Simplify c + 4c^2 - 2 + 2c

Suppose a production line operates with a mean filling weight of 16 ounces per container. Since over- or under-filling can be da

miss Akunina [59]

Answer:

From the question we are told that

The population mean is $\mu = 16$

The sample size is n = 30

The sample mean is $\= x = 16.32$

The population standard deviation is $\sigma = 0.8$

The level of significance is $\alpha = 0.10$

Step 1: State hypotheses:

The null hypothesis is $H_o : \mu = 16$

The alternative hypothesis is $H_a : \mu \ne 16$

Step 2: State the test statistic. Since we know the population standard deviation and the sample is large our test statistics is

$t = \frac{ \= x -\mu }{ \frac{\sigma}{ \sqrt{n} } }$

=> $t = \frac{ 16.32 -16 }{ \frac{0.8 }{ \sqrt{30} } }$

=> $t =2.191$

Generally the degree of freedom is mathematically represented as

$df = n - 1$

=> $df = 30 - 1$

=> $df =29$

Step 3: State the critical region(s):

From the student t-distribution table the critical value corresponding to $\alpha = 0.10$ is

$t = 1.311$

Generally the critical regions is mathematically represented as

$- 1.311 < T < 1.311$

Step 4: Conduct the experiment/study:

Generally the from the value obtained we see that the t value is outside the critical region so the decision is [Reject the null hypothesis ]

Step 5: Reach conclusions and state in English:

There is sufficient evidence to show that the filling weight has to be adjusted

Step 6: Calculate the p-value associated with this test. How does this the p-value support your conclusions in Step 5?

From the student t-distribution table the probability value to the right corresponding to $t =2.191$ at a degree of freedom of $df =29$ is

$P( t > 2.191) = 0.0183$

Generally the p-value is mathematically represented as

$p-value = 2 * P( t > 2.191 )$

=> $p-value = 2 * 0.0183$

=> $p-value = 0.0366$

Generally looking at the value obtained we see that $p- value < \alpha$ hence

The decision rule is

Reject the null hypothesis

Step-by-step explanation:

4 0

3 years ago

A company that makes hair-care products had 200 people try a new shampoo. Of the 200 people, 12 had a mild allergic reaction.

Likurg_2 [28]

Answer:

24

Step-by-step explanation:

first you multiply 200 by 12, Then you divide that by 100. your final product is 24, hope that helps

7 0

3 years ago

Read 2 more answers

Please help fast will give brainliest

Volgvan

Answer:

C

Step-by-step explanation:

Since the circle is inscribed inside of the square, its diameter is the same as the side length of the square, or 8. This means that the radius is 8/2=4. The formula for the volume of a circle is $\pi r^2$ , or in this case $3.14\cdot 4^2=50.24$ . The area of the square is 8*8=64 square feet. 50.24/64 is equal to 78.5%, or answer choice C. Hope this helps!

6 0

4 years ago

A driver averaged 53 miles per hour and took 13 hours to travel between two cities. What is the distance between the two cities

8_murik_8 [283]

Answer:

40 mph

MPH=miles per hour

Step-by-step explanation:

PLEASE GIVE BRAINLIEST

6 0

3 years ago

A certain manufactured product is supposed to contain 23% potassium by weight. A sample of 10 specimens of this product had an a

Taya2010 [7]

Answer:

(a) Null Hypothesis, $H_0$ : $\mu$ = 23%

Alternate Hypothesis, $H_A$ : $\mu$ $\neq$ 23%

(b) We conclude that the mean percentage is different from 23 and the manufacturing process will be re-calibrated.

(c) P-value is 0.6%.

Step-by-step explanation:

We are given that a certain manufactured product is supposed to contain 23% potassium by weight.

A sample of 10 specimens of this product had an average percentage of 23.2 with a standard deviation of 0.2.

Let $\mu$ = mean percentage of potassium by weight.

(a) Null Hypothesis, $H_0$ : $\mu$ = 23% {means that the mean percentage is equal to 23 and the manufacturing process will not be re-calibrated}

Alternate Hypothesis, $H_A$ : $\mu$ $\neq$ 23% {means that the mean percentage is different from 23 and the manufacturing process will be re-calibrated}

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

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean percentage = 23.2

s = sample standard deviation = 0.2

n = sample of specimens = 10

So, the test statistics = $\frac{23.2-23}{\frac{0.2}{\sqrt{10} } }$ ~ $t_9$

= 3.162

The value of t test statistic is 3.162.

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 value of -2.262 and 2.262 at 9 degree of freedom for two-tailed test.

(b) Since our test statistic doesn't lie within the range of critical values of t, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the mean percentage is different from 23 and the manufacturing process will be re-calibrated.

(c) The P-value of the test statistics is given by;

P-value = P( $t_9$ > 3.162) = 0.006 or 0.6%

3 0

3 years ago

Simplify c + 4c^2 - 2 + 2c - c^2 + 5