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It has been found that women are excluded from social events with coworkers. One study at Company X, found that women only recei

Natasha2012 [34]

Answer:

Option C is correct.

The p-value in this case is the probability of getting data like that observed or more extreme (i.e., getting an observed invitation rate of 33% or higher) assuming that;

The social strategies did not increase the rate of invitations for female colleagues.

Step-by-step explanation:

The p-value is actually defined as the probability of obtaining results as extreme as the observed results in the statistical analysis provided that the null hypothesis is true.

The standing condition for the p-value definition is the assumption that the null hypothesis is true, then it is the probability of getting the extreme result (usually not due to just random chance), that is usually observed, being tested for and contained in the alternative hypothesis.

Normally, in hypothesis testing, especially one comparing two sets of data, the null hypothesis plays the devil's advocate and usually takes the form of the opposite of the theory to be tested. It usually contains the signs =, ≤ and ≥ depending on the directions of the test. It usually maintains that, with random chance responsible for the outcome or results of any experimental study/hypothesis testing, its statement is true.

The alternative hypothesis usually confirms the theory being tested by the experimental setup. It usually contains the signs ≠, < and > depending on the directions of the test. It usually maintains that significant factors other than random chance, affect the outcome or results of the experimental study/hypothesis testing and result in its own statement.

Since this test aimed to test whether the rate of invitation of women to outside work events after using the social strategies, the null hypothesis for the question would be that there is no significant difference in the rate of invitation of the women after using the social strategies while the alternative hypothesis would be that the social strategies worked and the rate invitation of women to outside of work events increased.

So, the null hypothesis that we need to assume was true in the p-value definition is that; the social strategies did not increase the rate of invitations for female colleagues.

This is option C!

Hope this Helps!!!

8 0

3 years ago

Write an equation for the line in slope-intercept

Tamiku [17]

The equation would be: y=3x+5
The line crosses the y-axis at 5, so that’s the y-intercept.
The slope is 3 because it goes up by 3 and then 1 to the right, as a fraction it’s 3/1 which is just 3

5 0

3 years ago

Apply the distributive property to factor out the greatest common factor.

ruslelena [56]

80+32=112 so all you do is just add or use a brain .

4 0

3 years ago

A study by Consumer Reports showed that 64% of supermarket shoppers believe supermarket brands to be as good as national name br

natima [27]

Answer:

Step-by-step explanation:

Given that:

A study is conducted and the study by Consumer Reports showed that 64% of supermarket shoppers believe supermarket brands to be as good as national name brands.

a)

Thus; we formulate the null and the alternative hypotheses as follows:

The proportion is ; $\frac{64}{100}$ = 0.64

Null hypothesis: ${H_0}:p = 0.64$

The Null hypothesis states that there is no evidence that “the percentage of supermarket shoppers believes that the supermarket ketchup is good as the national brand ketchup differ form 64%”.

Alternative hypothesis: ${H_a}:p \ne 0.64H$

The Alternative hypothesis states that there is evidence that “the percentage of supermarket shoppers believes that the supermarket ketchup is good as the national brand ketchup differ form 64%”.

b)

The proportion of the population = 0.64

The sample size n = 100

The number of shoppers = 52 stating that the supermarket brand was as good as the national brand

The sample proportion $\hat p = \frac{52}{100}$

$\hat p = 0.52$

The z - value is calculated by the formula:

$z = \dfrac{\hat p-p}{\frac{\sqrt{ p(1-p)}}{100 }}$

$z = \dfrac{0.52-0.64}{\frac{\sqrt{ 0.64(1-0.64)}}{100 }}$

$z = \dfrac{-0.12}{0.048 }}$

z = -2.50

Since z = -2.50

the p-value = 2P(Z ≤ -2.50)

p-value = 2 × 0.0062

The p-value = 0.0124

c)

At significance level, ∝ = 0.05 ; The p-value = 0.0124

According to the rejection rule, if p-value is less than 0.05 then we will reject null hypothesis at ∝ = 0.05

Hence, the p-value =0.0124 < ∝ (=0.05)

According to the reject rule; reject null hypothesis.

Conclusion: There is no evidence at all that the percentage of supermarket shoppers believes that the supermarket ketchup is good as the national brand ketchup differ form 64%.

d) we know that the significance level of ∝ = 0.05

The value ${z__{0.05}}$ is obtained as:

$P\left( {\left| Z \right| \le z} \right) = 0.05$

Now; to determine Z ; we locate the probability value of 0.05 form the table of standard normal distribution. Then we proceed to the left until the first column is reached and the value is 1.90. Also , we move upward until we reach the top and the value is 0.06. Now; the intersection of the row and column results the area to the left of z

This implies that : $P(Z \leq -1.96)=0.05$

The critical value for left tail is -1.96 and the critical value for right tail is 1.96.

Conclusion:

The critical value is -1.96 and the value of test statistic is - 2.50. Here, we can see that the value of test statistic is lesser than the critical value. Hence, we can be concluded that there is evidence that reject the null hypothesis.

Therefore, Yes, the national brand ketchup manufacturer is pleased with the conclusion.

5 0

3 years ago

Does this look correct?

Neko [114]

That is not correct. I will help you. Begin by subtracting the 9x from both sides, like this: 9x - 9x + 2y = 6 - 9x. That simplifies to 2y = -9x + 6. Now divide both sides by 2 to get y all alone, like this: $\frac{2y}{2}=- \frac{9}{2}x+ \frac{6}{2}$ . That reduces down to $y=- \frac{9}{2}x+3$ . That is slope-intercept form. y = mx + b.

7 0

3 years ago