The U.S. Bureau of Labor Statistics reports that 11.3% of U.S. workers belong to unions. Suppose a sample of 400 U.S. workers is collected in 2014 to determine whether union efforts to organize have increased union membership at 0.025 level of significance. The sample results in a test statistic (z) of 2.2.

We conclude that union membership increased in 2014. (Enter 1 if the conclusion is correct. Enter 0 otherwise.)

Answer:

D)

Step-by-step explanation:

Remember, if a is critical point of f then f'(a)=0. And criterion of the second derivative says that if a is a critical point of f and

1. if [tex]f''(a)<0[/tex] then f has a relative maximum in (a,f(a)),

2. if [tex]f''(a)>0[/tex] then f has a relative minimum in (a,f(a)),

3. if [tex]f''(a)=0,[/tex] Then the criterion does not decide. That is,  f may have a relative maximum at a, a relative minimum at (a, f (a)) or neither.

Since [tex]f'(5)=0[/tex] then 5 is a critical point of f. Now we apply the second criterium:

since [tex]f''(5)=0[/tex] then the criterium doesn't decide, that means, more information is needed to determine if f has a maximum or minimum at x = 5.

Using the concept of critical point and the second derivative test, it is found that the correct option is:

D. More information is needed to determine if f has a maximum or minimum at x = 5.

Applying the second derivative test, we have that:

In this problem:

D. More information is needed to determine if f has a maximum or minimum at x = 5.

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Answer:

Both are binomials.

Step-by-step explanation:

Given that

a) X is the number of dots on the top face of fair die that is rolled.

When a fair die is rolled, there will be 1 to 6 numbers on each side with dots in that.  Each time a die is rolled the events are independent.  Hence probability of getting a particular number in the die is 1/6. There will be two outcomes either the number or not the number.  Hence X no of times we get a particular number of dots on the top face of fair die that is rolled is binomial with n = no of rolls, and p = 1/6

b) X is the number of defective parts in a sample of ten randomly selected parts coming from a manufacturing process in which 0.02% of all parts are defective.

Here X has two outcomes whether defective or non defective.  EAch part is independent of the other in the sense that the probability for each trial is constant with 0.02% =p and no of trials = n = 10.

Final answer:

The volume of the pancreas is estimated using the Midpoint Rule by adding the cross-sectional areas and multiplying by the distance between sections, resulting in an estimated volume of 100.5 cubic centimeters.

Explanation:

The question asks us to use the Midpoint Rule to estimate the volume of the human pancreas given its length (12 cm) and cross-sectional areas at 1 cm intervals. To do this, we add up the areas of the cross-sections (excluding the first and last since they're 0) and multiply by the distance between the sections (1 cm). This approach approximates the volume using cylindrical segments where each segment's volume is its cross-sectional area times the height (1 cm).

The given cross-sectional areas are 7.9, 15.3, 18.1, 10.2, 10.7, 9.5, 8.5, 7.8, 5.6, 4.2, and 2.7 square cm. Adding these gives a total of 100.5 square cm. Since the distance between each section is 1 cm, multiplying 100.5 by 1 cm gives an estimated pancreatic volume of 100.5 cubic centimeters.

This method, while not perfectly accurate due to it being an approximation, provides a useful estimate of the volume based on the available data. It's a practical application of the Midpoint Rule in estimating volumes from cross-sectional data.

Answer:

138

Step-by-step explanation:

Population standard deviation = [tex]\sigma = 120[/tex]

.95 probability of estimating the population mean monthly income within a margin of $20

So, Significance level = 1-0.95 = 0.05

α =0.05

Margin error = 20

[tex]ME =Z \times \frac{\sigma}{\sqrt{n}}[/tex]

Z at 0.05 = 1.96

[tex]20 =1.96 \times \frac{120}{\sqrt{n}}[/tex]

[tex]\sqrt{n} =1.96 \times \frac{120}{20}[/tex]

[tex]n =(1.96 \times \frac{120}{20})^2[/tex]

[tex]n =138.2976[/tex]

So, n = 138

Hence sample size should be 138 selected to obtain a .95 probability of estimating the population mean monthly income within a margin of $20

Answer:

  [tex]\dfrac{1}{x^2}+\dfrac{y^2}{x^4}[/tex]

Step-by-step explanation:

Separate the fraction(s) at the plus sign and simplify.

[tex]\dfrac{x^2+y^2}{x^4}=\dfrac{x^2}{x^4}+\dfrac{y^2}{x^4}\\\\=\dfrac{1}{x^2}+\dfrac{y^2}{x^4}[/tex]

Answer:

a. purple to orange  =[tex]\frac{5}{6}[/tex]

b. purple to all beads=[tex]\frac{5){11}[/tex]

c. all beads to orange=[tex]\frac{11}{6}[/tex]

Step-by-step explanation:

Given:

Colour of beads in jar = orange or purple

The ratio of orange to purple beads = 6 : 5

Solution:

Let number of orange beads be= 6x

Number of purple  beads=5x

Total no of beads= 6x+5x=11x

a. Ratio of  purple to orange.

Ratio of  purple to orange=[tex]\frac{\text{number of purple boads}}{\text{number of orange beads}}[/tex]

Ratio of  purple to orange=[tex]\frac{5x}{6x}[/tex]

Ratio of  purple to orange=[tex]\frac{5}{6}[/tex]

b. Ratio of purple to all beads

Ratio of  purple to all beads=[tex]\frac{\text{number of purple beads}}{\text{Total number of  beads}}[/tex]

Ratio of purple to all beads=[tex]\frac{5x}{11x}[/tex]

Ratio of purple to all beads=[tex]\frac{5}{11}[/tex]

c. Ratio of all beads to orange

Ratio of  purple to all beads=[tex]\frac{\text{Total number of  beads}}{\text{number of orange beads}}[/tex]

Ratio of purple to all beads=[tex]\frac{11x}{6x}[/tex]

Ratio of purple to all beads=[tex]\frac{11}{6}[/tex]

Answer:

a) 1.29

b) 1.29

c) 1.28

d) 1.23

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 60

Population standard Deviation = 10

a) Standard error with infinite population

[tex]\text{Standard error} = \displaystyle\frac{\sigma}{\sqrt{n}} = \frac{10}{\sqrt{60}} = 1.29[/tex]

For finite population with size N,

[tex]\text{Standard error} = \sqrt{\displaystyle\frac{N-n}{N-1}}\times \displaystyle\frac{\sigma}{\sqrt{n}}[/tex]

b) N = 60,000

[tex]\text{Standard error} = \sqrt{\displaystyle\frac{60000-50}{60000-1}}\times \displaystyle\frac{10}{\sqrt{60}} = 1.29[/tex]

c) N = 6,000

[tex]\text{Standard error} = \sqrt{\displaystyle\frac{6000-50}{6000-1}}\times \displaystyle\frac{10}{\sqrt{60}} = 1.28[/tex]

d) N = 600

[tex]\text{Standard error} = \sqrt{\displaystyle\frac{600-50}{600-1}}\times \displaystyle\frac{10}{\sqrt{60}} = 1.23[/tex]

To find the standard error of the mean, use the formula σx/√n for infinite population size and σx/√n * √((N - n)/(N - 1)) for finite population size. Use the finite population correction factor when the sample size is not negligible relative to the population size.

To find the value of the standard error of the mean in each case, we need to use the formula for the standard error of the mean, which is σx/sqrt(n), where σ is the population standard deviation and n is the sample size.

If the population size is infinite, we don't need to use the finite population correction factor. So, the standard error of the mean would be σ/√n.

In this case, the population size is finite, but the sample size is less than 5% of the population size. Therefore, we can still consider the population as effectively infinite. So, the standard error of the mean would be σ/√n.

In this case, the population size is finite and the sample size is not negligible relative to the population size. Therefore, we need to use the finite population correction factor, which is √((N - n)/(N - 1)). So, the standard error of the mean would be σ/√n * √((N - n)/(N - 1)).

In this case, the population size is finite and the sample size is not negligible relative to the population size. Therefore, we need to use the finite population correction factor. So, the standard error of the mean would be σ/√n * √((N - n)/(N - 1)).

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Answer:

(a) Hypotheses relevant to the research question are  

[tex]H_{0}: \mu_{1}-\mu_{2} = 0[/tex] vs [tex]H_{1}: \mu_{1}-\mu_{2} < 0[/tex] (lower-tail alternative) and

[tex]H_{0}: \mu_{1}-\mu_{2} = 0[/tex] vs [tex]H_{1}: \mu_{1}-\mu_{2} > 0[/tex] (upper-tail alternative).  

(b) There is no evidence that one mathematic learning technique is better than the other.

Step-by-step explanation:

Let's suppose that the data related to Technique A comes from population 1 and that the data related to Technique B comes from population 2. We have large sample sizes [tex]n_{1} = 90[/tex]  and [tex]n_{2} = 110[/tex]. The unbiased point estimate for [tex]\mu_{1}-\mu_{2}[/tex] is [tex]\bar{x}_{1} - \bar{x}_{2}[/tex], i.e., 26.1 - 27.9 = -1.8. The standard error is given by [tex]\sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}\approx[/tex] [tex]\sqrt{\frac{(7.2)^{2}}{90}+\frac{(12.4)^{2}}{110}}[/tex] = 1.4049.

(a) Hypotheses relevant to the research question are  

[tex]H_{0}: \mu_{1}-\mu_{2} = 0[/tex] vs [tex]H_{1}: \mu_{1}-\mu_{2} < 0[/tex] (lower-tail alternative) and

[tex]H_{0}: \mu_{1}-\mu_{2} = 0[/tex] vs [tex]H_{1}: \mu_{1}-\mu_{2} > 0[/tex] (upper-tail alternative).  

(b) The test statistic is [tex]Z = \frac{\bar{x}_{1} - \bar{x}_{2}-0}{\sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}}[/tex] and the observed value is [tex]z_{0} = \frac{-1.8}{1.4049} = -1.2812[/tex].  

The p-value for the lower-tail alternative is P(Z < -1.2812) =  0.1000617 and the p-value for the upper-tail alternative is P(Z > -1.2812) = 0.8999. The p-value is greater than 0.10 in both cases, therefore, we fail to reject the null hypotheses in both cases.

The null hypothesis states that the mean completion time using Technique A equals that using Technique B. The alternative hypothesis states that they are not equal. A t-test can be used to test these hypotheses. The test statistic and the P-value are calculated and compared with the given significance level (0.10) to make a conclusion.

Given the data provided for Techniques A and B, let’s formulate the hypothesis and carry out the test to determine the effectiveness of the techniques.

(a) State the Hypotheses

Here, the null hypothesis (H₀) is that the mean completion time using Technique A equals that using Technique B. The alternative hypothesis (H₁) is that the mean time using Technique A is not equal to that using Technique B.

(b) Hypothesis Testing

For the hypothesis test, we have to calculate the t-statistic and the P-value. Given that the data is normally distributed, we can use the two-sample t-test. The standard formula used to calculate the t-statistic is: (sample mean₁ - sample mean₂) / √[(sample variance₁/n₁) + (sample variance₂/n₂)]. After calculating these values, the resultant t-statistic can be compared to critical t values for the given level of significance (α=10%).

Next, compute the P-value using the t-statistic, which indicates the probability of observing such a difference in means under the null hypothesis. We reject the null hypothesis if the P-value is less than our significance level (α).

Without specific test statistics and P-value, we can’t make a definitive conclusion. In general, if we have a statistically significant result (P<0.10), we would reject the null hypothesis and conclude that there is a significant difference in the performance of Techniques A and B.

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Answer:

1044 sq. inches

Step-by-step explanation:

The frame adds 2 inches on each side of the window, so the width becomes 29 in. and the length becomes 36 in.

Area = width × length

So the total area of the window and the frame is:

A = 29 × 36 = 1044 in²

Answer:

t = 34.548

Step-by-step explanation:

Here  X₁ = Solid values / n

              = 26.3 +25.3+26.1+25.6+26.7+25.9/6

             =155.9 / 6

             =25.984

X₂ = Liquid values/n

    = 16.9 +16.6+16.5+17.4+17.4+17.2 / 6

    =102/6 = 17

Formula for the sample standard deviation is

[tex]s = \sqrt {Σ(x - mean )^2/n-1}[/tex]

we get

s₁ = 0.49967

s₂ = 0.3949

Construction of hypothesis

H₀ :μ₁ = μ₂

H₁ : μ₁ ≠ μ₂

Apply test statistic formula we get the value

[tex]t = X_{1}  - X_{2}  /\sqrt{S_{1}^2 /n + S_{2}^2 /n }[/tex]

putting all these values

t = 25.984 - 17 / √ (o.49967)²/6 + (0.39497)²/6

t = 34.548

To determine the test statistic for the difference in the average amount of saturated fat in solid and liquid fats, a two-sample t-test would be performed. Without the specific data, the precise value cannot be calculated. The formula for the test statistic in a two-sample t-test is (mean1 - mean2) / sqrt[(sd1^2/n1) + (sd2^2/n2)].

To answer your question, we want to perform a two-sample t-test to determine if there is a significant difference in the average amount of saturated fat in solid and liquid fats. However, without the specific data of saturated fat percentage in both types of fats, we can't calculate the exact test statistic. The test statistic in a two-sample t-test is calculated using the difference between the two sample means divided by the pooled standard error of the means. The formula is as follows:

Test statistic (t) = (mean1 - mean2) / sqrt[(sd1^2/n1) + (sd2^2/n2)]

Where 'mean1' and 'mean2' are the sample means, 'sd1' and 'sd2' are the sample standard deviations, and 'n1' and 'n2' are the sample sizes of the two sets of data respectively.

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a. Using Hamilton's method, 4 counselors should be assigned to Lowell, and 9 counselors should be assigned to Fairview.

b. The same divisor, 2 counselors should be hired for the new school.

Hamilton's method is used to allocate resources, such as counselors, among several entities based on a certain divisor. The divisor is typically calculated by dividing the total number of students by the total number of counselors available.

Let's start by determining the divisor and assigning counselors to each school based on the given information:

1. Calculate the divisor:

[tex]\[ \text{Divisor} = \frac{\text{Total Students}}{\text{Total Counselors}} \]\\\text{Divisor} = \frac{3584 + 6816}{13} = \frac{10400}{13} \approx 800 \][/tex]

2. Assign counselors to each school:

[tex]\[ \text{Counselors Assigned to Lowell} = \frac{\text{Students Enrolled in Lowell}}{\text{Divisor}} \][/tex]

[tex]\[ \text{Counselors Assigned to Fairview} = \frac{\text{Students Enrolled in Fairview}}{\text{Divisor}} \][/tex]

[tex]\[ \text{Counselors Assigned to Lowell} = \frac{3584}{800} = 4.48 \approx 4 \][/tex]

[tex]\[ \text{Counselors Assigned to Fairview} = \frac{6816}{800} = 8.52 \approx 9 \][/tex]

So, using Hamilton's method, 4 counselors should be assigned to Lowell, and 9 counselors should be assigned to Fairview.

3. New School:

  If a new school with 1824 students is opened, we can use the same divisor to determine how many additional counselors should be hired for the new school:

[tex]\[ \text{Counselors for New School} = \frac{\text{Students in New School}}{\text{Divisor}} \][/tex]

[tex]\[ \text{Counselors for New School} = \frac{1824}{800} = 2.28 \approx 2 \][/tex]

So, using the same divisor, 2 counselors should be hired for the new school.

Answer:

The correct option is b) 1.70

Step-by-step explanation:

Consider the provided information.

The weekly sulfur dioxide emissions follow a normal distribution with a mean of 1000 ppm (parts per million) and a standard deviation of 25.

Thus, μ=1000 and σ = 25

The CEO wants to know if the mean level of emissions is different from 1000.

Therefore the null and alternative hypothesis is:

[tex]H_0:\mu =1000[/tex] and [tex]H_a:\mu \neq1000[/tex]

The sample size is n = 50 and [tex]\bar x=1006[/tex] ppm.

Now calculate the test statistic by using the formula: [tex]z=\frac{\bar x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

Substitute the respective values in the above formula.

[tex]z=\frac{1006-1000}{\frac{25}{\sqrt{50}}}[/tex]

[tex]z=\frac{6}{\frac{25}{\sqrt{50}}}=1.697\approx 1.70[/tex]

Hence, the correct option is b) 1.70

Answer:

1.25 liters of oil

Step-by-step explanation:

Volume in Beaker A = 1 L

Volume of Oil in Beaker A = 1*0.3 = 0.3 L

Volume of Vinegar in Beaker A = 1*0.7 = 0.7 L

Volume in Beaker B = 2 L

Volume of Oil in Beaker B = 2*0.4 = 0.8 L

Volume of Vinegar in Beaker B = 1*0.6 = 1.2 L

If half of the contents of B are poured into A and assuming a homogeneous mixture, the new volumes of oil (Voa) and vinegar (Vva) in beaker A are:

[tex]V_{oa} = 0.3+\frac{0.8}{2} \\V_{oa} = 0.7 \\V_{va} = 0.7+\frac{1.2}{2} \\V_{va} = 1.3[/tex]

The amount of oil needed to be added to beaker A in order to produce a mixture which is 60 percent oil (Vomix) is given by:

[tex]0.6*V_{total} = V_{oa} +V_{omix}\\0.6*(V_{va}+V_{oa} +V_{omix}) = V_{oa} +V_{omix}\\0.6*(1.3+0.7+V_{omix})=0.7+V_{omix}\\V_{omix}=\frac{0.5}{0.4} \\V_{omix}=1.25 \ L[/tex]

1.25 liters of oil are needed.

There is 1.25 litres of oil that should now be added to A to produce a mixture that is 60 per cent oil.

Given

Beaker A contains 1 litre which is 30 per cent oil and the rest is vinegar, thoroughly mixed up.

Beaker B contains 2 litres which are 40 per cent oil and the rest vinegar, completely mixed up.

Half of the contents of B are poured into A, then completely mixed up.

Contents in beaker A implies;

15% of oil in 1 litre = 0.15 litre of oil

So that, there are 0.15 litres of oil and 0.85 litres of vinegar in beaker A.

Contents in beaker B implies:

55% of oil in 2 litres = 1.1 litres of oil

So that, thee are 1.1 litres of oil and 0.9 litres of vinegar in beaker B.

Half of the contents of B poured into A implies that beaker A now contains:

0.15 litres + 0.55 litres = 0.7 litres of oil

0.85 litres + 0.45 litres = 1.3 litres of vinegar

Then,

The percentage of oil in A is;

[tex]=\dfrac{0.7}{2} \times 100\\\\= 35[/tex]

To increase the percentage of oil to 60%, then:

0.7 litres + 1.25 litres = 1.95 litres of oil

And The new total litres of the content in beaker A = 3.25 litres

[tex]=\dfrac{1.95}{3.25}\times 100\\\\=60 \rm \ percent[/tex]

Hence, 1.25 litres of oil should now be added to A to produce a mixture that is 60 per cent oil.

To know more about percentages click the link given below.

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Answer:

[tex](0.2935, 0.4265)[/tex]

Step-by-step explanation:

Given that a random sample of 200 books purchased at a local bookstore showed that 72 of the books were murder mysteries.

Sample proportion p follows the following characteristics

p 0.36

q 0.64

pq/n 0.001152

Std dev 0.033941125

For 95% confidence level we have critical value as

Z critical = 1.96

Hence margin of error = 1.96*std dev

Confidence interval = (sample proportion - margin of error, sample proportion + margin of error)

=[tex](0.2935, 0.4265)[/tex]

To construct the 95% confidence interval for the proportion of murder mystery books sold at a bookstore, we use the sample proportion, Z-score for 95% confidence, and sample size to find the margin of error and then add and subtract this from the sample proportion. The 95% confidence interval is approximately (0.2935, 0.4265).

To construct a 95% confidence interval for the true proportion of books sold by the store that are murder mysteries, we can use the formula for a confidence interval for a proportion, which is [tex]p \pm Z*(\sqrt{\frac{p(1-p)}{n}})[/tex]

Sample proportion (p) = 72÷200 = 0.36

Sample size (n) = 200

For a 95% confidence interval, we use the Z-score corresponding to 95% confidence level from the standard normal distribution table, which is approximately 1.96.

The margin of error (E) is calculated as follows:

[tex]E = 1.96 * \sqrt{\frac{0.36(1-0.36)}{200}}\\E = 1.96 * \sqrt{0.001152}[/tex]

E = 1.96 * 0.033941

E ≈ 0.0665

Now we can construct the confidence interval:

The lower boundary is p - E = 0.36 - 0.0665 ≈ 0.2935

The upper boundary is p + E = 0.36 + 0.0665 ≈ 0.4265

Therefore, the 95% confidence interval is (0.2935, 0.4265).

Answer: No, these response does not provide strong evidence that the 34% figure is not accurate for this region.

Step-by-step explanation:

Since we have given that

p = 0.34

x= 228

n = 750

So, [tex]\hat{p}=\dfrac{x}{n}=\dfrac{228}{750}=0.304[/tex]

So, hypothesis would be

[tex]H_0:p=\hat{p}\\\\H_a:p\neq \hat{p}[/tex]

So, test statistic value would be

[tex]z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}\\\\z=\dfrac{0.304-0.38}{\sqrt{\dfrac{0.38\times 0.62}{750}}}\\\\z=\dfrac{-0.076}{0.0177}\\\\z=-4.293[/tex]

At 95% confidence , z = 1.96

So, 1.96>-4.293.

So, we accept the null hypothesis.

No, these response does not provide strong evidence that the 34% figure is not accurate for this region.

The observed data doesn't provide enough evidence to suggest that the 34% figure reported by the CDC is inaccurate for this region.

let's break down the hypothesis test step by step with more detail.

1. Setting up the Hypotheses:

- Null Hypothesis (H0):The true proportion of obese adults in the region is 34%.

- Alternative Hypothesis (H1):The true proportion of obese adults in the region is not 34%.

2. Calculating the Expected Number of Obese Individuals:

To calculate the expected number of obese individuals in the sample under the assumption that the true proportion is 34%, we use the formula:

[tex]\[ \text{Expected number of obese individuals} = \text{Proportion} \times \text{Sample size} \][/tex]

So,

[tex]\[ \text{Expected number of obese individuals} = 0.34 \times 750 = 255 \][/tex]

3. Checking Conditions:

- The sample is randomly selected.

- The sample size is large enough for the Central Limit Theorem to apply (n = 750).

- Since the population size isn't provided, we'll assume it's much larger than the sample size (which is often the case with populations of adults in countries).

4. Calculating the Z-Score:

The formula for the z-score is:

[tex]\[ z = \frac{{\text{Observed proportion} - \text{Expected proportion}}}{{\sqrt{\frac{{\text{Expected proportion} \times (1 - \text{Expected proportion})}}{{\text{Sample size}}}}}} \][/tex]

So,

[tex]\[ z = \frac{{\frac{228}{750} - \frac{255}{750}}}{{\sqrt{\frac{255}{750} \times \frac{495}{750}}}} \]\[ z \approx \frac{{0.304 - 0.34}}{{\sqrt{\frac{191.25}{750}}}} \]\[ z \approx \frac{{-0.036}}{{\sqrt{0.255}}} \]\[ z \approx \frac{{-0.036}}{{0.505}} \]\[ z \approx -0.071 \][/tex]

5. Finding Critical Z-Value:

For a two-tailed test with a significance level of 0.05, the critical z-values are approximately ±1.96.

6. Making a Decision:

Since the calculated z-score (-0.071) falls within the range (-1.96, 1.96), we fail to reject the null hypothesis. This means there is not enough evidence to conclude that the true proportion of obese adults in the region is different from 34%.

In conclusion, based on the provided sample data, we cannot confidently claim that the 34% figure reported by the CDC is inaccurate for this region.

Answer:

Step-by-step explanation:

a) For a 95% confidence level: [tex]\( \$194,120 \) to \( \$205,880 \)[/tex]

b) For a 97% confidence level: [tex]\( \$193,490 \) to \( \$206,510 \)[/tex]

To solve this problem, we'll use the information provided and apply the formula for a confidence interval estimate for the population mean when the population standard deviation is known.

Given data:

- Sample size n: 36 days

- Sample mean [tex](\( \bar{x} \))[/tex]: $200,000

- Population standard deviation [tex](\( \sigma \)): $18,000[/tex]

(a) 95% Confidence Interval

For a 95% confidence interval, the critical value [tex]\( z^* \)[/tex] from the standard normal distribution is approximately 1.96.

The formula for the confidence interval is:

[tex]\[ \text{CI} = \bar{x} \pm z^* \cdot \frac{\sigma}{\sqrt{n}} \][/tex]

Calculate the standard error:

[tex]\[ SE = \frac{\sigma}{\sqrt{n}} = \frac{18,000}{\sqrt{36}} = \frac{18,000}{6} = 3,000 \][/tex]

Now, construct the confidence interval:

[tex]\[ \text{CI} = 200,000 \pm 1.96 \cdot 3,000 \][/tex]

[tex]\[ \text{CI} = 200,000 \pm 5,880 \][/tex]

Therefore, the 95% confidence interval estimate for the average daily sale is approximately [tex]\( \$194,120 \) to \( \$205,880 \)[/tex].

(b) For a 97% confidence interval, the critical value [tex]\( z^* \)[/tex] from the standard normal distribution is approximately 2.17.

Calculate the confidence interval using the same formula with the updated [tex]\( z^* \):[/tex]

[tex]\[ \text{CI} = \bar{x} \pm z^* \cdot \frac{\sigma}{\sqrt{n}} \][/tex]

[tex]\[ \text{CI} = 200,000 \pm 2.17 \cdot 3,000 \][/tex]

[tex]\[ \text{CI} = 200,000 \pm 6,510 \][/tex]

Therefore, the 97% confidence interval estimate for the average daily sale is approximately [tex]\( \$193,490 \) to \( \$206,510 \)[/tex].

Answer: The value of test statistic is 0.696.

Step-by-step explanation:

Since we have given that

[tex]n_1=1399\\\\x_1=411\\\\p_1=\dfrac{x_1}{n_1}=\dfrac{411}{1399}=0.294[/tex]

Similarly,

[tex]n_2=759, x_2=211\\\\p_2=\dfrac{x_2}{n_2}=\dfrac{211}{759}=0.278[/tex]

At 0.01 level of significance.

Hypothesis are :

[tex]H_0:P_1=P_2=0.5\\\\H_a:P_1\neq P_2[/tex]

So, the test statistic value would be

[tex]z=\dfrac{(p_1-p_2)-(P_1-P_2)}{\sqrt{\dfrac{P_1Q_1}{n_1}+\dfrac{P_2Q_2}{n_2}}}\\\\z=\dfrac{0.294-0.278}{\sqrt{0.5\times 0.5(\dfrac{1}{1399}+\dfrac{1}{759})}}\\\\z=\dfrac{0.016}{0.023}=0.696[/tex]

Hence, the value of test statistic is 0.696.

Answer:

Step-by-step explanation:

Hello!

You have the following experiment, a random sample of 106 people was made and they were asked if they dreamed in color. 80 persons of the sample reported dreaming in color.

The historical data from the 1940s informs that the population proportion of people that dreams in color is 0.21(usually when historical data is given unless said otherwise, is considered population information)

Your study variable is a discrete variable, you can define as:

X: Amount of people that reported dreaming in colors in a sample of 106.

Binomial criteria:

1. The number of observation of the trial is fixed (In this case n = 106)

2. Each observation in the trial is independent, this means that none of the trials will have an effect on the probability of the next trial (In this case, the fact that one person dreams in color doesn't affect or modify the probability of the next one dreaming in color)

3. The probability of success in the same from one trial to another (Or success is dreaming in color and the probability is 0.21)

So X~Bi(n;ρ)

In order to be able to run a proportion Z-test you have to apply the Central Limit Theorem to approximate the distribution of the sample proportion to normal:

^ρ ≈ N(ρ; (ρ(1-ρ))/n)

With this approximation, you can use the Z-test to run the hypothesis.

Now what the investigators want to know is if the proportion of people that dreams in color has change since the 1940s, so the hypothesis is:

H₀: ρ = 0.21

H₁: ρ ≠ 0.21

α: 0.10

The test is two-tailed,

Left critical value: [tex]Z_{\alpha/2} = Z_{0.95} = -1.64[/tex]

Right critical value: [tex]Z_{1-\alpha /2} = Z_{0.95} = 1.64[/tex]

If the calculated Z-value ≤ -1.64 or ≥ 1.64, the decision is to reject the null hypothesis.

If -1.64 < Z-value < 1.64, then you do not reject the null hypothesis.

The sample proportion is ^ρ= x/n = 80/106 = 0.75

Z=     0.75 - 0.21     = 12.83

    √[(0.75*0.25)/106]

⇒Decision: Reject the null hypothesis.

p-value < 0.00001 is less than 0.10

If you were to conduct a one-tailed upper test (H₀: ρ = 0.21 vs H₁: ρ > 0.21) with the information of this sample, at the same level 10%, the critical value would be [tex]Z_{0.90}[/tex]1.28 against the 12.83 from the Z-value, the decision would be to reject the null hypothesis (meaning that the proportion of people that dreams in colors has increased.)

I hope this helps!

Answer:

C. n=423

Step-by-step explanation:

1) Notation and important concepts

Margin of error for a proportion is defined as "percentage points your results will differ from the real population value"

Confidence=90%=0.9

[tex]\alpha=1-0.9=0.1[/tex] represent the significance level defined as "a measure of the strength of the evidence that must be present in your sample before you will reject the null hypothesis and conclude that the effect is statistically significant".

[tex]\hat p=0.5[/tex] represent the sample proportion of consumers in the Oconee County area who would react favorably to a marketing campaig. For this case since we don't have enough info we use the value of 0.5 since is equiprobable the event analyzed.

[tex]z_{\alpha/2}[/tex] represent a quantile of the normal standard distribution that accumulates [tex]{\alpha/2}[/tex] on each tail of the distribution.

2) Formulas and solution for the problem

For this case the margin of error for a proportion is given by this formula

[tex]ME=z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]   (1)

For this case the confidence level is 90% or 0.9 so then the significance would be

[tex]\alpha=1-0.9=0.1[/tex] and [tex]\alpha/2=0.05[/tex]

With [tex]\alpha/2=0.05[/tex], we can find the value for [tex]z_{\alpha/2}[/tex] using the normal standard distribution table or excel.

The calculated value is [tex]z_{\alpha/2}=1.644854[/tex]

Now from equation (1) we need to solve for n in order to answer the question.

[tex]\frac{ME}{z_{\alpha/2}}=\sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]  

Squaring both sides:

[tex](\frac{ME}{z_{\alpha/2}})^2=\frac{\hat p(1-\hat p)}{n}[/tex]

And solving for n we got:

[tex]n=\frac{\hat p(1-\hat p)}{(\frac{ME}{z_{\alpha/2}})^2}[/tex]

Now we can replpace the values

[tex]n=\frac{0.5(1-0.5)}{(\frac{0.04}{1.644854})^2}=422.739[/tex]

And rounded up to the nearest integer we got:

n=423

Correct graph is C

Step-by-step explanation:

Given two inequalities are:

1. [tex]y\leq -3x+1[/tex]

2.[tex]y\leq x+3[/tex]

Step1 :  Remove the inequalities

1. [tex]y =-3x+1[/tex]

2.[tex]y = x+3[/tex]

Step2 :  Finding intersection points of equations

By solving linear equation

[tex]y =-3x+1 =x+3 [/tex]

[tex]-3x+1 =x+3 [/tex]

[tex] x = -0.5[/tex]

Replacing value of x in any equations

we get,

[tex]y = x+3[/tex]

[tex]y =-0.5+3[/tex]

[tex]y = 2.5[/tex]

Therefore, Point of intersection is (-0.5,2.5)

Step3: Test of origin (0,0)

Here, If inequalities holds true for origin then, shades the graph towards the origin.

For equation 1.

[tex]y\leq -3x+1[/tex]

[tex]0\leq -3(0)+1[/tex]

[tex]0\leq +1[/tex]

True, Shade graph towards origin.

For equation 2.

[tex]y\leq x+3[/tex]

[tex]0\leq 0+3[/tex]

[tex]0\leq 3[/tex]

True, Shade graph towards origin.

Thus, Correct graph is C

Answer:

c

Step-by-step explanation:

took test

Answer:

  3

Step-by-step explanation:

The slope (m) is the ratio of the difference in y-values to the corresponding difference in x-values:

  m = (y2 -y1)/(x2 -x1)

  m = (9 -0)/(4 -1) = 9/3

  m = 3

The slope of the line is 3.

Answer:

yes

Step-by-step explanation:

Answer:

565 robots

Step-by-step explanation:

We can do this numerically, month by month:

At t = 0, sale is 10 units. Growth rate is 5 robots per month.

At t = 1, sale is 15 units. Growth rate is 7 robots per month.

At t = 2, sale is 22 units. Growth rate is 9 robots per month.

At t = 3, sale is 31 units. Growth rate is 11 robots per month.

At t = 4, sale is 42 units. Growth rate is 13 robots per month.

At t = 5, sale is 55 units. Growth rate is 15 robots per month.

At t = 6, sale is 70 units. Growth rate is 17 robots per month.

At t = 7, sale is 87 units. Growth rate is 19 robots per month.

At t = 8, sale is 106 units. Growth rate is 21 robots per month.

At t = 9, sale is 127 units. Growth rate is 23 robots per month.

So the total robots we can expect to sell by the end of tenth month is

565 robots.

Answer:

a then b

Step-by-step explanation:

a then b because you could mess up if you didn't draw a decision tree

Answer:

a) The probability that a pregnancy will last 308 days or longer is 0.0038

b) Babies who are born on or before 256 days are considered prematures.

Step-by-step explanation:

Let X be the random variable that represents the length of a pregnancy. Then, X is normally distributed with a mean of 268 days and a standard deviation of 15 days.  

a) The z-score related to 308 days is z = (308-268)/15 = 2.6667, so, the probability of a pregnancy lasting 308 days or longer is P(Z > 2.6667) = 0.0038

b) We are looking for a value q such that P(X < q) = 0.22, i.e., P((X-268)/15 < (q-268)/15) = 0.22, here, Z =  (X-268)/15 is a standard normal random variable and z = (q-268)/15 is the 22nd quantile of the standard normal distribution, i.e., z = -0.7722 =  (q-268)/15 and (-0.7722)(15) + 268 = q, i.e., q = 256.417, so, babies who are born on or before 256 days are considered premature.

Answer:

a) P(X>308) = 0.00383

b) 256.42 days

Step-by-step explanation:

Population mean (μ) = 268 days

Standard deviation (σ) = 15 days

a) P(X>308)?

The z-score for any length of pregnancy 'X' is given by:

[tex]z=\frac{X-\mu}{\sigma}[/tex]

The z-score for X=308 is:

[tex]z=\frac{308-268}{15}\\z=2.6667[/tex]

A z-score of 2.6667 is equivalent to the 99.617-th percentile of a normal distribution. Thus, the probability of a pregnancy lasting 208 days or longer is:

[tex]P(X>308)=1-0.99617\\P(X>308) = 0.00383[/tex]

b) X at which P(X) < 0.22?

At the 22-nd percentile, a normal distribution has an equivalent z-score of -0.772. Therefore, the length of pregnancy, X, that separates premature babies from those who are not premature is:

[tex]-0.772=\frac{X-268}{15}\\X= 256.42[/tex]

Answer:

  190

Step-by-step explanation:

C(n, k) = n!/(k!(n -k)!)

C(190, 1) = 190!/(1!(189!)) = 190/1 = 190

The number of combinations of 190 things taken 1 at a time is 190.

Answer:

a) New mean 78.07692308 thousands of dollars

b) The median does not vary

c) New standard deviation 150.1793799 thousands of dollars

Step-by-step explanation:

We are working in units of $1,000

a)What is the value of the mean after the change?

Let  

[tex]s_1,s-2,...,s_38[/tex]  

be the salaries of the employees that earn less than 100 units.

The mean of the 39 salaries is 55 units so

[tex]\displaystyle\frac{s_1+s_2+...+s_{38}+100}{39}=55[/tex]

and

[tex]s_1+s_2+...+s_{38}+100=55*39=2145[/tex]

By accident, the 100 on the left is changed to 1,000

[tex]s_1+s_2+...+s_{38}+100+900=55*39=2145+900\Rightarrow \\\\\Rightarrow s_1+s_2+...+s_{38}+1000=3045[/tex]

Dividing by 39 both sides, we get the new mean

[tex]\displaystyle\frac{s_1+s_2+...+s_{38}+1000}{39}=\displaystyle\frac{3045}{39}=78.07692308[/tex]

b)What is the value of the median after the change?

Since the number of data does not change and only the right end of the range of salaries is changed, the median remains the same; 55

c)What is the value of the standard deviation after the change?

The variance is 400, so

[tex]\displaystyle\frac{(s_1-70)^2+(s_2-70)^2+...+(s_{38}-70)^2+(100-70)^2}{39}=400\Rightarrow\\\\\Rightarrow (s_1-70)^2+(s_2-70)^2+...+(s_{38}-70)^2+(100-70)^2=400*39=15600[/tex]

Adding 864,000 to both sides we get

[tex](s_1-70)^2+(s_2-70)^2+...+(s_{38}-70)^2+(100-70)^2+864000=15600+864000\Rightarrow\\\\\Rightarrow (s_1-70)^2+(s_2-70)^2+...+(s_{38}-70)^2+(1000-70)^2=879600[/tex]

Dividing by 39 and taking the square root we get the new standard deviation

[tex]\displaystyle\frac{(s_1-70)^2+(s_2-70)^2+...+(s_{38}-70)^2+(1000-70)^2}{39}=\displaystyle\frac{879600}{39}=22553.84615\Rightarrow\\\\\Rightarrow \sqrt{\displaystyle\frac{(s_1-70)^2+(s_2-70)^2+...+(s_{38}-70)^2+(1000-70)^2}{39}}=150.1793799[/tex]

Final answer:

a) The new mean after the change is approximately $66,923,000. b) The new median after the change is approximately $60,961,000. c) The new standard deviation after the change is approximately $234,358,000.

Explanation:

a) To find the new value of the mean, we need to subtract the original largest number ($100,000) and add the new largest number ($1,000,000) to the sum of the salaries. The sum of the salaries can be calculated by multiplying the mean ($70,000) by the number of employees (39). So, the new sum of the salaries is $(70,000 × 39) - $100,000 + $1,000,000 = $2,610,000. Since there are still 39 employees, the new mean is $2,610,000 divided by 39, which is approximately $66,923. The value of the mean after the change is $66,923,000 (in units of $1000).

b) The median is the middle value in a list of numbers when they are ordered from smallest to largest. After the change, the values in the list would be: $55,000, $55,000, $55,000, ..., $55,000, $66,923, $66,923, ..., $1,000,000. Since there are 39 employees, the middle two values would be $55,000 and $66,923. To find the median, we take the average of these two values: ($55,000 + $66,923) / 2 = $60,961. The value of the median after the change is $60,961,000 (in units of $1000).

c) To find the new standard deviation, we need to recalculate it using the new values. First, we need to find the squared differences between each salary and the new mean. The sum of these squared differences is $(39 × ($66,923 - $66,923)^2) + $1,000,000. Then, we divide this sum by 39 and take the square root to find the new standard deviation. The value of the standard deviation after the change is approximately $234,358,000 (in units of $1000).

Answer:

The answer is  b

Step-by-step explanation:

cause we have the equation y- y1=m(x-x1)

so y-3=4/11(x-11)

Answer:

a) 0.9292

b)0.1515

Step-by-step explanation:

Explained in attachment.

Answer:

mwah, thank you so much for posting this question! you just saved my life!! :,)

Step-by-step explanation:

thank you thank you.

Mwah!!!!

Answer:

[tex]\mu_{\hat{p}}=0.19[/tex]

[tex]\sigma_{\hat{p}}=0.0785[/tex]

Step-by-step explanation:

We know that the mean and the standard error of the sampling distribution of the sample proportions will be :-

[tex]\mu_{\hat{p}}=p[/tex]

[tex]\sigma_{\hat{p}}=\sqrt{\dfrac{p(1-p)}{n}}[/tex]

, where p=population proportion and n= sample size.

Given : The proportion of students at a college who have GPA higher than 3.5 is 19%.

i.e. p= 19%=0.19

The for sample size n= 25

The mean and the standard error of the sampling distribution of the sample proportions will be :-

[tex]\mu_{\hat{p}}=0.19[/tex]

[tex]\sigma_{\hat{p}}=\sqrt{\dfrac{0.19(1-0.19)}{25}}\\\\=\sqrt{0.006156}=0.0784601809837\approx0.0785[/tex]

Hence , the mean and the standard error of the sampling distribution of the sample proportions :

[tex]\mu_{\hat{p}}=0.19[/tex]

[tex]\sigma_{\hat{p}}=0.0785[/tex]