What is a counter example for, when it rains, it pours?

The CEO of the Jen and Benny's ice cream company is concerned about the net weight of ice cream in their 50 ounce ice cream tubs

MissTica

Answer:

$t=\frac{54.98-52}{\frac{8.43}{\sqrt{26}}}=1.8025$

We need to find the degrees of freedom given by:

$df = n-1= 26-1 = 25$

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

$p_v =P(t_{25}>1.8025)=0.0418$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the average is higher than 52 at 5% of signficance.

Step-by-step explanation:

Data given and notation

$\bar X=54.98$ represent the sample mean

$s=8.43$ represent the sample deviation

$n=26$ sample size

$\mu_o =52$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is higher than 52, the system of hypothesis would be:

Null hypothesis: $\mu \leq 52$

Alternative hypothesis: $\mu > 52$

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

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

$t=\frac{54.98-52}{\frac{8.43}{\sqrt{26}}}=1.8025$

P-value

We need to find the degrees of freedom given by:

$df = n-1= 26-1 = 25$

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

$p_v =P(t_{25}>1.8025)=0.0418$

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the average is higher than 52 at 5% of signficance.

7 0

3 years ago

Read 2 more answers

11-15 pls help thanks

AlekseyPX

What do I need to solve :)

7 0

3 years ago

The diagram shows several planes, lines, and points.

zhannawk [14.2K]

I think the answer is C...

7 0

4 years ago

Read 2 more answers

John is twice as old as his son. In 42 years, the ratio of their ages will be 4:3. What is the son's current age?

dsp73

Answer: Jhon is 42 years old, and his son is 21 years old.

Step-by-step explanation:

Let's define:

J = John's age

S = Age of the son.

We know that:

"John is twice as old as his son"

J = 2*S

"In 42 years, the ratio of their ages will be 4:3. What is the son's current age?"

In 42, their ages will be:

J + 42 and S + 42.

And the ratio 4:3 means that:

(J + 42) = (4/3)*(S + 42)

Then we have the system of equations:

J = 2*S

(J + 42) = (4/3)*(S + 42)

To solve it we can first replace the first equation into the second one, to get:

(2*S + 42) = (4/3)*(S + 42)

2*S - (4/3)*S = (4/3)*42 - 42

S*(6/3 - 4/3) = (1/3)*42

S*(2/3) = (1/3)*42

S = (3/2)*(1/3)*42 = 21.

And we can find Jonh's age if we use the first equation:

J = 2*S = 2*21 = 42

Jhon is 42 years old, and his son is 21 years old.

8 0

3 years ago

Sketch a graph of the polynomial function f(x) =x3− 2x2. Use it to complete the following:

AlladinOne [14]

Answer:

We have the function:

f(x) = x^3 - 2*x^2

To sketch this, we need to graph some points, and then just draw a line that passes through the points.

The graph of this equation is shown below.

Now we can complete the question.

If the graph is below the x-axis in some interval, the function is negative in that interval

If the graph is above the x-axis in some interval, the function is positive in that interval.

If the graph goes up in a interval, then the function is increasing in that interval

If the graph goes down on an interval, then the function is decreasing in that interval.

Then:

1) f is------ on the intervals (−∞, 0) and (0, 2).

Here we can see that the graph is below the x-axis in those intervals, then here we have:

f is negative on the intervals (−∞, 0) and (0, 2).

2) f is------ on the interval (2,∞)

Here the answer is positive:

f is positive on the interval (2,∞)

3) fi is ------ on the interval (0, 4/3)

In the graph, you can see that the graph goes down in that interval, then the correct answer here is:

f is decreasing on the interval (0, 4/3)

4) f is------ on the intervals (−∞, 0) and (4/3, ∞).

In this case, we can see that the graph goes up in these intervals, then the correct answer here is:

f is increasing on the intervals (−∞, 0) and (4/3, ∞).

5 0

3 years ago