A cereal manufacturer is concerned that the boxes of cereal not be under filled or overfilled. Each box of cereal is supposed to contain 13 ounces of cereal. A random sample of 36 boxes is tested. The sample average weight is 12.85 ounces and the sample standard deviation is 0.75 ounces. Use the critical-value approach to test whether the population average weight of a cereal box differs from 13 ounces. The critical-value should be:

We conclude that the population average weight of a cereal box is equal to 13 ounces.

Step-by-step explanation:

We are given that a cereal manufacturer is concerned that the boxes of cereal not be under filled or overfilled.

A random sample of 36 boxes is tested. The sample average weight is 12.85 ounces and the sample standard deviation is 0.75 ounces.

Let = population average weight of a cereal box

So, Null Hypothesis, : = 13 ounces    {means that the population average weight of a cereal box is equal to 13 ounces}

Alternate Hypothesis, : 13 ounces    {means that the population average weight of a cereal box differs from 13 ounces}

The test statistics that will be used here is One-sample t test statistics as we don't know about population standard deviation;

                                  T.S.  =  ~

where, = sample average weight = 12.85 ounces

            = sample standard deviation = 0.75 ounces

            n = sample of boxes = 36

So, test statistics  =    ~

                               =  -1.20

The value of the test statistics is -1.20.

Since, in the question we are not given with the level of significance so we assume it to be 5%. Now at 5% significance level, the t table gives critical values between -2.03 and 2.03 at 35 degree of freedom for two-tailed test.

Since our test statistics lies within the range of critical values of t, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the population average weight of a cereal box is equal to 13 ounces.

Función de beneficios = P(x) =-0.01x² + 3x - 100

Ganancias máximos = 125

Y ocurre cuando x = 150 unidades

Profits Function = P(x) = -0.01x² + 3x - 100

Maximum profits = 125

And it occurs when x = 150 units

Step-by-step explanation:

Ganancias = Ingresos - Costos

Ingresos = R (x) = 5x - 0.01x²

Costos = C (x) = 100 + 2x

Función de ganancias = P (x) = R (x) - C (x)

= (5x - 0.01x²) - (100 + 2x)

= 5x - 0.01x² - 100 - 2x

= 3x - 0.01x² - 100

P (x) = -0.01x² + 3x - 100

Para obtener las máximas ganancias, utilizamos el análisis de diferenciación para la función de ganancias.

En el nivel de Ganancias máximo, (dP / dx) = 0 y (d²P / dx²) < 0

P(x) = -0.01x² + 3x - 100

(dP/dx) = -0.02x + 3

Al máximo Ganancias, (dP/dx) = 0

(dP/dx) = -0.02x + 3 = 0

x = (3/0.02) = 150 unidades

Este es el nivel de producto que corresponde a las ganancias máximas.

Para verificar si este es realmente el punto máximo de la función de ganancias,

(dP/dx) = -0.02x + 3

(d²P/dx²) = -0.02 < 0 (lo que demuestra que realmente es el punto de máximo ganancias).

Por lo tanto, la ganancia máxima ocurre cuando x = 150 unidades

P(x) = -0.01x² + 3x - 100

P(x) = -0.01(150²) + 3(150) - 100 = 125

¡¡¡Espero que esto ayude!!!

English Translation

The profits of a business can be determined by subtracting the costs from the income. Suppose that the revenue of a business is represented by the function:

R(x) = 5x - 0.01x²

and the manufacturing costs of the product are represented by

C(x) = 100 + 2x

where x is the number of units of the product. Determine the profits function. Find a function P (x) that represents the profits of the firm and determine the maximum profit.

Solution

Profits = Revenue - Costs

Revenue = R(x) = 5x - 0.01x²

Costs = C(x) = 100 + 2x

Profits function = P(x) = R(x) - C(x)

= (5x - 0.01x²) - (100 + 2x)

= 5x - 0.01x² - 100 - 2x

= 3x - 0.01x² - 100

P(x) = -0.01x² + 3x - 100

To obtain maximum profits, we use differentiation analysis for the profit function.

At the maximum profit level, (dP/dx) = 0 and (d²P/dx²) < 0

P(x) = -0.01x² + 3x - 100

(dP/dx) = -0.02x + 3

At maximum profit, (dP/dx) = 0

(dP/dx) = -0.02x + 3 = 0

x = (3/0.02) = 150 units

This is the product level that corresponds to maximum profits.

To check if this is truly the maximum point of the profit function,

(dP/dx) = -0.02x + 3

(d²P/dx²) = -0.02 < 0 (which shows that it truly is the maximum profit point.

Hence, maximum profit occurs when x = 150 units

P(x) = -0.01x² + 3x - 100

P(x) = -0.01(150²) + 3(150) - 100 = 125

Hope this Helps!!!