Mathematics, 12.11.2020 21:20 amoryfe28p0vpwo

Suppose you are the manager of Speedy Oil Change which claims that it will change the oil in customers’ cars in less than 30 minutes on average. Further suppose that several complaints have been filed from customers stating that their oil change took longer than 30 minutes. Upper-level management at Speedy Oil Change headquarters has requested that you investigate the complaints. To begin your investigation, you randomly audit 36 oil changes performed by Speedy Oil Change and record the time each customer waited for the oil change. The number of minutes to complete each of the 36 oil changes is reported below. Suppose you are the manager of Speedy Oil Change which claims that it will change the oil in customers’ cars in less than 30 minutes on average. Further suppose that several complaints have been filed from customers stating that their oil change took longer than 30 minutes. Upper-level management at Speedy Oil Change headquarters has requested that you investigate the complaints. To begin your investigation, you randomly audit 36 oil changes performed by Speedy Oil Change and record the time each customer waited for the oil change. The number of minutes to complete each of the 36 oil changes is reported below.

42
29
19
11
10
27
41
27
22
26
28
32
17
15
25
35
31
22
13
31
17
37
33
25
18
24
28
21
40
19
33
30
14
23
22
10

1. Write the null and alternative hypothesis. Label each accordingly.
2. Calculate the Test Statistic and the P-value. Label each accordingly, and report the P-value to 4 decimal places.
3. Using the P-value calculated in #2, explain why you would or would not reject the null hypothesis? Answer using a complete sentence, proper grammar and correct spelling. Do NOT write a conclusion.
4. Based on your answer to #3, write a final conclusion for the test of hypotheses, and explain if the company is fulfilling its promise to “change the oil in customers’ cars in less than 30 minutes on average?” Answer using a complete sentence, proper grammar and correct spelling.

Answers: 3

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Answer from: calebwoodall6477

Using the t-distribution, we have that:

The null hypothesis is $H_0: \mu \geq 30$

The alternative hypothesis is $H_1: \mu < 30$

2. The test statistic is t = 3.51.

3. The p-value is of 0.0063.

4. Since the p-value is of less than the standard significance level of 0.05, we reject the null hypothesis, and can conclude that the company is fulfilling its promise.

Item 1:

At the null hypothesis, we test if the mean is of at least 30 minutes, that is:

$H_0: \mu \geq 30$

At the alternative hypothesis, we test if the mean is of less than 30, that is:

$H_1: \mu < 30$

Item 2:

We have the standard deviation for the sample, thus, the t-distribution is used. The test statistic is given by:

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

The parameters are:

$\overline{x}$ is the sample mean. $\mu$ is the value tested at the null hypothesis. s is the standard deviation of the sample. n is the sample size.

For this problem:

$\mu = 30, n = 36$ Using a calculator, $\overline{x} = 24.92, s = 8.68$

The test statistic is:

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

$t = \frac{24.92 - 30}{\frac{8.68}{\sqrt{36}}}$

[tx]t = 3.51[/tex]

The test statistic is t = 3.51.

Item 3:

The p-value is found using a left-tailed test, as we are testing if the mean is less than a value, with t = 3.51 and 36 - 1 = 35 df.

Using a calculator, this p-value is of 0.0063.

Item 4:

Since the p-value is of less than the standard significance level of 0.05, we reject the null hypothesis, and can conclude that the company is fulfilling its promise.

A similar problem is given at link

Suppose you are the manager of Speedy Oil Change which claims that it will change the oil in custome

Answer from: griseldaguerree

Explained below.

Step-by-step explanation:

(1)

The hypothesis can be defined as follows:

H₀: The Speedy Oil Change will change the oil in customers’ cars in more than 30 minutes on average, i.e. μ > 30.

Hₐ: The Speedy Oil Change will change the oil in customers’ cars in less than 30 minutes on average, i.e. μ ≤ 30.

(2)

Use Excel to compute the sample mean and standard deviation as follows:

$\bar x=\text{AVERAGE(array)}=24.917\\\\s=\text{STDEV.S(array)}=8.683$

Compute the test statistic as follows:

$t=\frac{\bar x-\mu}{s/\sqrt{n}}=\frac{24.917-30}{8.683/\sqrt{36}}=-3.51$

The degrees of freedom is:

df = n - 1

= 36 - 1

= 35

Compute the p-value as follows:

$p-value=P(t_{35}$

(3)

The p-value = 0.0006 is very small.

The null hypothesis will be rejected at any of the commonly used significance level.

(4)

There is sufficient evidence to support the claim that the Speedy Oil Change will change the oil in customers’ cars in less than 30 minutes on average.

Answer from: Quest

answer: 40.95

step-by-step explanation:

Answer from: Quest

Here for a good time of free question aha

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