Hypothesis Test of Mean for Normal Distribution (Sigma, σ, is Known) - One Sample
Example: A sample of size 200 has a mean of 20. Assume the population standard deviation is 6.  Use the TI-83 calculator to test the hypothesis that the population mean is not different from 19.2 with a level of significance of α = 5%. Solution: “The population mean is not different from 19.2” means the same as “the population mean is equal to 19.2.”  Therefore, the null and alternate hypotheses are H0:   µ = 19.2 and H1: µ ≠ 19.2, respectively.  Follow the steps below to solve the problem using the TI-83.  [NOTE: If the p-value < α, reject the null hypothesis; otherwise, do not reject the null hypothesis.]
Press STAT and the right arrow twice to select TESTS.  To select the highlighted 1:Z-Test… Press ENTER.

Use right arrow to select Stats (summary values rather than raw data) and Press ENTER. Use the down arrow to Enter the hypothesized mean, population standard deviation, sample mean, and sample size. Select alternate hypothesis. Press down arrow to select Calculate and press ENTER.

Results: Since the p-value is 0.1, do not reject the null hypothesis with an α (alpha) value of 0.10 or smaller (10% level of significance or smaller). [In this example, α = 0.05.]

Hypothesis Test of Mean
for Normal Distribution (Sigma, σ, is Known) - Two Samples

Example: Two samples were taken, one from each of two populations.  Use the TI-83 calculator to test the hypothesis that the two population means are not different with a level of significance of α = 5%.

Solution: For the two samples, we have the following summary data:

n1 = 38	n2 = 35
σ1 = 5	σ2 = 7
H0: µ1 = µ2
H1: µ1 ≠ µ2
Use α = 5%

“The two population means are not different” means the same as “the two population means are equal.”  Therefore, the null and alternate hypotheses are H₀: µ₁ = µ₂ and H₁: µ₁ ≠ µ₂, respectively.  Follow the steps below to solve the problem using the TI-83.  [NOTE: If the p-value < α , reject the null hypothesis; otherwise, do not reject the null hypothesis.

Press STAT and the right arrow twice to select TESTS. 

Use the down arrow to select
3:2-SampZTest…
Press ENTER.

Use right arrow to select Stats (summary values rather than raw data). Enter standard deviations, mean and sample size for samples 1 and 2. Select alternate hypothesis. Press down arrow to select Calculate and press ENTER.

Results: Since the p-value is 0.0186, reject the null hypothesis with an alpha value of 0.05 or larger (5% level of significance or larger). Conclude that the two population means are not different.

Hypothesis Test of Proportion for Normal Distribution - One Sample
Example: In sampling 200 people, we found that 30% of them favored a certain candy.  Use α = 10% to test the hypothesis that the proportion of people who favored that candy is less than 35%.  Solution: This represents a one-sample test of proportion.  So we use the "1-PropZTest" function. The sample proportion is 30% or p = 0.30, and the hypotheses are H0: p ≥ 0.35 and H1: p < 0.35 (claim). Hypothesized value is 0.35.
Press STAT and the right arrow twice to select TESTS.  Use the down arrow to select 5:1-PropZTest… Press ENTER.
Enter hypothesized proportion, number of favorable outcomes, x, sample size, n, and select the alternate hypothesis. Use down arrow to select Calculate and press ENTER.
Results: Since the p = 0.069 is less than α = 0.10, reject the null ypothesis.  Conclude that the sample proportion of 0.30 is significantly less than the hypothesized proportion of 0.35.

Hypothesis Test of Proportion for Normal Distribution - Two Samples
Example: In sampling 200 freshman college students (Sample 1), we found that 61 of them earned an A in statistics.  A sample of 250 sophomore college students (Sample 2) had 60 people who earned an A in statistics.  Test the hypothesis that the proportion of freshmen that earned an A in statistics is greater than the proportion of sophomores that earned an A in statistics. Solution: This represents a two-sample test of proportion.  We use the "2-PropZTest" function. The hypotheses are H0: p1 ≤ p2 and H1: p1 > p2 (claim)
Press STAT and the right arrow twice to select TESTS.  Use the down arrow to select 6:2-PropZTest… Press ENTER.
Enter number of favorable outcomes and sample size of samples 1 and 2.  Select the alternate hypothesis. Use down arrow to select Calculate and press ENTER.
Results: Since the p = 0.061, reject the null hypothesis for values of a > 0.061.  Conclude that the sample 1 proportion of 0.305 is significantly greater than sample 2 proportion of 0.24 when a > 0.061.