The average local cell phone call length was reported to be 2.27 minutes. A random sample of 20 phone calls showed an average of 2.98 minutes in length with a standard deviation of 0.98 minutes.
At α = 0.05 can it be concluded that the average differs from the population average?
Report your answer to two decimal places.
Note: Find the test statistic ONLY to test this claim using an appropriate formula. You must show your work.

We are given that the average local cell phone call length was reported to be 2.27 minutes. A random sample of 20 phone calls showed an average of 2.98 minutes in length with a standard deviation of 0.98 minutes.

Let [tex]\mu[/tex] = population average local cell phone call length

SO, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] = 2.27 minutes {means that the average is same as that of population average}

Alternate Hypothesis, [tex]H_a[/tex] : [tex]\mu[/tex] [tex]\neq[/tex] 2.27 minutes {means that the average differs from the population average}

The test statistics that will be used here is One-sample t test statistics because we don't know about population standard deviation;

T.S. = [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]

where, [tex]\bar X[/tex] = sample average call length = 2.98 minutes

s = sample standard deviation = 0.98 minutes

n = sample of phone calls = 20

So, test statistics = [tex]\frac{2.98-2.27}{\frac{0.98}{\sqrt{20} } }[/tex] ~ [tex]t_1_9[/tex]

= 3.24

Now, at 0.05 level of significance the t table gives critical values between -2.093 and 2.093 at 19 degree of freedom for two-tailed test. Since our test statistics does not lie within these range of critical values so we have sufficient evidence to reject our null hypothesis as test statistics will fall in the rejection region.

Therefore, we conclude that the average differs from the population average.