9-17 = 23 q= what does q equal?

true or false If x represents a random variable with mean 114 and standard deviation 40, then the standard deviation of the samp

Goshia [24]

Answer:

$\mu = 114, \sigma = 40$

We also know that we select a sample size of n =100 and on this case since the sample size is higher than 30 we can apply the central limit theorem and the distribution for the sample mean would be given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And the standard deviation for the sampling distribution would be:

$\sigma_{\bar X}= \frac{40}{\sqrt{100}}= 4$

So then the answer is TRUE

Step-by-step explanation:

Let X the random variable of interest and we know that the true mean and deviation for this case are given by:

$\mu = 114, \sigma = 40$

We also know that we select a sample size of n =100 and on this case since the sample size is higher than 30 we can apply the central limit theorem and the distribution for the sample mean would be given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And the standard deviation for the sampling distribution would be:

$\sigma_{\bar X}= \frac{40}{\sqrt{100}}= 4$

So then the answer is TRUE

6 0

3 years ago

Suppose you are the manager of Speedy Oil Change which claims that it will change the oil in customers’ cars in less than 30 mi

Veronika [31]

Answer:

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.