• The sample size is large enough to ensure that the sampling distribution is normal.
• The are data selected at random such that each data point is independent of the one-another.
• The sample is random, representative and non-bias.

• H₀: μ = $5,400
• H_A: μ > $5,400

We want to prove that the mean income after graduation is higher than $5,400. For this reason, we need our sample result to be significantly higher than $5,400. The rejection region would be the upper 5% right tail. The original mean of $5,400 is what we are comparing to and is marked in the middle of the graph.

This is a right-tailed test. In order to prove that the mean starting salary is higher than $5,400, we will need our sample result to be far enough above the original mean of $5,400 and be, therefore, in the upper right tail (upper rejection region).

$\begin{align} t_{test} &= \frac{\bar{x}-\mu}{\frac{s}{\sqrt{n}}} \\ &= \frac{\$6,500-\$5,400}{\frac{\$3,200}{\sqrt{100}}} \\ &= 3.4375 \end{align}$

Because we are performing a right-tailed test on a mean, we will use Excel’s T.DIST.RT(t_test, n-1) function:

$p\text{-value}=\text{T.DIST.RT}(3.4375, 99) = 0.00043$

Because the p-value = 0.00043 (0.043%) is less than (<) the level of significance of 5% (0.05), we reject H₀.

Because we rejected H₀, there is sufficient evidence to conclude that the college’s business diploma grads earn higher than $5,400 upon graduation.

• The sample size is large enough to ensure that the sampling distribution is normal.
• The are data selected at random such that each data point is independent of the one-another.
• The sample is random, representative and non-bias.

• H₀: μ = 0.625
• H_A: μ ≠ 0.625

We want to prove that (or check if) the average screw length from this batch of screws is not 5/8 inches = 0.625 inches in length. When proving that the screws are not equal to 0.625 inches, we could prove this by having a batch of screws with an average length much longer or much shorter than 0.625 inches. For this reason, extreme values in either tail would prove this. This means we have a two-tailed problem:

This is a two-tailed test, as described in the previous section. We always have a two-tailed test when the alternate hypothesis (H_A) contains a ‘not equal’ (≠) symbol.

$\begin{align} t_{test} &= \frac{\bar{x}-\mu}{\frac{s}{\sqrt{n}}} \\ &= \frac{0.6245-0.625}{\frac{0.0124}{\sqrt{100}}} \\ &= -0.4032 \end{align}$

Because we are performing a right-tailed test on a mean, we will use Excel’s T.DIST.RT(t_test, n-1) function:

$p\text{-value}=\text{T.DIST.2T}(abs(-0.4032), 99) = 0.68765$

Because the p-value = 0.68765 (68.765%) is more than (>) the level of significance of 1% (0.01), we do not reject H₀.

Because we failed to reject H₀, there is not sufficient evidence to conclude that the batch of hexagon head cap screws do not have an average length of 5/8 inches.

• The sample size is large enough to ensure that the sampling distribution is normal.
• The are data selected at random such that each data point is independent of the one-another.
• The sample is random, representative and non-bias.

• H₀: μ ≥ 100,000 or μ = 100,000 (we can always use an ‘=’ symbol for H₀)
• H_A: μ < 100,000

We want to prove that the average number of kilometers the tires will last is less than 100,000 kilometers (km). If the sample of 100 tires last much less than 100,000 km, there is sufficient evidence to conclude that the real average duration of the tires is less than 100,000 km. This means we need a result in the left-tail or far below 100,000.

This is a left-tailed test. In order to prove that the mean number of kilometers that tires last is lower than 100,000 km, we will need our sample result to be far enough below the stated mean of 100,000 and be, therefore, in the lower left tail.

$\begin{align} t_{test} &= \frac{\bar{x}-\mu}{\frac{s}{\sqrt{n}}} \\ &= \frac{98,000-100,000}{\frac{9,434}{\sqrt{100}}} \\ &= -2.1200 \end{align}$

Because we are performing a left-tailed test on a mean, we will use Excel’s T.DIST(t_test, n-1, TRUE) function:

$p\text{-value}=\text{T.DIST}(-2.12, 99, \text{TRUE}) = 0.01825$

Because the p-value = 0.01825 (1.825%) is less than (<) the level of significance of 5% (0.05), we reject H₀.

Because we rejected H₀, there is sufficient evidence to conclude that the salesmen’s claim regarding the life expectancy of XYZ tires is an exaggeration.