Question

A random sample of 32 salespersons with a degree had an average weekly sale of $3599 last year, while 35 salespersons without a college degree averaged $3311 in weekly sales. The standard deviations were $468 and $642 respectively.

Required:
Is there evidence at the 5% level to support the retailer's belief?

Answer 1

There is enough evidence to support the claim that the average sales are significantly greater for salespersons with a college degree than for salespersons without it.

Explanation:

The question is incomplete:

"A national computer retailer believes that the average sales are greater for salespersons with a college degree. A random sample of 32 salespersons with a degree had an average weekly sale of $3599 last year, while 35 salespersons without a college degree averaged $3311 in weekly sales. The standard deviations were $468 and $642 respectively.

Required:

Is there evidence at the 5% level to support the retailer's belief?"

This is a hypothesis test for the difference between populations means.

The claim is that the average sales are significantly greater for salespersons with a college degree than for salespersons without it.

Then, the null and alternative hypothesis are:

`H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2> 0`

The significance level is 0.05.

The sample 1 (college degree), of size n1=32 has a mean of 3599 and a standard deviation of 468.

The sample 2 (non-college degree), of size n2=35 has a mean of 3311 and a standard deviation of 642.

The difference between sample means is Md=288.

`M_d=M_1-M_2=3599-3311=288`

The estimated standard error of the difference between means is computed using the formula:

`s_(M_d)=\sqrt{(\sigma_1^2)/(n_1)+(\sigma_2^2)/(n_2)}=\sqrt{(468^2)/(32)+(642^2)/(35)}\\\\\\s_(M_d)=√(6844.5+11776.114)=√(18620.614)=136.4574`

Then, we can calculate the t-statistic as:

`t=(M_d-(\mu_1-\mu_2))/(s_(M_d))=(288-0)/(136.4574)=(288)/(136.4574)=2.11`

The degrees of freedom for this test are:

`df=n_1+n_2-2=32+35-2=65`

This test is a right-tailed test, with 65 degrees of freedom and t=2.11, so the P-value for this test is calculated as (using a t-table):

`\text{P-value}=P(t>2.11)=0.019`

As the P-value (0.019) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the average sales are significantly greater for salespersons with a college degree than for salespersons without it.