Two-fifths of one less than a number is less than three-fifths of one more than that number. What numbers are in the solution set of this problem

A.x<-5
B.x>-5
C.x>-1
D.x<-1

Answer:

97% Confidence interval = (12.62, 18.98)

Step-by-step explanation:

Complete Question

Recorded here are the germination times (in days) for ten randomly chosen seeds of a new type of bean: 18, 12, 20, 17, 14, 15, 13, 11, 21, 17. Assume that the population germination time is normally distributed. Find the 97% confidence interval for the mean germination time.

Solution

We first compute the sample mean and standard deviation for this sample distribution

Sample mean = (Σx)/N = (158/10) = 15.8

Standard deviation = √{[Σ(x - xbar)²]/(N-1)} = 3.3598941782278 = 3.36

Confidence Interval for the population mean is basically an interval of range of values where the true population mean can be found with a certain level of confidence.

Mathematically,

Confidence Interval = (Sample mean) ± (Margin of error)

Sample Mean = 15.8

Margin of Error is the width of the confidence interval about the mean.

It is given mathematically as,

Margin of Error = (Critical value) × (standard Error of the mean)

Critical value will be obtained using the t-distribution. This is because there is no information provided for the population mean and standard deviation.

To find the critical value from the t-tables, we first find the degree of freedom and the significance level.

Degree of freedom = df = n - 1 = 10 - 1 = 9

Significance level for 97% confidence interval

(100% - 97%)/2 = 1.5% = 0.015

t (0.015, 9) = 2.9982 (from the t-tables)

Standard error of the mean = σₓ = (σ/√n)

σ = standard deviation of the sample = 3.36

n = sample size = 10

σₓ = (3.36/√10) = 1.0625

97% Confidence Interval = (Sample mean) ± [(Critical value) × (standard Error of the mean)]

CI = 15.8 ± (2.9982 × 1.0625)

CI = 15.8 ± 3.1809882246

97% CI = (12.6190117754, 18.9809882246)

97% Confidence interval = (12.62, 18.98)

Hope this Helps!!!

Answer:

X= -3

Y= 7

a= -6

b = -28

C = 9

D= -7

Step-by-step explanation:

2x – 4y = -34

-3x - y = 2

2x – 4y = -34

-12x - 4y = 8

-14x = 42

X= 42/-14

X= -3

-3x - y = 2

-3(-3) -y = 2

9-y = 2

9-2= y

Y= 7

a = 2x

a= 2*-3

a= -6

b = -4y

b = -4*7

b = -28

C= -3x

C= -3*-3

C = 9

D= -y

D= -7

Answer:

A= 2 B=-4

C= -3 D= -1

Step-by-step explanation:

Took it on Edge

Answer:

I am assuming the underlined value is 25%. It is a parameter

Step-by-step explanation:

The value is is a parameter. This is because the parameter is a value that describes the population.

The survey carried out was a national survey of which there were 25% respondents who reported that someone had offered, sold, or given them an illegal drug on school property. It is not a statistics because a sample was not taken out of the population and a survey made on the sample.

The underlined 25% value is the value that summarizes the entire population of high school students

Answer:

x = 3.4

Step-by-step explanation:

Step 1: Isolate x

0.5x = 1.7

Step 2: Divide both sides by 0.5

x = 3.4

And we have our final answer!

Answer: x=3.4

Step-by-step explanation:

[tex]0.5x+4.2=5.9[/tex]

multiply both sides by 10

[tex]5x+42=59[/tex]

subtract 42 on both sides

[tex]5x=17[/tex]

divide 5 on both sides

[tex]x=\frac{17}{5}[/tex] or

Simplify

x= 3.4

Answer:

[tex]t=\frac{5.2-5}{\frac{0.8}{\sqrt{40}}}=1.58[/tex]    

The degrees of freedom are given by:

[tex]df=n-1=40-1=39[/tex]  

Thep value for this case would be given by:

[tex]p_v =P(t_{(39)}>1.58)=0.061[/tex]  

Since the p value is higher than the significance level of 0.05 we have enough evidence to FAIL to reject the null hypothesis and we can conclude that true mean is not significantly higher than 5.

Step-by-step explanation:

Information provided

[tex]\bar X=5.2[/tex] represent the sample mean

[tex]s=0.8[/tex] represent the sample standard deviation

[tex]n=40[/tex] sample size  

[tex]\mu_o =5[/tex] represent the value to verify

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

t would represent the statistic

[tex]p_v[/tex] represent the p value

Hypothesis to test

We want to verify if the true mean is higher than 5, the system of hypothesis would be:  

Null hypothesis:[tex]\mu \leq 5[/tex]  

Alternative hypothesis:[tex]\mu > 5[/tex]  

The statistic is given by:

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

Replacing the info given we got:

[tex]t=\frac{5.2-5}{\frac{0.8}{\sqrt{40}}}=1.58[/tex]    

The degrees of freedom are given by:

[tex]df=n-1=40-1=39[/tex]  

Thep value for this case would be given by:

[tex]p_v =P(t_{(39)}>1.58)=0.061[/tex]  

Since the p value is higher than the significance level of 0.05 we have enough evidence to FAIL to reject the null hypothesis and we can conclude that true mean is not significantly higher than 5.

Answer:

after the 1st year

Step-by-step explanation:

$2300 × 75% = $1725.00

$2300-$1725= $575

Answer:

Height of tomato plant is the dependent variable

Presence of walnut leaves in the soil is the independent variable

Tomato plants grown without walnut leaves is the control

Step-by-step explanation:

An independent variable is the variable in an experiment that can be altered to test for a certain result. It is independent, or does not change with change in other factors in the experiment. In this case, the presence or absence, or quantity of walnut available in the soil is the independent variable in the experiment.

A dependent variable varies, and depends on the independent variable. It is what is measured in the experiment. In this case, the height of the tomato plants is the dependent variable that depends on the presence, absence or quantity of walnut in the soil.

A control in an experiment, is a replicate experiment, that is manipulated in order to be able to test a single variable at a time. Controls are variables are held constant so as to minimize their effect on the system under study. In this case, some of the tomato plants are planted without walnut in the soil, to test the effect of the absence of the walnut in the soil.

The Laplace transform of the function [tex]\frac{1}{2} (\frac{1}{s} - \frac{s}{s^2 + 4w^2} )[/tex] .

The Laplace transform exist when s > 0 .

Here, the given function is f(t) = sin²(wt) .

The Laplace transform of the the function f(t),

F(s) = f(t) = { [tex]{\frac{1}{2} \times 2sin^2(wt) }[/tex] }

F(s) = { [tex]\frac{1}{2} \times (1- cos2wt)[/tex] }

F(s) = { [tex]\frac{1}{2} - \frac{1}{2} \times cos(2wt)\\[/tex] }

F(s) = [tex]\frac{1}{2} (\frac{1}{s} - \frac{s}{s^2 + 4w^2} )[/tex]

Next,

The above Laplace transform exist if s > 0 .

Know more about Laplace transform,

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#SPJ4

Answer:

Option A is correct.

A uniform distribution.

Step-by-step explanation:

Complete Question

T-Mobile sells 6 different models of cell phones and have found that they sell an equal number of each model. The probability distribution that would describe this random variable is called:

A) Uniform Distribution

B) Continuous Distribution

C) Poisson Distribution

D) Relative Frequency Distribution

Solution

A uniform distribution is one in which all the variables have the same probability of occurring.

It is also known as a rectangular distribution, as every portion of the sample space has an equal chance of occurring, with equal length on the probability curve, leading to a rectangular probability curve.

And for this question, 6 different models of phones sell an equal number, hence, the probability of selling each model is equal to one another, hence, this is evidently a uniform distribution.

Hope this Helps!!!

Answer:

  17

Step-by-step explanation:

Let x represent the smaller integer. Then we have ...

  (x +1)² +(x +2)² = 40x +5

  2x² +6x +5 = 40x +5

  x² -17x = 0 . . . . . subtract (40x+5), divide by 2

  x(x -17) = 0 . . . . . factor

The solution of interest is x = 17.

The smaller integer is 17.

Answer:

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

And replacing we got:

[tex] SE=\sqrt{\frac{0.75*(1-0.75)}{900}}= 0.014[/tex]

Step-by-step explanation:

For this case we have the following info given:

[tex] n=900[/tex] represent the sample size selected

[tex]p = 0.75[/tex] represent the population proportion

We want to find the standard error and we can use the distribution for the sample proportion and for this case since the sample size is large enough and we satisfy np>10 and n(1-p) >10 we have:

[tex] \hat p \sim N (p,\sqrt{\frac{p(1-p)}{n}})[/tex]

And the standard error is given;

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

And replacing we got:

[tex] SE= \sqrt{\frac{0.75* (1-0.75)}{900}}= 0.014[/tex]

Answer:

(a) The fraction of the calls last between 4.50 and 5.30 minutes is 0.3729.

(b) The fraction of the calls last more than 5.30 minutes is 0.1271.

(c) The fraction of the calls last between 5.30 and 6.00 minutes is 0.1109.

(d) The fraction of the calls last between 4.00 and 6.00 minutes is 0.745.

(e) The time is 5.65 minutes.

Step-by-step explanation:

We are given that the mean length of time per call was 4.5 minutes and the standard deviation was 0.70 minutes.

Let X = the length of the calls, in minutes.

So, X ~ Normal([tex]\mu=4.5,\sigma^{2} =0.70^{2}[/tex])

The z-score probability distribution for the normal distribution is given by;

                           Z  =  [tex]\frac{X-\mu}{\sigma}[/tex]  ~ N(0,1)

where, [tex]\mu[/tex] = population mean time = 4.5 minutes

           [tex]\sigma[/tex] = standard deviation = 0.7 minutes

(a) The fraction of the calls last between 4.50 and 5.30 minutes is given by = P(4.50 min < X < 5.30 min) = P(X < 5.30 min) - P(X [tex]\leq[/tex] 4.50 min)

    P(X < 5.30 min) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{5.30-4.5}{0.7}[/tex] ) = P(Z < 1.14) = 0.8729

    P(X [tex]\leq[/tex] 4.50 min) = P( [tex]\frac{X-\mu}{\sigma}[/tex] [tex]\leq[/tex] [tex]\frac{4.5-4.5}{0.7}[/tex] ) = P(Z [tex]\leq[/tex] 0) = 0.50

The above probability is calculated by looking at the value of x = 1.14 and x = 0 in the z table which has an area of 0.8729 and 0.50 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.8729 - 0.50 = 0.3729.

(b) The fraction of the calls last more than 5.30 minutes is given by = P(X > 5.30 minutes)

    P(X > 5.30 min) = P( [tex]\frac{X-\mu}{\sigma}[/tex] > [tex]\frac{5.30-4.5}{0.7}[/tex] ) = P(Z > 1.14) = 1 - P(Z [tex]\leq[/tex] 1.14)

                                                              = 1 - 0.8729 = 0.1271

The above probability is calculated by looking at the value of x = 1.14 in the z table which has an area of 0.8729.

(c) The fraction of the calls last between 5.30 and 6.00 minutes is given by = P(5.30 min < X < 6.00 min) = P(X < 6.00 min) - P(X [tex]\leq[/tex] 5.30 min)

    P(X < 6.00 min) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{6-4.5}{0.7}[/tex] ) = P(Z < 2.14) = 0.9838

    P(X [tex]\leq[/tex] 5.30 min) = P( [tex]\frac{X-\mu}{\sigma}[/tex] [tex]\leq[/tex] [tex]\frac{5.30-4.5}{0.7}[/tex] ) = P(Z [tex]\leq[/tex] 1.14) = 0.8729

The above probability is calculated by looking at the value of x = 2.14 and x = 1.14 in the z table which has an area of 0.9838 and 0.8729 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.8729 = 0.1109.

(d) The fraction of the calls last between 4.00 and 6.00 minutes is given by = P(4.00 min < X < 6.00 min) = P(X < 6.00 min) - P(X [tex]\leq[/tex] 4.00 min)

    P(X < 6.00 min) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{6-4.5}{0.7}[/tex] ) = P(Z < 2.14) = 0.9838

    P(X [tex]\leq[/tex] 4.00 min) = P( [tex]\frac{X-\mu}{\sigma}[/tex] [tex]\leq[/tex] [tex]\frac{4.0-4.5}{0.7}[/tex] ) = P(Z [tex]\leq[/tex] -0.71) = 1 - P(Z < 0.71)

                                                              = 1 - 0.7612 = 0.2388

The above probability is calculated by looking at the value of x = 2.14 and x = 0.71 in the z table which has an area of 0.9838 and 0.7612 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.2388 = 0.745.

(e) We have to find the time that represents the length of the longest (in duration) 5 percent of the calls, that means;

            P(X > x) = 0.05            {where x is the required time}

            P( [tex]\frac{X-\mu}{\sigma}[/tex] > [tex]\frac{x-4.5}{0.7}[/tex] ) = 0.05

            P(Z > [tex]\frac{x-4.5}{0.7}[/tex] ) = 0.05

Now, in the z table the critical value of x which represents the top 5% of the area is given as 1.645, that is;

                      [tex]\frac{x-4.5}{0.7}=1.645[/tex]

                      [tex]{x-4.5}{}=1.645 \times 0.7[/tex]

                       x = 4.5 + 1.15 = 5.65 minutes.

SO, the time is 5.65 minutes.

Answer:

r = 7.5

Step-by-step explanation:

Circle equation: [tex](x - h)^2 + (y - k)^2 = r^2[/tex]

Since we are already give r², we simply just take the square root of 56.25, and we should get 7.5 as our final answer!

Answer:

(x-4) (x-4)

Step-by-step explanation:

x^2-8x+16

What 2 numbers multiply to 16 and add to -8

-4*-4 = 16

-4+-4 = -8

(x-4) (x-4)

Answer:

correct ^

Step-by-step explanation:

Answer:

2/5

Step-by-step explanation:

Answer: Rs. 40,000

Step-by-step explanation:

Let say Marked price of the Watch  

= M  Rs

Discount = 20 %

Discount = (20/100)M = 0.2M   Rs

Price after Discount = M - 0.2M =

Rs 0.8M

13 % Value added tax

=> VAT  = (13/100) * 0.8M  = 0.104M  Rs

Value of Watch = 0.8M + 0.104M    = 0.904M Rs

0.904M = 36160

=> M = 40000 

Marked Price of Watch = Rs 40,000

Answer: B.

x ≈2.5

Step-by-step explanation:

[tex]-\left(u\right)^{-1}-6=-u+10[/tex]

[tex]u=8-\sqrt{65},\:u=8+\sqrt{65}[/tex]

[tex]x=\frac{\ln \left(8+\sqrt{65}\right)}{\ln \left(3\right)}[/tex]

x=2.52...

Answer:

x=2.5

Step-by-step explanation:

Answer:

the gram

Step-by-step explanation:

casegunnell is my gram, follow if your a real g

Following are the calculation to the given points:

When the ABCD is parallelogram:

The properties of parallelogram:

[tex]\to \angle CBD = \angle ADB \\\\[/tex]

[tex]\to \angle BCA = \angle DAC \\\\[/tex]

When the Two-lines are parallel and alternate interior angles are equal:

[tex]\to \Delta BEC \cong \Delta AED \ \ \ \ \ \ \ \{ASA\} \\\\\to \overline {AE} \cong \overline{CE}\\\\ \to \overline{BE} \cong \overline{ED} \\\\[/tex]

When the properties of congruent triangle.

Learn more:

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Answer:

[0, positive infinity)

Step-by-step explanation:

The domain is all x values a graph inputs. In a square root function, you cannot have negative inputs as it will turn out imaginary numbers. Therefore, your domain is all values of x above and including 0.

Answer: d on Ed

Step-by-step explanation:

Just took the test

Answer:

[tex] \: \: \: \: \: \: \: \: \: \: \dfrac{x}{5} = 7 \\ \implies \: x = 7 \times 5 \\ \implies \: x = 35[/tex]

So,b. multiplication

Answer:

A. division

Step-by-step explanation:

[tex]x/5=7[/tex]

[tex]x[/tex] is being divided by an integer.

[tex]x=35[/tex]

[tex]35/5=7[/tex]

35 divided by 5 is equal to 7.

Answer:

domain = {-5, -3, -2, 4, 8, 12}

Step-by-step explanation:

The domain is the set containing the x-coordinates of all ordered pairs.

domain = {-2, -3, 12, 8, 4, -5}

If you'd like, you can put the numbers in ascending order:

domain = {-5, -3, -2, 4, 8, 12}

Answer:

Step-by-step explanation:

The question is incomplete. The missing information is the group of answer choices. The group of answer choices are

a) none of the above

b) Data provides sufficient evidence, at 1% significance level, to reject the expert's claim. In addition the p-value (or the observed significance level) is equal to P( T < -3.281).

c) Data provides insufficient evidence, at 1% significance level, to support the expert's claim. In addition the p-value (or the observed significance level) is equal to P( Z > 2.896).

d) Data provides sufficient evidence, at 1% significance level, to support the expert's claim. In addition the p-value (or the observed significance level) is equal to P( T >3.355).

e) Data provides insufficient evidence, at 1% significance level, to support the researcher's claim. In addition the p-value (or the observed significance level) is equal to P(Z > 2.896).

Solution:

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

H0: µ ≥ 10

For the alternative hypothesis,

µ < 10

This is a left tailed test.

Since the number of samples is small and the population standard deviation is not given, the distribution is a student's t.

Since n = 9,

Degrees of freedom, df = n - 1 = 9 - 1 = 8

t = (x - µ)/(s/√n)

Where

x = sample mean = 13.5

µ = population mean = 10

s = samples standard deviation = 3.2

t = (13.5 - 10)/(3.2/√9) = 3.28

Since α = 0.01, the critical value is determined from the t distribution table. Recall that this is a left tailed test. Therefore, we would find the critical value corresponding to 1 - α and reject the null hypothesis if the test statistic is less than the negative of the table value.

1 - α = 1 - 0.01 = 0.99

The negative critical value is - 2.896

Since - 3.28 is lesser than - 2.896, then we would reject the null hypothesis.

By using probability value,

We would determine the p value using the t test calculator. It becomes

p = 0.0056

Level of significance = 1%

Since alpha, 0.01 > than the p value, 0.0056, then we would reject the null hypothesis. Therefore, At a 1% level of significance, the sample data showed significant evidence that the average useful lifetime of a typical car transimssion which comes with ten years warranty is significantly less than 10 years

The correct option is

a) none of the above

Answer:

50.40% probability that it weighs more than 0.8544 g.

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

[tex]\mu = 0.8551, \sigma = 0.0518[/tex]

If 1 candy is randomly​ selected, find the probability that it weighs more than 0.8544 g.

This is 1 subtracted by the pvalue of Z when X = 0.8544. So

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

[tex]Z = \frac{0.8544 - 0.8551}{0.0518}[/tex]

[tex]Z = -0.01[/tex]

[tex]Z = -0.01[/tex] has a pvalue of 0.4960

1 - 0.4960 = 0.5040

50.40% probability that it weighs more than 0.8544 g.

Answer:

a. 0.0668

b. 0.9545

Step-by-step explanation:

We have the following information:

mean (m) = 10000

standard deviation (sd) = 100

(a)

We must calculate the proportion of the components exceed 10150 kilograms per square centimeter in tensile strength as follows:

P (x> 10150) = P [(x - m) / sd> (10150 - 1000 /) 100]

P (x> 10150) = P (z> 1.5)

P (x> 10150) = 1 - P (z <1.5)

P (x> 10150) = 1 - 0.9332 (attached table)

P (x> 10150) = 0.0668

Therefore the proportion of the components exceed 10150 kilograms per square centimeter in tensile strength is 0.0668

(b)

We must calculate the proportion of all components has tensile strength between 9800 and 10200, as follows:

P (9800 <x <10200) = P [(9800 - 1000 /) 100 <(x - m) / sd <(10200 - 1000 /) 100]

P (9800 <x <10200) = P (-2 <z <2)

P (9800 <x <10200) = P (z <2) - P (z <-2)

P (9800 <x <10200) = 0.9773 - 0.0228 (attached table)

P (9800 <x <10200) = 0.9545

the proportion of pieces that would expect to scrap is 0.9545

Answer:

Check below, please

Step-by-step explanation:

Step-by-step explanation:

1.For which values of x is f '(x) zero? (Enter your answers as a comma-separated list.)

When the derivative of a function is equal to zero, then it occurs when we have either a local minimum or a local maximum point. So for our x-coordinates we can say

 [tex]f'(x)=0\: at \:x=2, and\: x=-2[/tex]

2. For which values of x is f '(x) positive?

Whenever we have  

 [tex]f'(x)>0[/tex]

then function is increasing. Since if we could start tracing tangent lines over that graph, those tangent lines would point up.

 [tex]f'(x)>0 \:at [-4,-2) \:and\:(2, \infty)[/tex]

3. For which values of x is f '(x) negative?  

On the other hand, every time the function is decreasing its derivative would be negative. The opposite case of the previous explanation. So

 [tex]f'(x) <0 \: at\: [-2,2][/tex]

4.What do these values mean?

 [tex]f(x) \:is \:increasing\:when\:f'(x) >0\\\\f(x)\:is\:decreasing\:when f'(x)<0[/tex]

5.(b) For which values of x is f ''(x) zero?

In its inflection points, i.e. when the concavity of the curve changes. Since the function was not provided. There's no way to be precise, but roughly

at x=-4 and x=4

Answer:

I answered this one for you already I think.

Answer:

The probability that the proportion of rooms booked in a sample of 423 rooms would differ from the population proportion by less than 6% is 0.9946.

Step-by-step explanation:

According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.

The mean of this sampling distribution of sample proportion is:

 [tex]\mu_{\hat p}=p[/tex]

The standard deviation of this sampling distribution of sample proportion is:

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

The information provided here is:

p = 0.27

n = 423

As n = 423 > 30, the sampling distribution of sample proportion can be approximated by the Normal distribution.

The mean and standard deviation of the sampling distribution of sample proportion are:

[tex]\mu_{\hat p}=p=0.27\\\\\sigma_{\hat p}=\sqrt{\frac{\hat p(1-\hat p)}{n}}=\sqrt{\frac{0.27\times(1-0.27)}{423}}=0.0216[/tex]

Compute the probability that the proportion of rooms booked in a sample of 423 rooms would differ from the population proportion by less than 6% as follows:

[tex]P(|\hat p-p|<0.06)=P(p-0.06<\hat p<p+0.06)[/tex]

                           [tex]=P(0.27-0.06<\hat p<0.27+0.06)\\\\=P(0.21<\hat p<0.33)\\\\=P(\frac{0.21-0.27}{0.0216}<\frac{\hat p-\mu_{\hat p}}{\sigma_{\hat p}}<\frac{0.33-0.27}{0.0216})\\\\=P(-2.78<Z<2.78)\\\\=P(Z<2.78)-P(Z<-2.78)\\\\=0.99728-0.00272\\\\=0.99456\\\\\approx 0.9946[/tex]

*Use a z-table.

Thus, the probability that the proportion of rooms booked in a sample of 423 rooms would differ from the population proportion by less than 6% is 0.9946.

Answer:

a) Null hypothesis (H0): the mean guest bill for a weekend is $600.

Alternative hypothesis (Ha): the mean guest bill for a weekend is significantly bigger than $600.

b) When H0 can not be rejected, the conclusion is that there is no enough evidence to claim that the mean guest bill had increased from $600.

c) When the H0 is rejected, they have enough evidence to claim that the mean guest bill is significantly bigger than $600.  

Step-by-step explanation:

a) The accountant, as he wants to see if there is evidence to support the claim that the mean guest bill has increased significanty, should write the hypothesis like that:

Null hypothesis (H0): the mean guest bill for a weekend is $600.

Alternative hypothesis (Ha): the mean guest bill for a weekend is significantly bigger than $600.

A sample of bills of the period in study needs to be taken in order to have a representation of the actual population of bills and then perform a t-test, as the sample mean and standard deviation will be used to perform the test.

b) When H0 can not be rejected, the conclusion is that there is no enough evidence to claim that the mean guest bill had increased from $600. If the P-value was low but not enough, they may take another sample to perform the test again or leave it like that.

c) When the H0 is rejected, they have enough evidence to claim that the mean guest bill is significantly bigger than $600.