Cortez, Leah-Faith T.
Z-test
The mean Verbal SAT score for the population of first students at Mind College is 520. The standard
deviation of scores in this population is 95. An investigator believes that the mean Verbal NMAT of
first year psychology majors is significantly different from the mean score of the population. The mean
of a sample of 36 first year psychology majors is 548. Test the investigator's prediction using an alpha
level of .05. Use the step by step approach (15 points).
Given: µ = 520, σ = 95, n = 36, M= 548, α = .05
Answer the following.
1. Is the problem directional or non-directional?
Non-directional (two-tailed)
2. State the null and alternative hypothesis.
H0: µ = 520
H1: µ ≠ 520
3. State alpha.
α = .05
z = ± 1.96
4. State the decision rule.
If –1.96 ≥ z ≥ +1.96, reject the null hypothesis, otherwise retain the null hypothesis.
5. Calculate the test statistics.
To calculate the z statistic, first compute
the standard error (σM), which is the
denominator for the z statistic:
𝛔
𝛔𝑴 =
√𝐧
𝟗𝟓
σ = 95, n = 36
𝛔𝑴 =
𝟔
𝟗𝟓
𝛔𝑴
𝛔𝑴 =
= 𝟏𝟓. 𝟖𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑
√𝟑𝟔
Then compute the z statistic by substituting the
values.
𝒛𝒐𝒃𝒕 =
𝑴−𝝁
𝝈𝑴
M= 548, µ = 520
𝒛𝒐𝒃𝒕
𝟓𝟒𝟖 − 𝟓𝟐𝟎
=
𝟏𝟓. 𝟖𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑
𝒛𝒐𝒃𝒕
=
𝟐𝟖
𝟏𝟓. 𝟖𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑
𝒛𝒐𝒃𝒕 = 𝟏. 𝟕𝟔𝟖𝟒𝟐𝟏𝟎𝟓𝟐𝟔
𝒛𝒐𝒃𝒕 = 𝟏. 𝟕𝟕
6. State the result.
If –1.96 ≥ zobt ≥ +1.96, reject the null hypothesis, otherwise, retain the null hypothesis.
𝒛𝒐𝒃𝒕 = 𝟏. 𝟕𝟔𝟖𝟒𝟐𝟏𝟎𝟓𝟐𝟔
𝒛𝒐𝒃𝒕 = 𝟏. 𝟕𝟕
𝒛𝒐𝒃𝒕 ≠ 𝟏. 𝟗𝟔
𝒛𝒐𝒃𝒕 < 𝟏. 𝟗𝟔
∴ 𝑹𝒆𝒕𝒂𝒊𝒏 𝒕𝒉𝒆 𝒏𝒖𝒍𝒍 𝒉𝒚𝒑𝒐𝒕𝒉𝒆𝒔𝒊𝒔.
7. State the conclusion.
To find the p value for the z statistic, find its own probability (toward the tail) in the unit
normal table (APPENDIX B) and multiply this probability times the number of tails for alpha.
One-tailed (Directional) Two-tailed (Non-Directional)
Number of tails
1
2
Probability
p
p
p value calculation
1p
2p
𝒛𝒐𝒃𝒕 = 𝟏. 𝟕𝟕
Area beyond z in tail = .0384 (using appendix B)
Two-tailed
p = 2 x .0384
p = .0768
If the null hypothesis were true, then p = 0.768 that we could have selected this sample mean
from this population. The criteria that was set in step 3 was that the probability must be less
than 5% (α = .05) that we obtain a sample mean, if the null hypothesis were true. Since p is
greater than 5% we decide to retain the null hypothesis.
We conclude that the mean Verbal SAT score in this population is 520. (the value stated in the
null hypothesis)