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Firdavs [7]
2 years ago
11

What is the input for which g(x)=3 x=

Mathematics
1 answer:
garri49 [273]2 years ago
8 0
I think it's x=5 because if you go to "3" on the x line and find where the line touches a point, it's at 5. (wow that made no sense, but yeah)
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A manager is comparing wait times for customers in a coffee shop based on which employee is
anyanavicka [17]

Using the t-distribution, as we have the standard deviation for the sample, it is found that there is a significant difference between the wait times for the two populations.

<h3>What are the hypothesis tested?</h3>

At the null hypothesis, we test if there is no difference, that is:

H_0: \mu_A - \mu_B = 0

At the alternative hypothesis, it is tested if there is difference, that is:

H_1: \mu_A - \mu_B = 0

<h3>What are the mean and the standard error of the distribution of differences?</h3>

For each sample, we have that:

\mu_A = 73, s_A = \frac{2}{\sqrt{100}} = 0.2

\mu_B = 74, s_B = \frac{4}{\sqrt{100}} = 0.4

For the distribution of differences, we have that:

\overline{x} = \mu_A - \mu_B = 73 - 74 = -1

s = \sqrt{s_A^2 + s_B^2} = \sqrt{0.2^2 + 0.4^2} = 0.447

<h3>What is the test statistic?</h3>

It is given by:

t = \frac{\overline{x} - \mu}{s}

In which \mu = 0 is the value tested at the null hypothesis.

Hence:

t = \frac{\overline{x} - \mu}{s}

t = \frac{-1 - 0}{0.447}

t = -2.24

<h3>What is the p-value and the decision?</h3>

Considering a one-tailed test, as stated in the exercise, with 100 - 1 = 99 df, using a t-distribution calculator, the p-value is of 0.014.

Since the p-value is less than the significance level of 0.05, it is found that there is a significant difference between the wait times for the two populations.

More can be learned about the t-distribution at brainly.com/question/16313918

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