A supervisor records the repair cost for 11 randomly selected refrigerators. A sample mean of $82.43 and standard deviation of $13.96 are subsequently computed. Determine the 99% confidence interval for the mean repair cost for the refrigerators. Assume the population is approximately normal. Step 1 of 2 : Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.

Answer: 0.0668

Step-by-step explanation:

Given : The volume of soda a dispensing machine pours into a 12-ounce can of soda follows a normal distribution with a mean of 12.45 ounces and a standard deviation of 0.30 ounce.

i.e. [tex]\mu=12.45[/tex] and [tex]\sigma=0.30[/tex]

Let x denotes the volume of soda a dispensing machine pours into a 12-ounce can.

Then, the proportion of the soda cans contain less than the advertised 12 ounces of soda will be :-

[tex]P(x<12)=P(\dfrac{x-\mu}{\sigma}<\dfrac{12-12.45}{0.30})\\\\=P(z<-1.5)\ \ [\because z=\dfrac{x-\mu}{\sigma}]\\\\=1-P(z<1.5)\ \ [\because P(Z<-z)=1-P(Z<z)]\\\\=1-0.9332\ [\text{By z-table}]\\\\=0.0668[/tex]

Hence, the proportion of the soda cans contain less than the advertised 12 ounces of soda = 0.0668

Final answer:

The proportion of 12-ounce soda cans that contain less than the advertised amount of soda is approximately 6.68%, which is found by calculating the z-score of 12 ounces and looking up the corresponding proportion in the standard normal distribution table.

Explanation:

To find the proportion of soda cans that contain less than the advertised 12 ounces of soda, we need to calculate the z-score for 12 ounces using the mean and standard deviation of the soda volumes. The z-score tells us how many standard deviations away from the mean a certain value is.

The z-score formula is:

Z = (X - μ) / σ

Where X is the value (12 ounces), μ is the mean (12.45 ounces), and σ is the standard deviation (0.30 ounces).

Plugging in the values we get:

Z = (12 - 12.45) / 0.30
Z = -0.45 / 0.30
Z = -1.5

Once we have the z-score, we can use the standard normal distribution table to find the proportion of values that lie below this z-score. The table gives us the proportion of the distribution that is to the left of the z-score. A z-score of -1.5 corresponds to a proportion of approximately 0.0668.

Therefore, the proportion of soda cans containing less than 12 ounces is approximately 0.0668, or 6.68%.

Answer:

C. The parent population must be normally distributed

Step-by-step explanation:

The parent population must follow a normal distribution in order to use a t procedure

Answer:

[tex]z=\frac{0.881 -0.8}{\sqrt{\frac{0.8(1-0.8)}{110}}}=2.124[/tex]  

Null hypothesis:[tex]p\geq 0.8[/tex]  

Alternative hypothesis:[tex]p < 0.8[/tex]  

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

[tex]p_v =P(Z<2.124)=0.983[/tex]  

So the p value obtained was a very high value and using the significance level assumed [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.

Be Careful with the system of hypothesis!

If we conduct the test with the following hypothesis:

Null hypothesis:[tex]p\leq 0.8[/tex]  

Alternative hypothesis:[tex]p > 0.8[/tex]

[tex]p_v =P(Z>2.124)=0.013[/tex]  

So the p value obtained was a very low value and using the significance level assumed [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis.

Step-by-step explanation:

1) Data given and notation  

n=110 represent the random sample taken

X=97 represent the students who completed the modules before class

[tex]\hat p=\frac{97}{110}=0.882[/tex] estimated proportion of students who completed the modules before class

[tex]p_o=0.8[/tex] is the value that we want to test

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

z would represent the statistic (variable of interest)

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

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion is less than 0.8 or 80%:  

Null hypothesis:[tex]p\geq 0.8[/tex]  

Alternative hypothesis:[tex]p < 0.8[/tex]  

When we conduct a proportion test we need to use the z statistic, 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].

3) Calculate the statistic  

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

[tex]z=\frac{0.881 -0.8}{\sqrt{\frac{0.8(1-0.8)}{110}}}=2.124[/tex]  

4) Statistical decision  

It's important to refresh the p value method or p value approach . "This method 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 assumed [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.  

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

[tex]p_v =P(Z<2.124)=0.983[/tex]  

So the p value obtained was a very high value and using the significance level assumed [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.

Be Careful with the system of hypothesis!

If we conduct the test with the following hypothesis:

Null hypothesis:[tex]p\leq 0.8[/tex]  

Alternative hypothesis:[tex]p > 0.8[/tex]

[tex]p_v =P(Z>2.124)=0.013[/tex]  

So the p value obtained was a very low value and using the significance level assumed [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis.

Based on the given data, we do not reject the null hypothesis because the p-value (0.1308) is greater than the alpha level (0.05). Therefore, we conclude that there is insufficient evidence to suggest that fewer than 80% of ST 311 Hybrid students complete all modules before class.

To determine whether fewer than 80% of ST 311 Hybrid students complete all modules before class, we conducted a hypothesis test with the following hypotheses:

We collected data from a sample of 110 students, where 97 had completed the modules before class. The test resulted in a p-value of 0.1308.

Conclusion: There is insufficient evidence to conclude that fewer than 80% of the students complete all modules before class.

Answer:

$48.72

Step-by-step explanation:

Step 1: identify the given parameters

Sky's gross annual income = $36,192

Step 2: calculate Sky's weekly salary

= 36,192/52

=$696/week

Step 3: calculate Sky's weekly 4% deduction

= $696 X 4%

=$696 X 0.04 = $27.84

Step 4: calculate Sky's weekly 3% deduction

= $696 X 3%

=$696 X 0.03 = $20.88

Step 5: calculate how much is deposited into Skye’s 403(b) each payday

= $27.84 + $20.88

=$48.72

Answer: 48.72

Step-by-step explanation:

She will need to use 4/5 cups of flour

Step-by-step explanation:

First of all we have to calculate the 60% of the flour used in actual recipe then it will be added to the original amount,

So,

Given

Cups of flour used in actual recipe = 1/2

60% of 1/2:

[tex]= \frac{1}{2} * \frac{60}{100}\\= \fra{60}{200}\\=\frac{3}{10}[/tex]

So the total cups of flour for 60% more pancakes will be:

[tex]= \frac{1}{2} + \frac{3}{10}\\\\= \frac{5+3}{10}\\\\=\frac{8}{10}\\\\=\frac{4}{5}[/tex]

Hence,

She will need to use 4/5 cups of flour

Keywords: Percentage, percent

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Answer:

True, they form a proportion because their cross-multiplication products are equal.

Step-by-step explanation:

[tex]\frac{49}{21} =\frac{28}{12}[/tex] True or False?

If they are a proportion, the products you get from cross-multiplying are the same.

Multiply the left numerator by right denominator:

49 X 12 = 588

Multiply the left denominator by right numerator:

21 X 28 = 588

588 = 588

Therefore it is true, 49/21 and 28/12 form a proportion.

Area of Jhons planter is 164.89 square inches

Solution:

Given that,

Jhons garden has a big planter that measures [tex]18\frac{2}{3}[/tex] in by [tex]8 \frac{5}{6}[/tex] inches

To find: area of Jhons planter

From given information,

[tex]\text{ length } = 18\frac{2}{3} = \frac{3 \times 18 + 2}{3} = \frac{56}{3} \text{ inches }[/tex]

[tex]\text{ width } = 8\frac{5}{6} = \frac{6 \times 8 + 5}{6} = \frac{53}{6} \text{ inches }[/tex]

The area of planter is given as:

[tex]area = length \times width[/tex]

[tex]area = \frac{56}{3} \times \frac{53}{6} = \frac{2968}{18} = 164.89[/tex]

Thus area of Jhons planter is 164.89 square inches

Answer:

[tex]t=\frac{(950 -915)-(0)}{\sqrt{\frac{50^2}{30}}+\frac{45^2}{25}}=2.73[/tex]  

[tex]p_v =2*P(t_{53}>2.73) =0.0086[/tex]  

So with the p value obtained 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 reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (Philadelphia) is  significantly different than the mean for the group 2 (Baltimore).  

Step-by-step explanation:

The system of hypothesis on this case are:  

Null hypothesis: [tex]\mu_1 = \mu_2[/tex]  

Alternative hypothesis: [tex]\mu_1 \neq \mu_2[/tex]  

Or equivalently:  

Null hypothesis: [tex]\mu_1 - \mu_2 = 0[/tex]  

Alternative hypothesis: [tex]\mu_1 -\mu_2\neq 0[/tex]  

Our notation on this case :  

[tex]n_1 =30[/tex] represent the sample size for group 1  (Philadelphia)

[tex]n_2 =25[/tex] represent the sample size for group 2  (Baltimore)

[tex]\bar X_1 =950[/tex] represent the sample mean for the group 1  

[tex]\bar X_2 =915[/tex] represent the sample mean for the group 2  

[tex]s_1=50[/tex] represent the sample standard deviation for group 1  

[tex]s_2=45[/tex] represent the sample standard deviation for group 2  

If we see the alternative hypothesis we see that we are conducting a bilateral test or two tail.

Critical values

On this case since the significance level is 0.05 and we are conducting a bilateral test we have two critical values, and we need on each tail of the distribution [tex]\alpha/2 = 0.025[/tex] of the area.  

The distribution on this case since we don't know the population deviation for both samples is the t distribution with [tex]df=n_1+n_2 -2= 30+25-2=53[/tex] degrees of freedom.

We can use the following excel codes in order to find the critical values:

"=T.INV(0.025,53)", "=T.INV(1-0.025,53)"

And we got: (-2.01, 2.01)

Calculate th statistic

The statistic is given by this formula:  

[tex]t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{\sqrt{\frac{s^2_1}{n_1}}+\frac{S^2_2}{n_2}}[/tex]  

And now we can calculate the statistic:  

[tex]t=\frac{(950 -915)-(0)}{\sqrt{\frac{50^2}{30}}+\frac{45^2}{25}}=2.73[/tex]  

The degrees of freedom are given by:  

[tex]df=30+25-2=53[/tex]

P value

And now we can calculate the p value using the altenative hypothesis:  

[tex]p_v =2*P(t_{53}>2.73) =0.0086[/tex]  

So with the p value obtained 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 reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (Philadelphia) is  significantly different than the mean for the group 2 (Baltimore).  

Answer:

Step-by-step explanation:

a)  While the mean is below 4 the standard deviation tells us that there is a pretty high chance for the value to be above 4.  One standard deviation away is 1.79 - 5.65.  We are concerned with things over 4 so we'll look at the upper half, which is 3.72 - 5.65.  Being one standard deviation to the right means there is a 34.1% chance of the readings being in this range.  And there is even chances of it being slightly higher, though that is comparatively low.  But even as low as 10% is usually considered too high a chance to risk.  If you don't understand how the standard deviation got me those percents let me know.

b) alternative hypothesis is always the option where we want to prove it.  So we want to prove the concentration is 4 or above.  So the null is less than 4 and the alternative is greater than or equal to 4.  Do you know the correct symbols?  if not I can get those written out.  As for the p value we need the confidence level for the question, do you have that?

The two hypothesis:

And the reasoning of the inspector is incorrect because it ignores the large standard deviation.

We know that the mean radon concentration must be smaller than 4.0 pC/L.

In this particular house, the mean is 3.72 pC/L with a really large standard deviation of 1.93 pC/L.

And the inspector says that the radon abatement is not necessary, as the mean is smaller than 4.0 pC/L.

Now, as you can see, the standard deviation is really large. This means that over a given period of time, the mean concentration per liter may be larger than 4.0 pC (and then decreases). But this would imply that the exposure over large periods of times could be really large. This is why the reasoning is incorrect.

b) The null hypothesis is what we want to prove. In this case, is that the mean concentration is smaller than 4.0 pC/L.

The alternative hypothesis is the other option, in this case, that the concentration is equal or larger than 4.0 pC/L.

using the correct notation and defining C as the concentration we can write:

If you want to learn more about null hypothesis, you can read:

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Answer:

D) 9

Step-by-step explanation:

7 x 2 - 4 - 9x

14 - 4 - 9x

10 - 9x

You could easily see at first that the coefficient of x is 9 because the coefficient is the number that is before the variable, but you can also do the whole process.

Hope this was helpful :)

To test the minister's claim, the correct null and alternative hypotheses would be H0 : p = .6 and Ha: p > .6, which aligns with the suggestion that more than 60% of the population attends religious services. These hypotheses represent equality and inequality respectively, and are tested to determine if there is sufficient statistical evidence for the claim.

To test the minister's claim, Letter b is the correct choice. The notation 'p' is typically used to denote a proportion in a population. The null hypothesis, denoted H0, would be that the proportion of the population who attend services (p) is equal to 0.6. The alternative hypothesis, denoted Ha, would state that the proportion of the population who attend services (p) is greater than 0.6 which aligns with the minister's claim.

Therefore:
H0 : p = .6
Ha: p > .6

This is because the minister’s claim is a statement of inequality (more than 60%), so the alternative hypothesis must reflect this particular inequality. We then test the null hypothesis to determine whether there is enough statistical evidence to accept the alternative hypothesis.

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Final answer:

Option b is the correct answer; H0: p = 0.6 and Ha: p > 0.6. These hypotheses are set for a right-tailed test to assess the claim that more than 60% of adults attend religious services monthly.

Explanation:

The correct answer to the question of which null and alternative hypotheses should be used to test the minister's claim that more than 60% of the adult population attends a religious service at least once a month is option b. The null hypothesis (H0) signifies the statement that is presumed true before the data evidence is considered, while the alternative hypothesis (Ha) represents the claim to be tested. In this case, the null hypothesis should state that the population proportion (p) equals 0.6, or H0: p = 0.6, indicating that 60% of the adult population attends a religious service at least once a month.

The alternative hypothesis should indicate that the proportion is greater than 0.6, or Ha: p > 0.6, which aligns with the minister's claim and sets up the hypothesis test for a right-tailed test. The other options are incorrect as they either test for a mean instead of a proportion, use the wrong symbol for the hypotheses, or present the hypothesis in an incorrect format.

Answer:

There is no effect of the good intentions i.e. to restrain their diet out of concern about their weight, have on their eating habits.

Step-by-step explanation:

The question is incomplete as some information is not provided, please refer below the remaining information of the question.

Here are the data on grams of potato chips consumed (note that the study report gave the standard error of the mean rather than the standard deviation):

Group          n     bar{x}         s

Unrestrained  9     63.83      24.72

Restrained  11     34      39.8

a. The standard error of the difference between sample means  ( ± 0.0001 )  is:

b. The critical value  ( ± 0.001 )  from the t distribution for confidence interval 80 % using the conservative degrees of freedom is:

c. Give a 80 % confidence interval   ( ± 0.01 )  that describes the effect of restraint:

Answer:

a) The standard error of the difference between sample means:

S E  =  √ s 2 1 /  n 1  +  s 2 2 /n 2

    =  √ s e 2 1 +  s e 2 2

    =  √ 24.72 2 +  39.8 2

    =  46.85

b) The degrees of freedom:

D f  =  n 1  + n 2  − 2

=  9  +  11  −  2

= 18

The confidence level  =  0.80

The significance level,  α  =  0.20

The critical value from the t distribution for confidence interval 80%:

t c r i t i c a l  =  t α / 2 ,  d f  =  t 0.10 , 18  =  ±  1.33

c) The 80% confidence interval:

( μ 1 − μ 2 ) = ( ¯ x 1 − ¯ x 2 )  ±  t ⋅ S E

= ( 63.83 − 34 ) ±  1.33 × 46.85  

= 29.83 ± 62.31

− 32.48 < ( μ 1 − μ 2 ) <  92.14

As the interval contains the zero. So, it can be concluded that there is no effect of the good intentions i.e. to restrain their diet out of concern about their weight, have on their eating habits.

Good intentions regarding diet may impact eating habits among college students, but environmental influences often override intentions.

The effect of good intentions on eating habits, especially among those who are trying to restrain their diet due to concerns about weight, can vary greatly. Research shows that consumption patterns, such as an increase in eating fast food and sugary beverages, coupled with a decrease in physical activity, contribute to significant weight gain among college students. Implementing healthier eating and physical activity habits is crucial, but it is also important to acknowledge that neither changing diets nor exercise are effective on their own for preventing health issues. Both elements must be combined to reduce the risk of chronic diseases such as cardiovascular disease and cancers.

Good intentions may have some impact on eating habits, but the susceptibility to environmental influences, such as availability of unhealthy snacks, can override these intentions. Therefore, colleges and universities are actively pursuing comprehensive approaches to encourage both healthy eating and increased physical activity. For individuals concerned about their diet and weight, it is essential to engage in consistent healthy behaviors, rather than relying purely on intentions.

Answer:

3

Step-by-step explanation:

View Image

Answer:

Hence safely 9 watermelons can be placed in a single container.

Step-by-step explanation:

Given that the  weight (in pounds) of ``medium-size'' watermelons is normally distributed with mean 15 and variance 4.

X = weight in pounds of medium size watermelons is N(15, 2)

Let the water melons stored be n

Then the sample of n has a mean of (15) and std error = [tex]\frac{2}{\sqrt{n} }[/tex]

Capacity = 140

Hence we can say either 8 or 9

If n=9, we have weight = 15*9+1.96*2/3 = 136.40

Hence safely 9 watermelons can be placed in a single container.

Answer:

option C

Step-by-step explanation:

given,

[tex]\bar{x_1} = 158.706, \sigma_1 = 25.36 , n_1= 17[/tex]

[tex]\bar{x_1} = 122.471, \sigma_1 = 25.183 , n_2= 17[/tex]

α = 1 - 0.95 = 0.05

degree of freedom (df) = 17 -1 = 16

critical value[tex]= t_{\alpha/2},df = t_{0.025},16 =2.120[/tex] (from t-table)

margin of error = [tex]t_{\alpha/2}\sqrt{\dfrac{s_1^2}{n_1}+\dfrac{s_2^2}{n_2}}[/tex]

=2.120\times \sqrt{\dfrac{25.36^2}{17}+\dfrac{25.183^2}{17}}[/tex]

= 2.120 x 8.6467

= 18.33

Margin of error = 18.33

Point estimation of difference = [tex]\bar{x_1} - \bar{x_2}[/tex]

                                                  = 36.235

lower limit = 36.235 - 18.33 = 17.91

upper limit = 36.235 + 18.33 = 54.57

hence, the nearest option near to answer is option C

Probability of graduating is 73% = 0.73

Probability of not graduating = 1-73% = 0.27

Number of students (n) = 15

Number of graduates (x) = 8

Use the binomial probability formula:

P(x) = (n/x) *p^x * (1-p)^n-x

P(8) = 15/8 * 0.73^8 * 0.27^7

P(8) = 0.0543

Rounded to nearest hundredth = 0.05

Final answer:

To find the probability, use the binomial formula: [tex]\( P(X = 8) = \binom{15}{8} \times (0.73)^8 \times (0.27)^7 \).[/tex] Calculate to get approximately 0.195.

Explanation:

To calculate the probability of exactly 8 out of 15 freshmen graduating, we can use the binomial probability formula:

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

Where:

( n = 15 ) (total number of students)

( k = 8 ) (number of successes, i.e., number of students graduating)

( p = 0.73 ) (probability of success, i.e., the probability that a student graduates)

Using these values, we can plug them into the formula:

[tex]\[ P(X = 8) = \binom{15}{8} \times (0.73)^8 \times (1-0.73)^{15-8} \][/tex]

[tex]\[ P(X = 8) = \binom{15}{8} \times (0.73)^8 \times (0.27)^7 \][/tex]

Using a calculator or statistical software to compute the binomial coefficient, we find:

[tex]\[ \binom{15}{8} = 6435 \][/tex]

Plugging this into the formula:

[tex]\[ P(X = 8) = 6435 \times (0.73)^8 \times (0.27)^7 \][/tex]

[tex]\[ P(X = 8) \approx 0.195 \][/tex]

So, the probability that exactly 8 out of 15 freshmen graduate is approximately 0.195, rounded to the nearest hundredth.

Answer:

0.5 or 50%

Step-by-step explanation:

For any given value of 'x' representing the time between arrivals of two customers. If 0 < x <120, then the cumulative distribution function is:

[tex]\frac{x-0}{120-0}=\frac{x}{120}[/tex]

Therefore, the probability that the time between the arrivals of two customers will be more than 60 seconds is determined by:

[tex]P(X>60) = 1 -\frac{60}{120}\\P(X>60) = 0.5[/tex]

The probability is 0.5 or 50%.

Answer:

[tex]f(4) = 1501 + 9e^{-2*4} = 1501.00[/tex]

Step-by-step explanation:

f(4) is the value of f when x = 4.

We have that

[tex]f(x) = 1501 + 9e^{-2x}[/tex]

So

[tex]f(4) = 1501 + 9e^{-2*4} = 1501.00[/tex]

Answer:1501

Step-by-step explanation:

The given logistic model is expressed as

f(x)=1501+9e−2x

To evaluate the function at f(4), we would substitute x = 4 into the given logistic model. It becomes

f(4)=1501+9e−2 × 4

f(4)=1501+9e−8

Input 1501 plus 9 plus shift Ln plus -8 in a calculator. It becomes

1501 + 0.00302 = 1501.00302

Approximating to the nearest tenth, it becomes I501

Answer:

The expected number of tickets written would be 6.5 per day.

Step-by-step explanation:

Given that the Sutton police  department must write, on average, 6 tickets a day to keep department revenues at budgeted levels.

Suppose the number of tickets written per day follows a Poisson distribution with a mean of 6.5 tickets per day.

This means population parameter i.e. expected value of tickets a day is 6.5

Option I is wrong, because 6.5 is the parameter of Poisson distribution for number of tickets

Option III is wrong because mean is the overall average and hence need not be balanced by exactly half.

Option IV is wrong because mean is the expected value and so we cannot say more than 6.5 tickets

Only correct option is option 2.

The expected number of tickets written would be 6.5 per day.

Answer:

As the distance between the y-intercepts of f(x) and g(x) is 3.

So, the y-intercept of f(x) is 3 units below g(x), as shown in attached figure a.

Step-by-step explanation:

Let us consider the equation of straight line in the form of slope-intercept form such as:

[tex]y = mx + b[/tex]      

where m is the slope of the line and b is the y-intercept.

Determining the y-intercept of f(x) = x - 1

As the given function f(x) = x-1 is in the form of [tex]y = mx + b[/tex].

So, the y-intercept will be -1.

y-intercept can also be computed by putting x = 0, as at y-intercept the value of x is zero.

f(x) = x - 1

y = 0 - 1 ⇒ y = -1

Determining the y-intercept of g(x) = x + 2

As the given function g(x) = x + 2 is in the form of [tex]y = mx + b[/tex].

So, the y-intercept will be 2.

y-intercept can also be computed by putting x = 0, as at y-intercept the value of x is zero.

g(x) = x + 2

y = 0 + 2 ⇒ y = 2

Determining the distance between the y - intercepts of f(x) and g(x)

So, the distance between the y - intercepts of f(x) and g(x): 2 - (-1) = 3

So, the y-intercept of f(x) is 3 units below g(x), as shown in attached figure a.

Keywords: y-intercept, graph, function

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Answer:

There is an 8.22% probability that a randomly selected person has a birthday in November.

Step-by-step explanation:

The theoretical method to find the probability is the division of the number of desired outcomes by the number of total outcomes.

A randomly selected person has a birthday in November

There are 365 days in a year, so the number of total outcomes is 365.

There are 30 days in november, so the number of desired outcomes is 30.

So the probability is

[tex]P = \frac{30}{365} = 0.0822[/tex]

There is an 8.22% probability that a randomly selected person has a birthday in November.

Answer:

The probability of a randomly selected person has a birthday in November is [tex]8.2\%[/tex]

Explanation

A randomly selected person has a birthday in November

The probability of the event is

[tex]$P(B)=\frac{\text { No }of \text { days in november }}{No\ of \text { days in year }}$[/tex]

[tex]$P(B)=\frac{30}{365}$[/tex]

[tex]P(B)=8.2\%[/tex]

Therefore, the probability of a randomly selected person has a birthday in November is [tex]8.2\%[/tex]

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To calculate the power of a hypothesis test for an IQ test with a true population mean of 105, we find the Z-score for the sample mean, use it to get the cumulative probability, and subtract that from 1. This results in the power of the test to be 0.0475.

[tex](15 / sqrt(25) = 3)[/tex]

Let's illustrate with numerical probabilities, assuming the Z-score of 1.67 corresponds to a cumulative probability of approximately 0.9525. The power of the test is then 1 - 0.9525 = 0.0475 or 4.75%. The power is relatively low, indicating a high risk of Type II error (failing to reject the null hypothesis when it should be rejected).

Standard Error: [tex]\(SE = \frac{σ}{\sqrt{n}} = 3\).[/tex] Z-score: [tex]\(Z = \frac{X - μ}{SE} \approx 3.33\).[/tex] Power: Find [tex]\(P(Z > 3.33)\).[/tex]

let's go through the process step by step:

Step 1: Standard Error of the Mean (SE)

The standard error of the mean (SE) is calculated using the formula:

[tex]\[ SE = \frac{σ}{\sqrt{n}} \][/tex]

Where:

- [tex]\( σ \)[/tex] is the population standard deviation (given as 15),

- [tex]\( n \)[/tex] is the sample size (given as 25).

Plugging in the values:

[tex]\[ SE = \frac{15}{\sqrt{25}} = \frac{15}{5} = 3 \][/tex]

So, the standard error of the mean is 3.

Step 2: Z-score Calculation

The Z-score measures how many standard deviations a data point is from the mean. It's calculated using the formula:

[tex]\[ Z = \frac{X - μ}{SE} \][/tex]

Where:

- [tex]\( X \)[/tex] is the sample mean,

-[tex]\( μ \)[/tex] is the population mean under the null hypothesis (given as 100),

- [tex]\( SE \)[/tex] is the standard error of the mean (calculated as 3).

We want to find the Z-score for a sample mean of 110:

[tex]\[ Z = \frac{110 - 100}{3} = \frac{10}{3} \]\[ Z \approx 3.33 \][/tex]

Step 3: Finding the Power

The power of a statistical test is the probability of correctly rejecting the null hypothesis when the alternative hypothesis is true.

In this case, the alternative hypothesis is that the true population mean is 105.

We need to find the probability of getting a Z-score greater than 3.33 when the true population mean is 105. This probability represents the likelihood of correctly rejecting the null hypothesis.

[tex]\[ P(Z > 3.33) = 1 - P(Z \leq 3.33) \][/tex]

We look up the probability [tex]\( P(Z \leq 3.33) \)[/tex] in a standard normal distribution table or calculate it using a calculator. Then, we subtract this probability from 1 to find [tex]\( P(Z > 3.33) \)[/tex]. This value gives us the power of the test.

So, in detail:

1. Calculate the standard error of the mean (SE) using the formula and given values.

2. Calculate the Z-score using the formula and the SE calculated in step 1.

3. Find the probability [tex]\( P(Z > 3.33) \)[/tex]  by subtracting [tex]\( P(Z \leq 3.33) \)[/tex] from 1.

Answer:

Sample size should be atleast 60025

Step-by-step explanation:

Given that a school board wishes to know the current mean reading level of 6th graders throughout a very large school district. Past experience shows that the standard deviation of such reading scores is about 2.5

For 95% confidence we have significance level = 5%

Margin of error = 0.04

Margin of error should be less than 0.04

i.e. Z critical value for 95% *std error <0.04

[tex]1.96*\frac{2.5}{\sqrt{n} } <0.02\\\\n>60025[/tex]

Sample size should be atleast 60025

The school board needs to sample 151 6th graders to be 95% confident that the sample mean reading level is within 0.4 of the true mean.

The school board is interested in determining the required sample size needed to estimate the mean reading level of 6th graders. Given the standard deviation of 2.5 and the margin of error of 0.4 for a 95% confidence interval, we must use the formula for sample size in estimating a mean. The formula is:

n = (z*s/E)^2

where n is the sample size, z is the z-score corresponding to the desired confidence level, s is the population standard deviation, and E is the margin of error. For a 95% confidence level, the z-score is 1.96.

Therefore, the sample size n would be:

n = (1.96*2.5/0.4)^2

n = (4.9/0.4)^2

n = (12.25)^2

n = 150.0625

Since we cannot have a fraction of a sample, we round up to the nearest whole number. Thus, the school board needs to sample 151 6th graders to be 95% confident that the sample mean reading level is within 0.4 of the true mean.

Answer:

A. It's the amount of change in y when x increases by one unit

Step-by-step explanation:

For this case we have the following linear model adjusted:

[tex]\hat y = 36.5 +0.48 x[/tex]

Where y is the dependent variable, x the independent variable, 36.5 represent the intercept and 0.48 the slope.

We can analyze one by one the options to select the most appropiate.

A. It's the amount of change in y when x increases by one unit

True, for this case the slope is defined as:

[tex]m =0.48 \frac{\Delta y}{\Delta x}[/tex]

And is defined as the amount of change in y when x increase 1 unit.

B. It's the value of y when x = 0

False the value of y when x=0 is y= 36.5+0.48(0) = 36.5

C. It's the value of x when y = 0

False, when y=0 we have this:

[tex] 0=36.5 +0.48 x[/tex]

[tex]x= -\frac{36.5}{0.48}=-76.04[/tex]

D. It's the mean of the distribution for the response variable (y) around the explanatory variable (x)

False, the value 0.48 represent the slope obtained from an estimation of least squares and not represent the mean for the response variable y.

E. It's the amount of change in x when y increases by one unit.

False, is defined inverse as: the amount of change in y when x increase 1 unit.

The value .48 in the given linear equation represents the rate of change in y for each unit increase in x. It signifies the slope of the line, implying that the value of y increases by .48 for every unit increase in x.

In the equation y = 36.5 + .48x, the value .48 represents the slope of the line in the context of linear equation. The slope indicates the rate of change in y with respect to x. In this case,

.48 means that for each unit increase in the value of x (independent variable), the value of y (dependent variable) will increase by .48 units.

Therefore, the correct answer is A: It's the amount of change in y when x increases by one unit. This is a fundamental component of understanding linear equations, slopes and how they represent the relationship between two variables.

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Answer:

Step-by-step explanation:

Given that X and Y are independent random variables with the following distributions:

x -1 10 1 2 Total

p 0.3 0.1 0.5 0.1 1

xp -0.3 1 0.5 0.2 1.4

x^2p 0.3 10 0.5 0.4 11.2

Mean of X = 1.4

Var(x) = 11.2-1.4^2 =  9.24    

y 2 3 5  

p 0.6 0.3 0.1  1

yp 1.2 0.9 0.5 0 2.6

y^2p 2.4 2.7 2.5 0 7.6

Mean of Y = 2.6

Var(Y) = 11.2-1.4^2 =  0.84

3) W=3+2x

Mean of w =3+2*Mean of x = 7.2

Var (w) = 0+2^2 Var(x)= 36.96

Answer:

[tex]x(x+1)^2[/tex]

Step-by-step explanation:

Given:

The expression to factor is given as:

[tex]x^3+2x^2+x[/tex]

In order to factor it, we write the factors of each of the terms of the given polynomial. So,

The factors of the three terms are:

[tex]x^3=x\times x\times x\\\\2x^2=2\times x\times x\\\\x=x[/tex]

Now, 'x' is a common factor for all the three terms. So, we factor it out. This gives,

[tex]x(\frac{x^3}{x}+2\frac{x^2}{x}+\frac{x}{x})\\\\x(x^2+2x+1)[/tex]

Now, we know a identity which is given as:

[tex](a+b)^2=a^2+2ab+b^2[/tex]

Here, [tex]x^2+2x+1[/tex] can be rewritten as [tex]x^2+2(1)(x)+1^2[/tex]

So, [tex]a=x\ and\ b=1[/tex]

Thus, [tex]x^2+2(1)(x)+1^2= (x+1)^2[/tex]

Therefore, the complete factorization of the given expression is:

[tex]x^3+2x^2+x=x(x+1)^2[/tex]

Answer:

x(x+1)^2

Step-by-step explanation:

Answer:

6 people ride in each car.

Step-by-step explanation:

42 people / 7 cars = 6 people/car

In this high school level mathematics question, we determine that 6 people ride in one car in the amusement park ride with 7 cars and 42 people. The calculation involves dividing the total number of people by the total number of cars.

Mathematics - In this question, we are determining how many people ride in one car in an amusement park ride where there are 7 cars and 42 people in total. Since the number of people is divided equally among the cars, we can calculate the number of people in one car by dividing the total number of people by the total number of cars.

Calculation:

Total People = 42

Total Cars = 7

Number of People in One Car = Total People / Total Cars = 42 / 7 = 6

Therefore, 6 people ride in one car in this amusement park ride.

Answer:

Equation:

250 + 132.5y = x

Solution for solving for x:

x = 250 + 132.5y

Solution for solving for y:

y = (x - 250)/132.5

In this situation, the solutions mean that the equation is linear or first grade because the solutions can be calculated in any of the following four forms:

1. Adding the same number to both sides

2. Subtracting the same number from both sides

3. Multiplying both sides by the same number

4. Dividing both sides by the same number

Completing the question and statement correctly:

The equation  represents the relationship between the quantities in this situation, where x is the weight, in grams, of the filled box and y the number of shirts in the box.

Step-by-step explanation:

1. Let's review all the information provided to us to answer the question correctly:

Weight of an empty box = 250 grams

Weight of each T-shirt = 132.5 grams

2. Name two possible solutions to the equation that represent the relationship between the quantities in this situation, where x  is the weight, in grams, of the filled box and y the number of shirts in the box. What do the solutions mean in this situation?

x = weight, in grams, of the filled box

y = number of shirts in the box

Equation:

250 + 132.5y = x

Solution for solving for x:

x = 250 + 132.5y

Solution for solving for y:

y = (x - 250)/132.5

In this situation, the solutions mean that the equation is linear or first grade because the solutions can be calculated in any of the following four forms:

1. Adding the same number to both sides

2. Subtracting the same number from both sides

3. Multiplying both sides by the same number

4. Dividing both sides by the same number

Step-by-step explanation:

The two possible solution are 647.5 gm and 912.5 gm.

Let's use the following equation:

Weight of the filled box (W) = Weight of the empty box (250 grams) + (Number of T-shirts) * (Weight of each T-shirt, 132.5 grams)

So, the equation would be:

W = 250 + 132.5 * N

Where:

- W is the weight of the filled box in grams.

- N is the number of T-shirts in the box.

Now, let's find two possible solutions:

Solution 1:

Suppose the box is filled with 3 T-shirts.

W = 250 + 132.5 * 3

W = 250 + 397.5

W = 647.5 grams

Solution 2:

Suppose the box is filled with 5 T-shirts.

W = 250 + 132.5 * 5

W = 250 + 662.5

W = 912.5 grams

- Solution 1: If the box is filled with 3 T-shirts, it would weigh 647.5 grams in total.

This means that the combined weight of the 3 T-shirts added to the empty box's weight is 647.5 grams.

- Solution 2: If the box is filled with 5 T-shirts, it would weigh 912.5 grams in total.

This means that the combined weight of the 5 T-shirts added to the empty box's weight is 912.5 grams.

In general, the equation allows you to calculate the weight of the filled box based on the number of T-shirts you put in it. Each additional T-shirt adds 132.5 grams to the total weight, considering the empty box's weight of 250 grams.

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Answer:

Null hypothesis:[tex]\mu_{m} \geq \mu_{w}[/tex]

Alternative hypothesis:[tex]\mu_{m} < \mu_{w}[/tex]

[tex]p_v =P(t_{134}<-2.85)=0.0025[/tex]

D. Reject H0, there is sufficient evidence to conclude that the mean step pulse of men was less than the mean step pulse of women.

So on this case the 95% confidence interval would be given by [tex]-10.502 \leq \mu_{m} -\mu_w \leq -1.898[/tex]  

We are 95% confident that the mean difference is in the confidence interval.

Step-by-step explanation:

1) Data given and notation

[tex]\bar X_{m}=112.5[/tex] represent the mean for the sample of men

[tex]\bar X_{w}=118.7[/tex] represent the mean for the sample women

[tex]s_{m}=11.1[/tex] represent the sample standard deviation for the sample men

[tex]s_{w}=14.2[/tex] represent the sample standard deviation for the sample eomen

[tex]n_{m}=56[/tex] sample size for the group men

[tex]n_{w}=80[/tex] sample size for the group women

z would represent the statistic (variable of interest)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to check if the means for the two groups are the same, the system of hypothesis would be:

Null hypothesis:[tex]\mu_{m} \geq \mu_{w}[/tex]

Alternative hypothesis:[tex]\mu_{m} < \mu_{w}[/tex]

We don't have the population standard deviation's, so for this case is better apply a t test to compare means, and the statistic is given by:

[tex]t=\frac{\bar X_{m}-\bar X_{w}}{\sqrt{\frac{s^2_{m}}{n_{m}}+\frac{s^2_{w}}{n_{w}}}}[/tex] (1)

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

3) Calculate the statistic

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

[tex]t=\frac{112.5-118.7}{\sqrt{\frac{11.1^2}{56}+\frac{14.2^2}{80}}}}=-2.85[/tex]  

4) Statistical decision

The degrees of freedom are given by:

[tex]df=n_m +n_w -2= 56+80-2=134[/tex]

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

[tex]p_v =P(t_{134}<-2.85)=0.0025[/tex]

D. Reject H0, there is sufficient evidence to conclude that the mean step pulse of men was less than the mean step pulse of women.

5) Confidence interval

The confidence interval for the difference of means is given by the following formula:  

[tex](\bar X_m -\bar X_w) \pm t_{\alpha/2}\sqrt{(\frac{s^2_m}{n_m}+\frac{s^2_w}{n_w})}[/tex] (1)  

The point of estimate for [tex]\mu_1 -\mu_2[/tex] is just given by:  

[tex]\bar X_m -\bar X_w =112.5-118.7=-6.2[/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(0.025,134)".And we see that [tex]z_{\alpha/2}=1.98[/tex]  

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

[tex]-6.2-1.98\sqrt{\frac{11.1^2}{56}+\frac{14.2^2}{80}}}=-10.502[/tex]  

[tex]-6.2+1.98\sqrt{\frac{11.1^2}{56}+\frac{14.2^2}{80}}}=-1.898[/tex]  

So on this case the 95% confidence interval would be given by [tex]-10.502 \leq \mu_{m} -\mu_w \leq -1.898[/tex]  

We are 95% confident that the mean difference is in the confidence interval.

In the context of the study conducted by the physical therapist, the null hypothesis states that the mean step pulse of men equals the mean step pulse of women, and the alternative hypothesis states that the mean step pulse of men is less than that of women. With a t-test result that has a P-value less than 0.05, we reject the null hypothesis and accept the alternative hypothesis, indicating a significant difference between the mean step pulses of men and women.

The physical therapist's study is essentially a hypothesis testing problem. The null hypothesis (σ_0) is that the mean step pulse of men is equal to the mean step pulse of women, and the alternative hypothesis (σ_a) is that the mean step pulse of men is less than the mean step pulse of women.

For this problem, we can write the hypotheses as follows:
σ_0: μ1 = μ2
σ_a: μ1 < μ2

The student's available choices seem to reflect this. Choosing between 'H0: μ1 = μ2; Ha :μ1>μ2' and 'H0:μ1 = μ2;Ha::μ1 not equal μ2'. The correct choice is not listed because the alternative hypothesis is incorrectly defined in both cases. The correct alternative hypothesis would be 'μ1 < μ2', pointing that the mean step pulse of men is less than that of women.

The t-test result given in the question 'T = 2.85 P = 0.0025 DF = 132' indicates that the mean step pulse of men is not equal to that of women, with the P-value being less than 0.05. Thus, with this information, you would typically reject the null hypothesis and accept the alternative hypothesis, which suggests a significant difference between the mean step pulses of men and women.

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Answer:

The minimum sample size is N=1537.

Step-by-step explanation:

We want to know the sample size to estimate the proportion of voters with a margin of error of 0.05, with a 95% of confidence.

The margin of error can be defined as:

[tex]e=UL-LL=(p+z*\sqrt{\frac{p(1-p)}{N}}) -(p-z*\sqrt{\frac{p(1-p)}{N}})=2z\sqrt{\frac{p(1-p)}{N}}[/tex]

We can calculate N from this

[tex]e=2z\sqrt{\frac{p(1-p)}{N}}\\\\\frac{p(1-p)}{N}=(\frac{e}{2z})^2\\\\N= (\frac{2z}{e})^2*p(1-p)[/tex]

The sample size will be calculated for p=0.5, which is the proportion that requires the larger sample size (maximum variance).

The z-value for a 95% CI is z=1.96.

The minimum sample size is then

[tex]N= (\frac{2z}{e})^2*p(1-p)=(\frac{2*1.96}{0.05})^2*0.5(1-0.5)=6146.56*0.5*0.5=1536.64[/tex]

The minimum sample size is N=1537.