A professional employee in a large corporation receives an average of $\mu =41.7$ e-mails per day. An anti-spam protection program was installed in the company’s server, and one month later, a random sample of 45 employees showed that they were receiving an average of $x̄=36.2$ e-mails per day. Assume that $\sigma =18.45$. Use a 5% level of significance to test whether there has been a change (either way) in the average number of emails received per day per employee.

a.) What is α? State the null and alternate hypotheses. Will you use a left-tailed, right-tailed, or two-tailed test?
$\alpha$ = ___
$H₀:\mu$ = ___
$H₁:\mu$ ___
The test is a ___ test.

b.) Identify the Sampling Distribution you will use. What is the value of the test statistic?
The best sampling distribution to use is the ___ distribution.
The test statistic (z or t value) is = ___

c.) Find or estimate the P-value for the test.
The p-value is = ___

d.) Conclude the test.
Based on this, we will ___ the null hypothesis.

Answer :

Given :

Sample mean $(x\overline{)}=36.2$

Population standard deviation $(\sigma )=18.45$

Sample size $(n)=45$

Significance level $(\alpha )=0.05$

Solution :

The significance level, $\alpha =0.05$

The null and alternative hypothesis,

$$H_0:\mu =41.7$$ $$H_1:\mu \ne 41.7$$

The test is a two-tailed test

(b) Since the sample size is greater than 30, the sampling distribution of sample mean would be approximately normal.

∴ The best sampling distribution to use is the normal distribution.

The test statistic,

$$z=\frac{\overline{x}–\mu }{\frac{\sigma }{\sqrt{n}}}=\frac{36.2–41.7}{\frac{18.45}{\sqrt{45}}}=-2$$

(c) The p-value can be obtained as,

$$\text{p-value}=P(z<-2\text{ or }z>2)$$ $$=2\times P(z<-2)$$ $$=2\times 0.0228$$ $$=0.0456$$

(d) As the p-value is less than the significance level of 0.02, we reject the null hypothesis.

Decision: Reject the null hypothesis.

Conclusion: There is sufficient evidence to support the claim that there has been a change (either way) in the average number of emails received per day per employee