I don’t understand the question 2 for this math equation.

Suppose that in a random selection of 100 colored candies, 21% of them are blue. The candy company claims that the percentage of

zhannawk [14.2K]

Answer:

The pvalue of the test is 0.0784 > 0.01, which means that at this significance level, we do not reject the null hypothesis that the percentage of blue candies is equal to 29%.

Step-by-step explanation:

The candy company claims that the percentage of blue candies is equal to 29%.

This means that the null hypothesis is:

$H_{0}: p = 0.29$

We want to test the hypothesis that this is true, so the alternate hypothesis is:

$H_{a}: p \neq 0.29$

The test statistic is:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, $\sigma$ is the standard deviation and n is the size of the sample.

0.29 is tested at the null hypothesis:

This means that $\mu = 0.29, \sigma = \sqrt{0.29*0.71}$

Suppose that in a random selection of 100 colored candies, 21% of them are blue.

This means that $n = 100, X = 0.21$

Value of the test statistic:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{0.21 - 0.29}{\frac{\sqrt{0.29*0.71}}{\sqrt{100}}}$

$z = -1.76$

Pvalue of the test:

The pvalue of the test is the probability of the sample proportion differing at least 0.21 - 0.29 = 0.08 from the population proportion, which is 2 multiplied by the pvalue of Z = -1.76.

Looking at the z-table, z = -1.76 has a pvalue of 0.0392

2*0.0392 = 0.0784

The pvalue of the test is 0.0784 > 0.01, which means that at this significance level, we do not reject the null hypothesis that the percentage of blue candies is equal to 29%.

5 0

3 years ago

PLEASE HELP: A standard six-sided die is rolled.

amm1812

Answer in fraction form is 1/3

Answer in decimal form is 0.3333

Pick one answer only.

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Explanation:

The sample space is the set of all possible outcomes. In this case, the outcomes consist of values between 1 and 6

S = sample space

S = {1,2,3,4,5,6}

There are 6 items here. Let B = 6.

We want to roll a number greater than 4, so the event space we're after is

E = {5,6}

which consists of 2 items. Let A = 2.

The probability we want is A/B = 2/6 = 1/3 = 0.3333

So if you go with the fraction option, then you'll type in 1/3

If you go with the decimal option, then you'll type in 0.3333

4 0

3 years ago

What is 5 to the power of 6 over 5 to the power of 2

AlekseyPX

Answer:

625

Step-by-step explanation: 5^6/5^2

multiply 5^6 and than devide your answer with 5^2

6 0

2 years ago

Read 2 more answers

Tim has m dollars . Rebecca has $6 dollars less than 1 over 10 times the number of dollars that Tim has which expression represe

grin007 [14]

y= 1/10x - 6

Which if you plug in 70 for x, y equals 1.

Rebecca has $1 and Tim has $70, together they have $71.

8 0

3 years ago

A researcher is going to estimate the average typing speed of students of a college. He selects a random sample of 20 students a

Stella [2.4K]

Answer:

The margin of error of the 90% confidence interval of a student's average typing speed is of 1.933 wpm.

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 20 - 1 = 19

90% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 19 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.9}{2} = 0.95$ . So we have T = 1.7291

The margin of error is:

$M = T\frac{s}{\sqrt{n}}$

In which s is the standard deviation of the sample and n is the size of the sample. For this question, we have $s = 5, n = 20$ . So

$M = T\frac{s}{\sqrt{n}}$

$M = 1.7291\frac{5}{\sqrt{20}}$

$M = 1.933$

The margin of error of the 90% confidence interval of a student's average typing speed is of 1.933 wpm.

8 0

3 years ago