/tex]  

So the p value obtained was a very high value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of orders that were not accurate is not significant different from 0.1 or 10% .  

Final answer:

The slightly higher proportion of Americans who indicated they are afraid to fly in a random sample is not necessarily evidence of an increased fear of flying overall. This variation could be due to sample variability, and without statistical analysis like hypothesis testing or confidence interval construction, one cannot conclude a genuine increase in fear of flying among Americans.

Explanation:

A study conducted by an organization found that the proportion of Americans who were afraid to fly in 2006 was 0.10. When a random sample of 1,400 Americans results in 154 individuals indicating that they are afraid to fly, it might initially seem like evidence that the proportion of Americans afraid to fly has increased. However, this is not necessarily indicative of an overall trend. To understand why we need to consider statistical variability and the concept of a confidence interval.

Statistical fluctuations in samples can lead to results that differ from the actual population proportion. In this case, the sample proportion of individuals afraid to fly is roughly 0.11 (154/1400), only slightly higher than the reported 0.10 from 2006. Without conducting a hypothesis test or constructing a confidence interval around the sample proportion, it's not possible to definitively say whether this difference is statistically significant or just due to random chance.

Furthermore, sample size plays a crucial role in the reliability of estimates. Although a sample size of 1,400 may seem large, the inherent randomness in sample selection could still lead to results differing from the true population proportion. In summary, a slight increase in the sample proportion of individuals afraid to fly, compared to a previous study, is not conclusive evidence of a trend without further statistical analysis to support such a claim.

The data from a sample of 1,400 Americans revealing 154 individuals who are afraid to fly does not necessarily indicate an increase in the proportion of Americans with this fear compared to 2006. Statistically significant evidence through a hypothesis test is required to make such a conclusion since sample data can fluctuate due to random chance.

A random sample of 1,400 Americans resulted in 154 indicating that they are afraid to fly, which might suggest an increase from the 2006 study that showed a 0.10 proportion of Americans had this fear. However, this is not necessarily evidence that the proportion of Americans who are afraid to fly has increased due to potential sampling error or the natural fluctuation inherent in sample data. To determine if there is a statistically significant increase, one would need to perform a hypothesis test comparing the sample proportion to the known proportion of 0.10.

Variability can occur in survey results, and a single sample might not represent the entire population accurately. Also, the difference observed might be a result of random chance. Therefore, it's essential to conduct statistical tests to infer whether the observed change is meaningful and not due to sampling variability.

The P-value is 0.0000; we reject the null hypothesis and conclude males consume more energy drinks than females.

We will conduct a hypothesis test for the difference between two means. Here are the steps we'll follow:

1. **State the Hypotheses**:

2. **Check the Conditions**:

3. **Calculate the Test Statistic**:

  We'll use the formula for the test statistic for two independent samples (assuming equal variances are not assumed):

  [tex]\[ t = \frac{(\bar{x}_1 - \bar{x}_2)}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \][/tex]

  where:

4. **Calculate the Degrees of Freedom**:

  We'll use the Welch-Satterthwaite equation to approximate the degrees of freedom for the t-distribution:

[tex]\[ df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1}} \][/tex]

5. **Find the P-value**:

  Since we are conducting a one-tailed test, we will find the area to the right of our calculated t-value in the t-distribution with the calculated degrees of freedom.

6. **State the Conclusion**:

  We will compare the P-value to the significance level (usually 0.05) and decide whether to reject the null hypothesis.

Let's perform the calculations.

The test statistic (t-value) for the difference in the mean number of energy drinks consumed per month between male and female college students is approximately 4.3285. The degrees of freedom for this test is approximately 720.78.

The P-value for this one-tailed test is approximately 0.00000857 (rounded to four decimal places, this is 0.0000). Since the P-value is significantly less than the common alpha level of 0.05, we reject the null hypothesis.

**Conclusion**: We can conclude that there is significant evidence at the 0.05 level to suggest that the mean number of energy drinks consumed by male college students is greater than that consumed by female college students.

Answer:

  9.5 ft wide by 24 ft long

Step-by-step explanation:

A graphing calculator shows there are two solutions to the system of equations ...

  2x + y = 43 . . . . . . fence length when x= width

  xy = 228 . . . . . . .  area

The solutions are ...

  (width, length) = (9.5, 24)   or   (12, 19)

Since we want the length as long as possible, the choice of dimensions is ...

  length = 24 feet; width = 9.5 feet.

For the first one, since the slope is 0 the line must be just a horizontal line that passes through the given point. The equation would be y=-4 because the y coordinate for the point is -4.

For the second problem, utilize y=mx+b, the parent equation for slope intercept form (m-slope, b-y intercept).

y=mx+b

-4=-1/3(5)+b

b=-2 1/3

equation: y=-1/3x - 2 1/3

Answer:

the answer is in the attached image below

Step-by-step explanation:

The sum of independent normally distributed random variable is also a normally distributed random variable. The needed probabilities are given as:

Suppose that a random variable X is formed by n mutually independent and normally distributed random variables such that:

[tex]X_i = N(\mu_i , \sigma^2_i) ; \: i = 1,2, \cdots, n[/tex]

And if  

[tex]X = X_1 + X_2 + \cdots + X_n[/tex]

Then, its distribution is given as:

[tex]X \sim N(\mu_1 + \mu_2 + \cdots + \mu_n, \: \: \sigma^2_1 + \sigma^2_2 + \cdots + \sigma^2_n)[/tex]

For the given case, let we take:

Then, we have:

[tex]X_1 \sim N(\mu = 75, \sigma^2 = (8)^2 = 64)\\\\X_2 \sim N(\mu = 70, \sigma^2 = (12)^2 = 144)\\[/tex]

Then, The random variable [tex]-X_2[/tex] has all negative values than of [tex]X_2[/tex], so  its mean will also become negative, but standard deviation would be same(since its measure of spread which would be same for [tex]-X_2[/tex] .

Thus, [tex]-X_2 \sim N(-70, 12^2) = N(-70, 144)[/tex]

Thus, we get:

[tex]X = X_1 - X_2\\\\X \sim N(75 -70, \sigma^2 = 64 + 144)\\\\X \sim N(5, 208)[/tex]

Thus, mean of X is 5, and standard deviation is

[tex]\sqrt{208} \approx 14.42[/tex]

(positive root since standard deviation is non negative quantity)

Thus, calculating the needed probability with the use  of z-scores:

P(X > 4)

Converting to standard normal distribution, we get

[tex]Z = \dfrac{X - \mu}{\sigma} \approx \dfrac{X - 5}{14.42}\\\\P(X > 4) \approx P(Z > \dfrac{4-5}{14.42}) \approx P(Z > -0.07) = 1-P(Z \leq -0.07)\\[/tex]

Using the z-tables, the p value for z = -0.07 is 0.4721

p value for Z = z gives [tex]P(Z \leq z) = p[/tex]

And therefore,

[tex]P(Z \leq -0.07) = 0.4721\\\\\rm and\: thus\\\\P(X > 4) = P(Z > -0.07) = 1 - P(Z \leq -0.07) = 1 - 0.4721 = 0.5279[/tex]

P( 3.5 < X1 − X2 < 5.5 ) = P( 3.5 < X < 5.5)

Using z scores, we get:

[tex]P( 3.5 < X < 5.5) = P(X < 5.5) - P(X < 3.5) \\\\P(3.5 < X < 5.5) \approx P(Z < \dfrac{5.5-5}{14.2}) - P(Z < \dfrac{3.5 - 5}{14.2})\\\\P(3.5 < X < 5.5) \approx P(Z < 0.113) - P(Z < -0.104)[/tex]

Using the z-tables, p value for z = 0.113 is 0.5438

and p value for z = -0.104 is 0.4602
Thus,

[tex]P(3.5 < X < 5.5) \approx P(Z < 0.113) - P(Z < -0.104)\\P(3.5 < X < 5.5) \approx 0.5438 - 0.4602 = 0.0836[/tex]

Learn more about standard normal distribution here:

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To answer this question, we must calculate the percentage of the total population in each SAT score category that was admitted to the university. We find that 55% of admitted applicants had a Math SAT scores of 700 or more.

First, multiply the overall percentage of applicants by the percentage admitted for each SAT score category. For those with scores of 700 or more, you get 36% * 38% = 13.68%. For those with scores less than 700, it's 18% * 62% = 11.16%. Then, add these results together to get the total percentage of all applicants who were admitted, which is 24.84%.

The next step is to calculate what fraction of this combined admitted students group got a Math SAT of 700 or more. The percentage of admitted students who had a SAT score of 700 or more is the percentage of admitted students in that category divided by the total percentage of all admitted students. So, you get 13.68% ÷ 24.84% = 55.05%. This rounds to 55% when expressed as a percentage, so we can say that 55% of admitted applicants had a Math SAT of 700 or more.

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Final answer:

To find the percentage of admitted applicants with a Math SAT of 700 or more, calculate the overall admission rate (AR), and then determine Group A's contribution to this rate. The percentage of admitted applicants with an SAT score of 700 or more is approximately 55% after rounding to the nearest percentage point.

Explanation:

To find the percentage of admitted applicants who had a Math SAT score of 700 or more, we can use the information provided to set up a weighted average problem. Let's denote the applicants with a Math SAT of 700 or more as Group A and those with a Math SAT of less than 700 as Group B.

From the information given:

36% of Group A were admitted.

18% of Group B were admitted.

38% of all applicants are in Group A; hence, 62% are in Group B (100% - 38%).

We want to find the percentage of all admitted students that had a Math SAT of 700 or more. The overall admission rate (AR) can be calculated as follows:

AR = (Percentage of Group A × Admission rate of Group A) + (Percentage of Group B × Admission rate of Group B)

AR = (38% × 36%) + (62% × 18%)

AR = 13.68% + 11.16%

AR = 24.84%

Next, we calculate the contribution of Group A to the overall admission rate:

Contribution from Group A = Percentage of Group A × Admission rate of Group A

Contribution from Group A = 38% × 36%

Contribution from Group A = 13.68%

Now, we find the percentage of the total admissions that were applicants with a Math SAT of 700 or more:

Percentage of admitted applicants with SAT ≥ 700 = (Contribution from Group A ÷ Overall admission rate) × 100

Percentage of admitted applicants with SAT ≥ 700 = (13.68% ÷ 24.84%) × 100

Percentage of admitted applicants with SAT ≥ 700 ≈ 55% (rounded to nearest percentage point)

Answer: There are 2.25×10³⁴ bacteria at the end of 24 hours.

Step-by-step explanation:

Since we have given that

Number of bacteria initially = 1

It triples in number every 20 minutes.

So, [tex]\dfrac{20}{60}=\dfrac{1}{3}[/tex]

So, our equation becomes

[tex]y=y_0e^{\frac{1}{3}k}\\\\3=1e^{\frac{1}{3}k}\\\\\ln 3=\dfrac{1}{3}k\\\\k=\dfrac{1.099}{0.333}=3.3[/tex]

We need to find the number of bacteria that it will contain at the end of 24 hours.

So, it becomes,

[tex]y=1e^{24\times 3.3}\\\\y=e^{79.1}\\\\y=2.25\times 10^{34}[/tex]

Hence, there are 2.25×10³⁴ bacteria at the end of 24 hours.

The colony will contain approximately 3,486,784,401 bacteria at the end of 24 hours.

The bacteria colony starts with 1 bacterium and triples in number every 20 minutes. To determine the number of bacteria at the end of 24 hours, we need to calculate the number of 20-minute intervals in 24 hours. There are 24 hours in a day, so there are 24 intervals of 20 minutes in a day. Therefore, the bacteria would triple in number 24 times. Starting with 1 bacterium, after 24 intervals, the number of bacteria would be:

1 * 3^24 = 1 * 3,486,784,401

So, at the end of 24 hours, the colony would contain approximately 3,486,784,401 bacteria.

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Final answer:

The research question that can be answered using a chi-square goodness-of-fit test is option c: Is the distribution of political party affiliation for all registered voters in 2012 the same as stated for 2008?

Explanation:

The research question that can be answered using a chi-square goodness-of-fit test in this scenario is option c: Is the distribution of political party affiliation for all registered voters in 2012 the same as stated for 2008?

In a chi-square goodness-of-fit test, we compare the observed frequencies (in this case, the distribution of political party affiliation in 2008) with the expected frequencies (the distribution we are testing, which is the distribution of political party affiliation in 2012). If the observed frequencies differ significantly from the expected frequencies, we can conclude that there is a significant difference between the two distributions.

In this case, the observed distribution is 36% Democrats, 27% Republicans, and 37% Independents in 2008. We want to test if this distribution is the same as the distribution in 2012. By performing a chi-square goodness-of-fit test, we can determine if there is a significant difference between the two distributions.

Answer:

0.0406, 0.8284,0.7887

Step-by-step explanation:

Given that Mattel Corporation produces a remote-controlled car that requires three AA batteries

X is N(34, 5.5)

Hence sample size of 25 would follow a t distribution with df = 24

This is because sample size <30

t distribution with df 24 would be bell shaped symmetrical about the mean and unimodal.

Std error of sample mean = std dev /sqrt n=[tex]\frac{5.5}{5} \\=1.1[/tex]

Prob (X>36) = [tex]P(t>\frac{36-34}{1.1} ) = P(t>1.82)\\= 0.04063[/tex]

i.e nearly 4.1% of the sample would have a mean useful life of more than 36 hours

X>33.5 implies [tex]t>-0.45[/tex]

=0.82837

=0.8284 proportion will have a mean useful life greater than 33.5 hours

Proportion between 33.5 and 36 hours

= [tex]0.3284+0.4593=0.7887[/tex]

The shape of the distribution of the sample mean is approximately normal due to the Central Limit Theorem. The standard error of the distribution of the sample mean can be calculated as 1.1 hours. Proportions of the sample mean falling above and between specific time intervals can be determined using z-scores and the Z-table.

The shape of the distribution of the sample mean, in this case, is approximately normal. This is because the distribution of the battery lives closely follows the normal probability distribution. When taking a sample of 25 batteries, the Central Limit Theorem states that the distribution of the sample mean approaches a normal distribution regardless of the shape of the original distribution.

The standard error of the distribution of the sample mean can be calculated using the formula: standard deviation / square root of sample size. In this case, the standard deviation is 5.5 hours and the sample size is 25. Therefore, the standard error is 5.5 / √25 = 1.1 hours.

To determine the proportion of the samples that will have a mean useful life of more than 36 hours, we need to calculate the z-score first. The formula for z-score is: (sample mean - population mean) / standard error. Plugging in the given values, we get (36 - 34) / 1.1 = 1.82. By looking up the z-score in the Z-table, we find the corresponding proportion is approximately 0.0344.

Similarly, to find the proportion of the sample that will have a mean useful life greater than 33.5 hours, we calculate the z-score: (33.5 - 34) / 1.1 = -0.45. By looking up the z-score in the Z-table, we find the corresponding proportion is approximately 0.3264.

To determine the proportion of the sample that will have a mean useful life between 33.5 and 36 hours, we subtract the proportion of the sample that will have a mean useful life greater than 36 hours from the proportion of the sample that will have a mean useful life greater than 33.5 hours. Therefore, 0.3264 - 0.0344 = 0.2920.

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Answer:

a. t-test for the population mean

Step-by-step explanation:

given that the average US woman wears 515 chemicals on an average day from her makeup and toiletries

A random sample from California found that on average the California woman wears 325 chemicals per day with a standard deviation of 90.5.

Note that here we have only sample std deviation and not population.

So better to use t test here.

Sample size is not given so whatever be the sample size here t test would be more appropriate.

Here we are testing for mean.  So the correct answer is

a. t-test for the population mean

Answer:

20.73 ≈ 21 persons per year

Step-by-step explanation:

Data provided in the question:

Total population of Texas = 17.7 million

People having diabetes = 1.45 million

Number of new cases of diabetes diagnosed for the first time = 366,921

Now,

the incidence of diabetes in Texas

= [ Number of new cases diagnosed ] ÷ [ Total population ]

= 366,921 ÷ 17.7 million

= 366,921 ÷ 17,700,000

= 0.02073

thus,

the incidence of diabetes in Texas per 1000 persons

=  0.02073 × 1,000

= 20.73 ≈ 21 persons per year

[tex]\bf \cfrac{1}{x-3}-6\implies \stackrel{\textit{using the LCD of x-3}}{\cfrac{1-(x-3)6}{x-3}}\implies \cfrac{1-6x+18}{x-3}\implies \cfrac{-6x+19}{x-3}[/tex]

for the vertical asymptote, we simply zero out the denominator and solve for "x"

x - 3 = 0

x = 3   <---- that's the only vertical asymptote

for the horizontal asymptote, well, let's notice the degrees of the numerator and denominator, in this case, the degree of the numerator is 1, and the degree of the denominator is 1, thus when that occurs, the horizontal asymptote occurs at the fraction from the leading terms' coefficients.

[tex]\bf \cfrac{-6x+19}{1x-3}\implies \stackrel{\textit{horizontal asymptote}}{\cfrac{-6}{1}\implies -6 = y}[/tex]

Answer:

a) [tex]tc=\pm 1.9601[/tex]

b) [tex]m=1.9601 \frac{26.34}{\sqrt{2322}}=1.0714[/tex]

c) The 95% confidence interval is given by (63.09;65.23)

Step-by-step explanation:

1) Notation and definitions

n=2322 represent the sample size

[tex]\bar X= 64.16[/tex] represent the sample mean

[tex]s=26.34[/tex] represent the sample standard deviation

m represent the margin of error

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".  

2) Calculate the critical value tc

In order to find the critical value is impornta to mention that we don't know about the population standard deviation, so on this case we need to use the t distribution. Since our interval is at 95% of confidence, our significance level would be given by [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2 =0.025[/tex]. The degrees of freedom are given by:

[tex]df=n-1=2322-1=2321[/tex]

We can find the critical values in excel using the following formulas:

"=T.INV(0.025,2321)" for [tex]t_{\alpha/2}=-1.9601[/tex]

"=T.INV(1-0.025,2321)" for [tex]t_{1-\alpha/2}=1.9601[/tex]

The critical value [tex]tc=\pm 1.9601[/tex]

3) Calculate the margin of error (m)

The margin of error for the sample mean is given by this formula:

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

[tex]m=1.9601 \frac{26.34}{\sqrt{2322}}=1.0714[/tex]

4) Calculate the confidence interval

The interval for the mean is given by this formula:

[tex]\bar X \pm t_{c} \frac{s}{\sqrt{n}}[/tex]

And calculating the limits we got:

[tex]64.16 - 1.9601 \frac{26.34}{\sqrt{2322}}=63.09[/tex]

[tex]64.16 + 1.9601 \frac{26.34}{\sqrt{2322}}=65.23[/tex]

The 95% confidence interval is given by (63.09;65.23)

Answer:

The point estimate for the population standard deviation of the length of the walking canes is 0.16.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\frac{\sigma}{\sqrt{n}}[/tex].

In this problem

The point estimate of the standard deviation is the standard deviation of the sample.

In a sample of 83 walking canes, the average length was found to be 34.9in. with a standard deviation of 1.5. By the Central Limit Theorem, we have that:

[tex]s = \frac{1.5}{\sqrt{83}} = 0.16[/tex]

The point estimate for the population standard deviation of the length of the walking canes is 0.16.

Answer:

[tex]\hat p =\frac{26}{39}=0.667[/tex]

Step-by-step explanation:

Some important concepts

A proportion refers to "the fraction of the total that possesses a certain attribute"

The data det ordered from the smallest to the largest value is:

3.1 ,3.71 ,3.71 ,3.74 ,3.98 ,3.98 ,4.15 ,4.16 ,4.16 ,4.32 ,4.34 ,4.51 ,4.51 ,4.56 ,4.57 ,4.57 ,4.57 ,4.61 ,4.62 ,4.63 ,4.64 ,4.64 ,4.64 ,4.64 ,4.86 ,4.96 ,5.04 ,5.11 ,5.12 ,5.37 ,5.37 ,5.38 ,5.39 ,5.47 ,5.48 ,5.62, 5.65 ,5.65 ,5.7

If we are interesed on the sample values below 5.0 we need to count how many values are below this number. If we do this we got that 26 numbers are below 5.0 and the total of numbers are 39.

so the proportion on this case would be"

[tex]\hat p =\frac{26}{39}=0.667[/tex]

To determine the proportion of rain samples with a pH below 5, count the number of data points less than 5.0 and divide it by the total number of data points. The pH value of the environment can vary greatly due to factors such as the presence of pollution.

This question is about finding out the proportion of rain samples that have a pH below 5.0 in Ingham County, Michigan. To do this, you count the number of data points less than 5.0 and divide it by the total number of data points. The pH of the environment (rain, in this case) can vary greatly due to factors such as pollution, affecting the acidity or alkalinity of the rain. Normal rainwater has a pH between 5 and 6 due to the presence of dissolved CO₂ which forms carbonic acid. Anything below 7.0 is acidic and above 7.0 is alkaline.

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Answer:

67,757 errors per million opportunities

Step-by-step explanation:

Assuming that each customer can only make a single complaint (1 error opportunity per customer), the number of errors per million opportunities (EPMO) is given by:

[tex]EPMO = \frac{complaints}{guests}*1,000,000 \\EPMO = \frac{29}{428}*1,000,000 \\EPMO = 67,757[/tex]

Refington hotel should expect 67,757 complaints per million guests.

Answer:

z-score is 2.16

98.46% of men are SHORTER than 6 feet 3 inches

Step-by-step explanation:

The heights of adult men in America are normally distributed, with a mean of 69.3 inches and a standard deviation of 2.64 inches.

Mean = [tex]\mu = 69.3 inches[/tex]

Standard deviation = [tex]\sigma = 2.64 inches[/tex]

a) If a man is 6 feet 3 inches tall, what is his z-score (to two decimal places)

x = 6 feet 3 inches

1 feet =12 inches

6 feet = 12*6 = 72 inches

So, x = 6 feet 3 inches  = 72+3=75 inches

Formula : [tex]Z=\frac{x-\mu}{\sigma}[/tex]

[tex]Z=\frac{75-69.3}{2.64}[/tex]

[tex]Z=2.159[/tex]

So, his z-score is 2.16

No to find percentage of men are SHORTER than 6 feet 3 inches

We are supposed to find P(x< 6 feet 3 inches)

z-score is 2.16

Refer the z table

P(x< 6 feet 3 inches) =0.9846

So, 98.46% of men are SHORTER than 6 feet 3 inches

Answer:

  c.  20 baskets per hour

Step-by-step explanation:

In this context, "per" can be considered to mean "divided by." Then to find the unit rate in terms of baskets per hour, we compute ...

  rate = (baskets)/(hours) = (60 baskets)/(3 hours) = (60/3) baskets/hour

  rate = 20 baskets/hour . . . . matches choice C

Answer:

(a) 6.2843 billion

(b) 6.8716 billion

Step-by-step explanation:

Since there is a constant growth rate of 1.5% per year, and the population in 2002 was 6.1 billion people. The general equation for the total world population, in billions, after 2002 is:

[tex]P(t) = 6.1*(1+0.015)^t[/tex]

With t being the time, in years, after 2002.

a) What was the population in 2004?

[tex]t=2004-2002=2\\P(2) = 6.1*(1.015)^2\\P(2) = 6.2843 \ billion[/tex]

b) What was the population in 2010?

[tex]t=2010-2002=8\\P(8) = 6.1*(1.015)^8\\P(8) = 6.8716 \ billion[/tex]

Answer:

(x + 3)^2 + (y + 2)^2 = 98.

Step-by-step explanation:

(x - h)^2 + (y - k) ^2 = r^2   ( center is (h, k) and r = radius.)

So we have:

(x + 3)^2 + (y + 2)^2 = r^2

When x = 4 y = 5, so

(4 + 3)^2 + (5 + 2)^2 = r^2

r^2 =  7^2 + 7^2 =  98

.

Answer:

Step-by-step explanation:

(x+2)^2+(y+3)^2=100

Answer:

There is an association between drying choice and illness

a.True

Step-by-step explanation:

H_0: There is no association between the drying choice and ilness

H_a: There is an association between drying choice and illness

(Two tailed chi square test)

Observations are given as per the following table

Observed    

Sick Not sick  

Paper 19 27 46

Hot air 10 42 52

29 69 98

Expected row total*col total/grand total  

Paper 13.6122449 32.3877551 46

Hot air 15.3877551 36.6122449 52

29 69 98

Chi square =(observed-expected)^2/Expected    

Paper 2.132484778 0.896261718 3.028746496

Hot air 1.886428842 0.792846905 2.679275747

4.01891362 1.689108623 5.708022243

Thus chi square = 5.708

degrees of freedom = (r-1)(c-1) = 1

p value = 0.0167

Since p <0.05 our alpha we reject null hypothesis

At 5% significance level, we can say

There is an association between drying choice and illness

a.True

Answer:

The total number of ways are 168.

Step-by-step explanation:

Consider the provided information.

There are 7 junior and 3 senior coders in her group.

The first project can be written by any of the coders. The second project must be written by a senior person and the third project must be written by a junior person.

For second project we have 3 choices and for third project we have 7 choices.

Now there are 2 possible case:

Case I: If first and second coder is senior, then the total number of ways are:

[tex]3\times 2\times 7=42[/tex]

Case II: If first and third coder is junior, then the total number of ways are:

[tex]7\times 3\times 6=126[/tex]

Hence, the total number of ways are: 42+126=168

The manager has 180 different ways to assign the three coders to the three different projects based on the given conditions.

The question is about the number of ways to assign three coders to three different projects, respecting certain conditions. This is a combinatorics problem. The first coder can be selected from any coder in the group (10 coders) so we have ten options. The second coder must be a senior so we have 1×3 options (as we have already selected one coder for the first project and we can't assign them to more than one project). Similarly, the third coder must be a junior from the remaining 9 coders (3 seniors and 6 juniors), hence we have 1×6 options for this choice. So, the total number of ways to assign the coders is the product of these options: 10×3×6 = 180 different ways.

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Answer:

(4,8)

Step-by-step explanation:

We have the two equations:

[tex]$ 4x + 2y = 32 \hspace{15mm} .....(1) $[/tex] and

[tex]$ 5x + 2y = 36 \hspace{15mm}  .....(2) $[/tex]

Subtracting (1) from (2):

⇒ 5x + 2y -4x - 2y = 36 - 32

⇒                  x = 4

Substituting the value of x in (1), we get:

4(4) + 2y = 32

⇒ 2y = 16

⇒                      y = 8

We write the solution in ordered pair as (x,y) = (4,8).

Answer:

The answer is the ordered pairs (4,8) because we used the system equations to solve for x and y.

Step-by-step explanation:

Answer:the probability that 2 will stick with their tour group is 0.21

Step-by-step explanation:

The survey found that 23% of the respondents stick with their tour group. This means that the probability of success,

p is 23/100 = 0.23

Probability of failure, q = 1 - p = 1 - 0.23

q = 0.77

Number of international travellers sampled, n is 6

Assuming a binomial distribution for the responses, the formula for binomial distribution is

P(x = r) = nCr × q^(n-1) × p^r

To determine the probability that 2 will stick with their tour group, we would determine

P( x= r = 2). It becomes

6C2 × 0.77^(6-1) × 0.23^2

= 15 × 0.77^5 × 0.0529

= 0.21

The probability that exactly 2 out of 6 international travelers will stick with their tour group, given that the probability of an individual sticking with the group is 23%, is approximately 27.81%.

The problem described involves calculating the probability of a specific number of successes in a series of independent trials and is an example of a binomial probability problem. In this case, with the probability that a person sticks with their tour group being 23%, or 0.23, and a sample of 6 international travelers, we want to find the probability that exactly 2 out of these 6 travelers stick with their group. This can be calculated using the binomial probability formula:

P(X = k) = [tex]C(n, k) \times p^k \times (1-p)^{n-k}[/tex]

Where:

P(X = k) is the probability of k successes in n trials

C(n, k) is the combination of n things taken k at a time

p is the probability of success on a single trial

n is the number of trials

k is the number of successes in n trials

For our problem:

P(X = 2) = C(6, 2) x [tex]0.23^2[/tex]x[tex](1-0.23)^{6-2}[/tex]

Calculating the combination, we have C(6, 2) = 6! / (2! x (6-2)!).

Thus:

P(X = 2) = 15 x [tex](0.23^2)[/tex] x [tex](0.77^4)[/tex]

P(X = 2) = 15 x 0.0529 x 0.3515

P(X = 2) = 0.2781

Therefore, the probability that exactly 2 of the 6 international travelers will stick with their tour group is approximately 0.2781, or 27.81%.

Answer:

The average blood pressure were higher than after

Step-by-step explanation:

As twenty people are selected and their blood pressure measured. After one month as stress reduction program BP were measured .Confidence interval is 95% so we conduct that blood pressure were higher than after reduction program.

Answer:

The p-value should be smaller than 0.05.

Step-by-step explanation:

The only information we have is the 95% confidence interval for the difference of means. As the lower bound is positive, we are 95% confident that the reduction program had a effect in the blood pressure level.

Then, as the program had a effect in the blood pressure level, we know that the null hypothesis, that states that the blood pressure levels would no change significantly, is rejected.

If the significance level is 0.05, according to the confidence of the interval, to reject the null hypothesis, the p-value had to be lower than the significance level.

Then, the p-value should be lower than 0.05.

Answer:

X[bar]= 115

Step-by-step explanation:

Hello!

Every Confidence interval to estimate the population mean are constructed following the structure:

"Estimator" ± margin of error"

Wich means that the intervals are centered around the sample mean. To know the value of the sample mean you have to make the following calculation:

[tex]X[bar]= \frac{Upper bond + Low bond}{2}[/tex]

[tex]X[bar]= \frac{115.6+114.4}{2}[/tex] = 115

Since both intervals were calculated with the information of the same sample, you can choose either to calculate the sample mean.

I hope it helps!

Answer:

15th term =29/3

16th term = 31/3

Step-by-step explanation:

Given an arithmetic sequence with the first term a1 and the common difference d , the nth (or general) term is given by an=a1+(n−1)d .

First we find the 15th term

n=15

a1=1/3

d=1 - 1/3 = 2/3

Solution

1/3+(15-1)2/3

1/3+28/3

(1+28)/3

29/3

Lets find the 16th term

1/3+(16-1)2/3

1/3+30/3

(1+30)/3

31/3