Appropriate test statistic and interpretation of the results for a hypothesis test for population mean for known and unknown variance of a normally distributed population

Hypothesis Test Concerning the Population Mean

Population Variance Known

Population Variance Unknown

Distribution

Large Sample

Small Sample

Large Sample

Small Sample

Normal

z-test

z-test

t-test or z-test

t-test

Non-normal

z-test

Not Available

t-test or z-test

Not Available

Example 1: Normal Distribution with known Population Variance

The annual returns of a mutual fund follow a normal distribution. The sample mean of the annual returns is 8 percent. The population standard deviation for the returns is 12 percent. The sample size is 25. You want to test whether the annual returns of the mutual fund is positive.

(a) Formulate a null hypothesis and alternative hypothesis for the test.

(b) Identify the test statistic for conducting a test of the hypotheses in the above part.

(c) Identify the rejection points for the hypothesis tested in part (a) at the level of significance of 0.01 and 0.05.

(d) Conduct the test and state the result of the hypothesis test for both 0.01 and 0.05 levels of significance.

Solution:

(a) We want to check whether the return is positive i.e. R>0. Since the alternative hypothesis does not contain "equal to" sign, the alternative hypothesis is R>0 and the null hypothesis is opposite of it i.e. R≤0.

Null hypothesis, H₀: R≤0
Alternative hypothesis, H_a: R>0

(c) The hypothesis test is a one-tailed test. So, the entire area of 1 percent and 5 percent will lie on the right tail of the distribution for 0.01 and 0.05 level of significance respectively. The rejection points will be z_0.01 and z>sub>0.05. The values for those points are 2.33 and 1.645 respectively.

(d) z = (X ̅ - µ₀)/(σ/√n) = (0.05-0)/(0.12/√25) = 0.05/0.024 = 2.083.

The calculated z-statistic value of 2.083 is greater than 1.645 and lesser than 2.33. So, we reject the null hypothesis for 0.05 level of significance and fail to reject the null hypothesis for 0.01 level of significance.

The annual return of the mutual fund is statistically greater than zero at 0.05 level of significance.

The annual return of the mutual fund is statistically not greater than zero at 0.01 level of significance.

Example 2: Non-normal Distribution with unknown Population Variance

A pharmaceutical company wants to test a drug for its pH level. The pH levels of the drug do not follow the normal distribution. The pH levels range from 0 to 14 with seven considered neutral. A pH less than seven is said to be acidic and greater than seven is said to be basic or alkaline. The company wants to check whether the drug is neutral in pH level or not. The sample size is 36. The standard deviation of the pH level of the sample is 4 and a mean of 5.

(a) Formulate null and alternative hypothesis consistent with the verbal description of the test.

(b) Identify the test statistic for conducting a test of hypotheses in the above part.

(c) Conduct the hypotheses test at 0.01 level of significance and state the results.

Solution:

(a) The null hypothesis is pH equal to 7, and the alternative hypothesis is pH not equal to 7.

Null hypothesis, H₀: pH= 7
Alternative hypothesis, H_a: pH ≠ 7

It is a two-tailed test.

(b) The population distribution is non-normal, but the sample size is greater than 36. Since the sample size is large, the distribution of the sample mean is approximately normal per the central limit theorem. As the population variance is unknown, the t-test will be used with 36-1 = 35 degrees of freedom.

(c) t-statistic = (X ̅ - µ₀)/(s/√n) = (5 - 7)/(4/√36) = -2/(4/6) = -3
,br> The test is a two-tailed test. So for a 0.01 level of significance, 0.5 percent of the rejection area will lie on both the right tail and left tail. The t-critical values will be t_0.005 and -t_0.005. The t-critical values for 35 degrees of freedom are 2.724 and -2.724. The area between these two points belongs to the null hypothesis, and the remaining area belongs to the alternative hypothesis.

Since -3 does not lie between -2.724 and 2.724, the null hypothesis is rejected.

The pH level of the drug is significantly different than seven at 0.01 level of significance.