Mathematics, 07.07.2020 19:01 gracetay3721

The Rocky Mountain district sales manager of Rath Publishing Inc., a college textbook publishing company, claims that the sales representatives make an average of 41 sales calls per week on professors. Several reps say that this estimate is too low. To investigate, a random sample of 38 sales representatives reveals that the mean number of calls made last week was 42. The standard deviation of the sample is 3.9 calls. Using the 0.025 significance level, can we conclude that the mean number of calls per salesperson per week is more than 41?H0 : Âµ = 40 H1 : Âµ > 401. Compute the value of the test statistic. 2. What is your decision regarding H0?

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Answer from: bc25rodriguez

1. Test statistic t=1.581.

2. The null hypothesis H0 failed to be rejected.

There is not enough evidence to support the claim that the mean number of calls per salesperson per week is significantly more than 41.

NOTE: if the null hypothesis is µ = 40, there is enough evidence to support the claim that the mean number of calls per salesperson per week is significantly more than 40 (test statistic t=3.161).

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the mean number of calls per salesperson per week is significantly more than 41.

Then, the null and alternative hypothesis are:

$H_0: \mu=41\\\\H_a:\mu 41$

The significance level is 0.025.

The sample has a size n=38.

The sample mean is M=42.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=3.9.

The estimated standard error of the mean is computed using the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{3.9}{\sqrt{38}}=0.633$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{42-41}{0.633}=\dfrac{1}{0.633}=1.581$

The degrees of freedom for this sample size are:

df=n-1=38-1=37

This test is a right-tailed test, with 37 degrees of freedom and t=1.581, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=P(t1.581)=0.061$

As the P-value (0.061) is bigger than the significance level (0.025), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the mean number of calls per salesperson per week is significantly more than 41.

For µ = 40:

This is a hypothesis test for the population mean.

The claim is that the mean number of calls per salesperson per week is significantly more than 40.

Then, the null and alternative hypothesis are:

$H_0: \mu=40\\\\H_a:\mu 40$

The significance level is 0.025.

The sample has a size n=38.

The sample mean is M=42.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=3.9.

The estimated standard error of the mean is computed using the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{3.9}{\sqrt{38}}=0.633$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{42-40}{0.633}=\dfrac{2}{0.633}=3.161$

The degrees of freedom for this sample size are:

df=n-1=38-1=37

This test is a right-tailed test, with 37 degrees of freedom and t=3.161, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=P(t3.161)=0.002$

As the P-value (0.002) is smaller than the significance level (0.025), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the mean number of calls per salesperson per week is significantly more than 40.

Answer from: Quest

Iam thinking the answer is a.) knew

Answer from: Quest

Is there a picture of a graph or whatever that goes along with this?

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